algebraic geometry fall 2013looij101/ag2013.pdf · fall 2013 eduard looijenga. rings are always...

Algebraic Geometry IFall 2013

Eduard Looijenga

Rings are always supposed to possess a unit element 1 and a ring homomor-phism will always take unit to unit. We allow that 1 = 0, but in that case we getof course the zero ring 0. We assume a ring to be commutative unless the con-trary is stated. If R is a ring, then we denote the multiplicative group of invertibleelements (units) of R by R×. We say that R is a domain if R has no zero divisors,equivalently, if (0) is a prime ideal. An R-algebra is a ring A endowed with a ringhomomorphism φ : R→ A, but if φ is understood, then for every r ∈ R and a ∈ A,the product φ(r)a is often denoted by ra. We say that A is finitely generated asan R-algebra if we can find a1, . . . , an in A such that every element of A can bewritten as a polynomial in these elements with coefficients in R; in other words,the R-algebra homomorphism R[x1, . . . , xn]→ A which sends the variable xi to aiis onto. This is not to be confused with the notion of finite generation for an R-module M resp. which merely means the existence of a surjective homomorphismof R-modules Rn → M . Similarly, a field K is said to be finitely generated as afield over a subfield k if we can find elements b1, . . . , bn in K such that every ele-ment of K can be written as a fraction of two polynomials in these elements withcoefficients in k.

CHAPTER 1

Affine varieties

Throughout this chapter we fix an algebraically closed field k. Recall that thismeans that every polynomial f ∈ k[x] of positive degree has a root x1 ∈ k: f(x1) =0. This implies that we can split off the factor x − x1 from f with quotient apolynomial of degree one less than f . Continuing in this manner we find that fdecomposes simply as f(x) = c(x − x1) · · · (x − xd) with c ∈ k − 0, d = deg(f)and x1, . . . , xd ∈ k. Since an algebraic extension of k is obtained by adjoiningto k roots of polynomials in k[x], this also shows that the property in question isequivalent to: every algebraic extension of k is equal to k.

A first example you may think of is the field of complex numbers C, but as weproceed you should be increasingly be aware of the fact that there are many others:it is shown in a standard algebra course that for any field F an algebraic closure Fis obtained by adjoining to F the roots of every polynomial f ∈ F [x] (this can notalways be done in one step, and might involve an infinite process). So we couldtake for k an algebraic closure of the field of rational numbers Q, of the finite fieldFq, where q is a prime power1 or even of the quotient field of any domain such asC[x1, . . . , xr].

1. The Zariski topology

Any f ∈ k[x1, . . . , xn] determines in an evident manner a function kn → k.In such cases we prefer to think of kn not as vector space—its origin and vectoraddition will be irrelevant to us—but as a set with a weaker structure. We shallmake this precise later, but it basically amounts to only remembering that elementsof k[x1, . . . , xn] can be understood as k-valued functions on it. For that reason itis convenient to denote this set differently, namely as An (or as Ank , if we feel thatwe should not forget about the field k). We refer to An as the affine n-space overk. A k-valued function on An is then said to be regular if it is defined by somef ∈ k[x1, . . . , xn]. We denote the zero set of such a function by Z(f) and thecomplement (the nonzero set) by U(f) ⊂ An. If f is not a constant polynomial(that is, f /∈ k), then we call Z(f) a hypersurface of An.

EXERCISE 1. Prove that f ∈ k[x1, . . . , xn] is completely determined by the reg-ular function it defines. (Hint: do first the case n = 1.) So the ring k[x1, . . . , xn] canbe regarded as a ring of functions on An under pointwise addition and multiplica-tion. (This would fail to be so had we not assumed that k is algebraically closed:for instance the function on the finite field Fq defined by xq −x is identically zero.)

EXERCISE 2. Prove that a hypersurface is nonempty.

1Since the elements of any algebraic extension of Fq of degree n ≥ 2 are roots of x(qn) − x, we

only need to adjoin roots of such polynomials.

3

4 1. AFFINE VARIETIES

It is perhaps somewhat surprising that in this rather algebraic context, the lan-guage of topology proves to be quite effective: algebraic subsets of An shall appearas the closed sets of a topology, albeit a rather peculiar one.

LEMMA-DEFINITION 1.1. The collection U(f) : f ∈ k[x1, . . . , xn] is a basis of atopology on An, called the Zariski topology2. A subset of An is closed for this topologyif and only if it is an intersection of zero sets of regular functions.

PROOF. We recall that a collection Uαα of subsets of a set X is a basis for atopology if and only if its union is all of X and any intersection Uα1

∩Uα2is a union

of members of Uαα. This is here certainly the case, for we have U(0) = X andU(f1) ∩ U(f2) = U(f1f2). Since an open subset of An is by definition a union ofsubsets of the form U(f), a closed subset must be an intersection of subsets of theform Z(f).

EXAMPLE 1.2. The Zariski topology on A1 is the cofinite topology: its closedsubsets 6= A1 are the finite subsets.

EXERCISE 3. Show that the diagonal in A2 is closed for the Zariski topology,but not for the product topology (where each factor A1 is equipped with the Zariskitopology). So A2 does not have the product topology.

We will explore the mutual relationship between the following two basic maps:

subsets of An I−−−−→ ideals of k[x1, . . . , xn]

∪ ∩

closed subsets of An Z←−−−− subsets of k[x1, . . . , xn].where for a subset X ⊂ An, I(X) is the ideal of f ∈ k[x1, . . . , xn] with f |X = 0 andfor a subset J ⊂ k[x1, . . . , xn], Z(J) is the closed subset of An defined by ∩f∈JZ(f).Observe that

I(X1 ∪X2) = I(X1) ∩ I(X2) and Z(J1 ∪ J2) = Z(J1) ∩ Z(J2).

In particular, both I and Z are inclusion reversing. Furthermore, I defines a sectionof Z: if Y ⊂ An is closed, then Z(I(Y )) = Y . We also note that by Exercise 1I(An) = (0), and that any singleton p = (p1, . . . , pn) ⊂ An is closed, as it is thecommon zero set of the degree one polynomials x1 − p1, . . . , xn − pn.

EXERCISE 4. Prove that I(p) is equal to the ideal generated by these degreeone polynomials and that this ideal is maximal.

EXERCISE 5. Prove that the (Zariski) closure of a subset Y of An is equal toZ(I(Y )).

Given Y ⊂ An, then f, g ∈ k[x1, . . . , xn] have the same restriction to Y if andonly if f − g ∈ I(Y ). So the quotient ring k[x1, . . . , xn]/I(Y ) (a k-algebra) can beregarded as a ring of k-valued functions on Y . Notice that this k-algebra does notchange if we replace Y by its Zariski closure.

DEFINITION 1.3. Let Y ⊂ An be closed. The k-algebra k[x1, . . . , xn]/I(Y ) iscalled the coordinate ring of Y and we denote it by A(Y ). A k-valued function onY is said to be regular if it lies in this ring.

2We shall later modify the definition of both An and the Zariski topology.

1. THE ZARISKI TOPOLOGY 5

Given a closed subset Y ⊂ An, then for every subset X ⊂ An we have X ⊂ Y ifand only if I(X) ⊃ I(Y ), and in that case IY (X) := I(X)/I(Y ) is an ideal of A(Y ):it is the ideal of regular functions on Y that vanish on X. Conversely, an ideal ofA(Y ) is of the form J/I(Y ), with J an ideal of k[x1, . . . , xn] that contains I(Y ), andsuch an ideal defines a closed subset Z(J) contained in Y . So the two basic mapsabove give rise to such a pair on Y :

subsets of Y IY−−−−→ ideals of A(Y )

∪ ∩

closed subsets of Y ZY←−−−− subsets of A(Y ).

We ask: which ideals of k[x1, . . . , xn] are of the form I(Y ) for some Y ? Clearly, iff ∈ k[x1, . . . , xn] is such that some positive power vanishes on Y , then f vanisheson Y . In other words: if fm ∈ I(Y ) for some m > 0, then f ∈ I(Y ). This suggests:

PROPOSITION-DEFINITION 1.4. Let R be a ring (as always commutative and with1) and let J ⊂ R be an ideal. Then the set of a ∈ R with the property that am ∈ J forsome m > 0 is an ideal of R, called the radical of J and denoted

√J .

We say that J is a radical ideal if√J = J .

We say that the ring R is reduced if the zero ideal (0) is a radical ideal (in otherwords, R has no nonzero nilpotents: if a ∈ R is such that am = 0, then a = 0).

PROOF. We show that√J is an ideal. Let a, b ∈

√J so that am, bn ∈ J for

certain positive integers m,n. Then for every r ∈ R, ra ∈√J , since (ra)m =

rmam ∈ J . Similarly a−b ∈√J , for (a−b)m+n is a linear combination of monomials

that are multiples of am or bn and hence lie in J .

EXERCISE 6. Show that a prime ideal is a radical ideal.

Notice that J is a radical ideal if and only if R/J is reduced. The precedingshows that for every Y ⊂ An, I(Y ) is a radical ideal, so that A(Y ) is reduced. Thedictionary between algebra and geometry begins in a more substantial manner with

THEOREM 1.5 (Hilbert’s Nullstellensatz). For every ideal J ⊂ k[x1, . . . , xn] wehave I(Z(J)) =

√J .

We inclusion ⊃ is clear; the hard part is the opposite inclusion (which says thatif f ∈ k[x1, . . . , xn] vanishes on Z(J), then fm ∈ J for some positive integer m).We postpone its proof, but we discuss some of the consequences.

COROLLARY 1.6. Let Y ⊂ An be closed. Then the maps IY and ZY restrict tobijections:

closed subsets of Y ↔ radical ideals of A(Y ).These bijections are inclusion reversing and each others inverse.

PROOF. We first prove this for Y = An. We already observed that for everyclosed subset X of An we have Z(I(X)) = X . The Nullstellensatz says that for aradical ideal J ⊂ k[x1, . . . , xn], we have I(Z(J)) = J .

An ideal of A(Y ) is of the form J/I(Y ). This is a radical ideal if and only if Jis one. So the property also follows for an arbitrary closed Y .


Let m be a maximal ideal of k[x1, . . . , xn]. Such an ideal is certainly radicalas it is a prime ideal and so it is also maximal among the radical ideals that aredistinct from k[x1, . . . , xn]. Hence it is of the form I(Y ) for a closed subset Y . Sincethe empty subset of An is defined by the radical ideal k[x1, . . . , xn], Corollary 1.6implies that Y will be nonempty and minimal for this property. In other words,Y is a singleton y (and if y = (a1, . . . , an), then m = (x1 − a1, . . . , xn − an), byExercise 4). Thus the above correspondence provides a bijection between the pointsof An and the maximal ideals of An. If we are given a closed subset Y ⊂ An anda point y ∈ An, then y ∈ Y if and only if the maximal ideal defined by y containsI(Y ). But a maximal ideal containing I(Y ) is the preimage of a maximal ideal inA(Y ) = k[x1, . . . , xn]/I(Y ). We conclude that points of Y correspond to maximalideals of A(Y ). Via this (or a very similar) correspondence, algebraic geometryseeks to express geometric properties of Y in terms of algebraic properties of A(Y )and vice versa. In the end we want to forget about the ambient An completely.

2. Irreducibility and decomposition

We introduce a property which for most topological spaces is of little interest,but as we will see, is useful and natural for the Zariski topology.

DEFINITION 2.1. Let Y be a topological space. We say that Y is irreducible ifit is nonempty and cannot be written as the union of two closed subsets 6= Y . Anirreducible component of Y is a maximal irreducible subset of Y .

EXERCISE 7. Prove that an irreducible Hausdorff space must consist of a singlepoint. Prove also that an infinite set with the cofinite topology is irreducible.

EXERCISE 8. Let G1, . . . , Gs be closed subsets of a topological space Y . Provethat any irreducible subset of G1 ∪ · · · ∪Gs is contained in some Gi.

EXERCISE 9. This exercise has been scrapped.

LEMMA 2.2. Let Y be a topological space.(i) If Y is irreducible, then every nonempty open subset of Y is dense in Y and

irreducible.(ii) Conversely, if C ⊂ Y is an irreducible subspace, then C is also irreducible.

In particular, an irreducible component of Y is always closed in Y .

PROOF. (i) Suppose Y is irreducible and let U ⊂ Y be open and nonempty.Then Y is the union of the two closed subspaces Y −U and U . Since Y is irreducible,and Y − U 6= Y , we must have U = Y . So U is dense in Y . To see that U isirreducible, suppose that U is the union of two subsets that are both closed in U .These subsets will be of the form Gi ∩ U with Gi closed in Y . Then G1 ∪ G2 is aclosed subset of Y which contains U . Since U = Y , it follows that G1 ∪ G2 = Y .The irreducibility of Y implies that one of the Gi (say G1) equals Y and thenU = G1 ∩ U .

(ii) Let C ⊂ Y be irreducible (and hence nonempty). If C is written as a unionof two closed subsets G1, G2 of Y , then C is the union of the two subsets G1 ∩ Cand G2 ∩ C that are both closed in C, and so one of these, say G1 ∩ C, equals C.This means that G1 ⊃ C and hence G1 ⊃ C. So C is irreducible.

The following proposition tells us what irreducibility amounts to in the Zariskitopology.

2. IRREDUCIBILITY AND DECOMPOSITION 7

PROPOSITION 2.3. A closed subset Y ⊂ An is irreducible if and only if I(Y ) isa prime ideal (which, we recall, is equivalent to: A(Y ) = k[x1, . . . , xn]/I(Y ) is adomain).

PROOF. Suppose Y is irreducible and f, g ∈ k[x1, . . . , xn] are such that fg ∈I(Y ). Then Y ⊂ Z(fg) = Z(f) ∪ Z(g). Since Y is irreducible, Y is contained inZ(f) or in Z(g). So f ∈ I(Y ) or g ∈ I(Y ), proving that I(Y ) is a prime ideal.

Suppose I(Y ) is a prime ideal, but that Y is the union of two closed subsets Y1

and Y2 that are both 6= Y . Since Yi 6= Y , there exists a fi ∈ I(Yi) − I(Y ). Thenf1f2 vanishes on Y1 ∪ Y2 = Y , so that f1f2 ∈ I(Y ). The fact I(Y ) is a prime idealimplies that one of f1 and f2 is in I(Y ). Whence a contradiction.

One of our first aims is to prove that the irreducible components of any closedsubset Y ⊂ An are finite in number and have Y as their union. This may notsound very surprising, but we will see that this reflects some nonobvious algebraicproperties. Let us first consider the case of a hypersurface. Since we are goingto use the fact that k[x1, . . . , xn] is a unique factorization domain, we begin withrecalling that notion:

DEFINITION 2.4. We say that R is a unique factorization domain if R has nozero divisors and every principal ideal (a) := Ra in R which is neither the zeroideal nor all of R is in unique manner a product of principal prime ideals: (a) =(p1)(p2) · · · (ps) (so the ideals (p1), . . . , (ps) are unique up to order).

Observe that the principal ideal generated by p ∈ R is prime if and only if forany factoring of p, p = p′p′′, one of the factors must be a unit and that the identity(a) = (p1)(p2) · · · (ps) amounts to the statement that a is a unit times p1p2 · · · ps.The factors are then unique up to order and multiplication by a unit.

For a field (which has no proper principal ideals distinct from (0)) the imposedcondition is empty and hence a field is automatically unique factorization domain.A more substantial example (that motivated this notion in the first place) is Z: aprincipal prime ideal of Z is of the form (p), with p a prime number. Every integern ≥ 2 has a unique prime decomposition and so Z is a unique factorization domain.

A basic theorem in the theory of rings asserts that if R is a unique factorizationdomain, then so is its polynomial ring R[x]. This implies for instance (with induc-tion on n) that R[x1, . . . , xn] is one. Indeed, in the case when R is a field (think ofour k), then a nonzero principal ideal of this ring is prime precisely when it is gen-erated by an irreducible polynomial of positive degree and every f ∈ R[x1, . . . , xn]of positive degree then can be written as a product of irreducible polynomials:f = f1f2 · · · fs, a factorization that is unique up to order and multiplication of eachfi by a nonzero element of R.

The following proposition connects two notions of irreducibility.

PROPOSITION 2.5. If f ∈ k[x1, . . . , xn] is irreducible, then the hypersurface Z(f)it defines is irreducible. If f ∈ k[x1, . . . , xn] is of positive degree and f = f1f2 · · · fsis a factoring into irreducible polynomials, then Z(f1), . . . , Z(fs) are the irreduciblecomponents of Z(f) and their union equals Z(f) (but we are not claiming that theZ(fi)’s are pairwise distinct). In particular, a hypersurface is the union of its ir-reducible components; these irreducible components are hypersurfaces and finite innumber.


PROOF. If f ∈ k[x1, . . . , xn] is irreducible, then f generates a prime ideal andso Z(f) is an irreducible hypersurface by Proposition 2.3.

If f ∈ k[x1, . . . , xn] is of positive degree and f = f1f2 · · · fs is as in the propo-sition, then it is clear that Z(f) = Z(f1) ∪ · · · ∪ Z(fs) with each Z(fi) irreducible.To see that Z(fi) is an irreducible component of Z(f), suppose that C ⊂ Z(f) isirreducible. By Exercise 8, C is contained in some Z(fi). Since C is a maximalirreducible subset of Z(f), it follows that we then must have equality: C = Z(fi).It remains to observe that if any inclusion relation Z(fi) ⊂ Z(fj) is necessarily anidentity (prove this yourself) so that each Z(fi) is already maximal and hence inirreducible component.

The discussion of the general case begins with the rather formal

LEMMA 2.6. For a partially ordered set (A,≤) the following are equivalent:(i) (A,≤) satisfies the ascending chain condition: every ascending chain a1 ≤

a2 ≤ a3 ≤ · · · becomes stationary: an = an+1 = · · · for n sufficiently large.(ii) Every nonempty subset B ⊂ A has a maximal element, that is, an element

b0 ∈ B such that there is no b ∈ B with b > b0.

PROOF. (i)⇒(ii). Suppose (A,≤) satisfies the ascending chain condition andlet B ⊂ A be nonempty. Choose b1 ∈ B. If b1 is maximal, we are done. If not, thenthere exists a b2 ∈ B with b2 > b1. We repeat the same argument for b2. We cannotindefinitely continue in this manner because of the ascending chain condition.

(ii)⇒(i). If (A,≤) satisfies (ii), then the set of members of any ascending chainhas a maximal element, in other words, the chain becomes stationary.

If we replace ≤ by ≥, then we obtain the notion of the descending chain condi-tion and we find that this property is equivalent to: every nonempty subset B ⊂ Ahas a minimal element. These properties appear in the following pair of definitions.

DEFINITIONS 2.7. We say that a ring R is noetherian if its collection of idealssatisfies the ascending chain condition.

We say that a topological space Y is noetherian if its collection of closed subsetssatisfies the descending chain condition.

EXERCISE 10. Prove that a subspace of a noetherian space is noetherian. Provealso that a ring quotient of a noetherian ring is noetherian.

EXERCISE 11. Prove that a noetherian space is compact: every covering of sucha space by open subsets contains a finite subcovering.

The interest of the noetherian property is that it is one which is possessed byalmost all the rings we encounter and that it implies many finiteness propertieswithout which we are often unable to go very far. Let us give a nonexample first:the ring R of holomorphic functions on C is not noetherian: if Ik denotes the idealof f ∈ R vanishing on all the integers ≥ k, then I1 ⊂ I2 ⊂ · · · is a strictly ascendingchain of ideals in R.

Obviously a field is noetherian. The ring Z is noetherian: if I1 ⊂ I2 ⊂ · · · is anascending chain of ideals in Z, then ∪∞s=1Is is an ideal of Z, hence of the form (n)for some n ∈ Z. But if s is such that n ∈ Is, then clearly the chain is stationary asof index s. (This argument only used the fact that any ideal in Z is generated by asingle element, i.e., that Z is a principal ideal domain.) That most rings we knoware noetherian is a consequence of


THEOREM 2.8 (Hilbert’s basis theorem). If R is a noetherian ring, then so isR[x].

As with the Nullstellensatz, we postpone the proof and discuss some of itsconsequences first.

COROLLARY 2.9. If R is a noetherian ring (for example, a field) then so is everyfinitely generated R-algebra. Also, the space An (and hence any closed subset of An)is noetherian.

PROOF. The Hilbert basis theorem implies (with induction on n) that the ringR[x1, . . . , xn] is noetherian. By Exercise 10, every quotient ring R[x1, . . . , xn]/I isthen also noetherian. But a finitely generated R-algebra is (by definition actually)isomorphic to some such quotient and so the first statement follows.

Suppose An ⊃ Y1 ⊃ Y2 ⊃ · · · is a descending chain of closed subsets. Then(0) ⊂ I(Y1) ⊂ I(Y2) ⊂ · · · is an ascending chain of ideals. As the latter becomesstationary, so will become the former.

PROPOSITION 2.10. If Y is noetherian space, then its irreducible components arefinite in number and their union equals Y .

PROOF. Suppose Y is a noetherian space. We first show that every closed sub-set can be written as a finite union of closed irreducible subsets. First note thatthe empty set has this property, for a union with empty index set is empty. Let Bbe the collection of closed subspaces of Y for which this is not possible, i.e., thatcannot be written as a finite union of closed irreducible subsets. Suppose that Bis nonempty. According to 2.6 this collection has a minimal element, Z, say. ThisZ must be nonempty and cannot be irreducible. So Z is the union of two properclosed subsets Z ′ and Z ′′. The minimality of Z implies that neither Z ′ nor Z ′′ isin B and so both Z ′ and Z ′′ can be written as a finite union of closed irreduciblesubsets. But then so can Z and we get a contradiction.

In particular, there exist closed irreducible subsets Y1, . . . , Yk of Y such thatY = Y1 ∪ · · · ∪ Yk (if Y = ∅, take k = 0). We may of course assume that noYi is contained in some Yj with j 6= i (otherwise, omit Yi). It now remains toprove that the Yi’s are the irreducible components of Y , that is, we must show thatevery irreducible subset C of Y is contained in some Yi. But this follows from anapplication of Exercise 8.

If we apply this to An, then we find that every subset Y ⊂ An has a finitenumber of irreducible components, the union of which is all of Y . If Y is closedin An, then so is every irreducible component of Y and according to Proposition2.3 any such irreducible component is defined by a prime ideal. This allows us torecover the irreducible components of a closed subset Y ⊂ An from its coordinatering:

COROLLARY 2.11. Let Y ⊂ An be a closed subset. If C is an irreducible componentof Y , then the image IY (C) of I(C) in A(Y ) is a minimal prime ideal of A(Y ) andany minimal prime ideal of A(Y ) is so obtained: we thus get a bijective correspondencebetween the irreducible components of Y and the minimal prime ideals of A(Y ).

PROOF. Let C be a closed subset of Y and let IY (C) be the corresponding idealof A(Y ). Now C is irreducible if and only if I(C) is a prime ideal of k[x1, . . . , xn],or what amounts to the same, if and only if IY (C) is a prime ideal of A(Y ). It is


an irreducible component if C is maximal for this property, or what amounts to thesame, if IY (C) is minimal for the property of being a prime ideal of A(Y ).

EXAMPLE 2.12. First consider the set C := (t, t2, t3) ∈ A3 | t ∈ k. This is aclosed subset of A3: if we use (x, y, z) instead of (x1, x2, x3), then C is the commonzero set of y−x2 and z−x3. Now the inclusion k[x] ⊂ k[x, y, z] composed with thering quotient k[x, y, z] → k[x, y, z]/(y − x2, z − x3) is easily seen to be an isomor-phism. Since k[x] has no zero divisors, (y− x2, z− x3) must be a prime ideal. So Cis irreducible and I(C) = (y − x2, z − x3).

We now turn to the closed subset Y ⊂ A3 defined by xy−z = 0 and y3−z2 = 0.Let p = (x, y, z) ∈ Y . If y 6= 0, then we put t := z/y; from y3 = x2, it follows thaty = t2 and z = t3 and xy = z implies that x = t. In other words, p ∈ C in thatcase. If y = 0, then z = 0, in other words p lies on the x-axis. Conversely, any pointon the x-axis lies in Y . So Y is the union of C and the x-axis and these are theirreducible components of Y .

We briefly discuss the corresponding issue in commutative algebra. We beginwith recalling the notion of localization and we do this in the generality that isneeded later.

2.13. LOCALIZATION. LetR be a ring and let S be a multiplicative subset ofR: 1 ∈ S, 0 /∈S and S closed under multiplication. Then a ring S−1R, together with a ring homomorphismR→ S−1R is defined as follows: an element of S−1R is written as a formal fraction r/s, withr ∈ R and s ∈ S, with the understanding that r/s = r′/s′ if and only if s′′(s′r− sr′) = 0 forsome s′′ ∈ S. This is a ring indeed: multiplication and subtraction is defined as for ordinaryfractions: r/s.r′/s′ = (rr′)/(ss′) and r/s− r′/s′ = (s′r − sr′)/(ss′); it has 0/1 as zero and1/1 as unit element. Since 0 /∈ S, we have 0/1 6= 1/1. The homomorphism R → S−1Ris simply r 7→ r/1. Notice that it maps any s ∈ S to an invertible element of S−1R: theinverse of s/1 is 1/s. In a sense (made precise in part (b) of Exercise 12 below) the ringhomomorphism R → S−1R is universal for that property. This construction is called thelocalization away from S.3

It is clear that if S does not contain zero divisors, then r/s = r′/s′ if and only ifs′r − sr′ = 0; in particular, r/1 = r′/1 if and only if r = r′, so that R → S−1R is theninjective. If we take S maximal for this property, namely take it to be the set of nonzerodivisors of R (which is indeed multiplicative), then S−1R is called the fraction ring Frac(R)of R. In case R is a domain, S = R − 0 and so Frac(R) is a field, the fraction field ofR. This gives the following corollary, which hints to the importance of prime ideals in thesubject.

COROLLARY 2.14. An ideal p of a ring R is a prime ideal if and only if it is the kernel of aring homomorphism from R to a field

PROOF. It is clear that the kernel of a ring homomorphism from R to a field is alwaysa prime ideal. Conversely, if p is a prime ideal, then it is the kernel of the composite R →R/p → Frac(R/p).

But the case of interest here is when we are given some a ∈ R which is not nilpotent.Then we can take S = an |n ≥ 0 in which case we usually write R[1/a] for S−1R.

EXERCISE 12. Let R be a ring and let S be a multiplicative subset of R.(a) What is the the kernel of R→ S−1R?(b) Prove that a ring homomorphism φ : R → R′ with the property that φ(s)

is invertible for every s ∈ S factors in a unique manner through S−1R.

3It is sometimes useful to allow 0 ∈ S. We then stipulate that S−1R is the zero ring.


(c) Consider the polynomial ring R[xs : s ∈ S] and the homomorphism ofR-algebras R[xs : s ∈ S] → S−1R that sends xs to 1/s. Prove that thishomomorphism is surjective and that its kernel consists of the f ∈ R[xs :s ∈ S] which after multiplication by an element of S lie in the idealgenerated the degree one polynomials sxs − 1, s ∈ S.

EXERCISE 13. Let R be a ring and let p be a prime ideal of R.(a) Prove that the complement R − p is a multiplicative system. The result-

ing localization (R − p)−1R is called the localization at p and is usuallydenoted Rp.

(b) Prove that pRp is a maximal ideal of Rp and that it is the only maximalideal of Rp. (A ring with a unique maximal ideal is called a local ring.)

(c) Prove that the localization mapR→ Rp drops to an isomorphism of fieldsFrac(R/p)→ Rp/pRp.

(d) Work this out for R = Z and p = (p), where p is a prime number.(e) Same for R = k[x, y] and p = (x).

LEMMA 2.15. Let R be a ring. Then the intersection of all the prime ideals of Ris the ideal of nilpotents

√(0) of R. Equivalently, for every nonnilpotent a ∈ R, there

exists a ring homomorphism from R to a field that is nonzero on a.

PROOF. It is easy to see that a nilpotent element lies in every prime ideal. Nowfor nonnilpotent a ∈ R consider the homomorphism R → R[1/a]. The ring R[1/a]

has a maximal ideal4 and hence admits a ring homomorphism to some field F :φ : R[1/a] → F . Then the kernel of the composite R → R[1/a] → F is a primeideal and a is not in this kernel (for its image is invertible with inverse φ(1/a)).

EXERCISE 14. Let R be a ring. Prove that the intersection of all the maximalideals of a ring R consists of the a ∈ R for which 1 + aR ⊂ R× (i.e., 1 + ax isinvertible for every x ∈ R). You may use the fact that every proper ideal of R iscontained in a maximal ideal.

We can do better if R is noetherian. The following proposition is the algebraiccounterpart of Proposition 2.10. Note the similarity between the proofs.

PROPOSITION 2.16. Let R be a noetherian ring. Then any radical ideal of R isan intersection of finitely many prime ideals. Also, the minimal prime ideals of R arefinite in number and their intersection is equal to the ideal of nilpotents

√(0).

PROOF. Let B be the collection of the radical ideals I ( R that can not be writ-ten as an intersection of finitely many prime ideals and suppose that B is nonempty.Since R is noetherian, B contains a maximal member I0. We derive a contradictionas follows.

Since I0 cannot be a prime ideal, there exist a1, a2 ∈ R − I0 with a1a2 ∈ I0.Consider the radical ideal Ji :=

√I0 +Rai. We claim that J1 ∩ J2 = I0. The

inclusion⊃ is obvious and⊂ is seen as follows: if a ∈ J1∩J2, then for i = 1, 2, thereexists an ni > 0 such that ani ∈ I0+Rai. Hence an1+n2 ∈ (I0+Ra1)(I0+Ra2) = I0,so that a ∈ I0. Since Ji strictly contains I0, it does not belong to B. In other words,

4Every ring has a maximal ideal. For noetherian rings, which are our main concern, this is obvious,but in general this follows with transfinite induction, the adoption of which is equivalent to the adoptionof the axiom of choice.


Ji is an intersection of finitely many prime ideals. But then so is J1 ∩ J2 = I0 andwe get a contradiction.

We thus find that√

(0) = p1 ∩ · · · ∩ ps for certain prime ideals pi. We mayof course assume that no pi contains some pj with j 6= i (otherwise, omit pi). Itnow remains to prove that every prime ideal p of R contains some pi. If that isnot the case, then for i = 1, . . . , s there exists a ai ∈ pi − p. But then a1a2 · · · as ∈p1 ∩ · · · ∩ ps =

√(0) ⊂ p and since p is a prime ideal, some factor ai lies in p. This

is clearly a contradiction.

EXERCISE 15. Let J be an ideal of the ring R. Show that√J is the intersection

of all the prime ideals that contain J . Prove that in case R is noetherian, the primeideals that are minimal for this property are finite in number and that their commonintersection is still

√J . What do we get for R = Z and J = Zn?

3. Finiteness properties and the Hilbert theorems

The noetherian property in commutative algebra is best discussed in the con-text of modules, even if one’s interest is only in rings. We fix a ring R and first recallthe notion of an R-module.

The notion of an R-module is the natural generalization of a K-vector space (where Kis some field). Let us observe that if M is an (additively written) abelian group, then theset End(M) of group homomorphisms M → M is a ring for which subtraction is pointwisedefined and multiplication is composition (so if f, g ∈ End(M), then f − g : m ∈ M 7→f(m) − g(m) and fg : m 7→ f(g(m))); clearly the zero element is the zero homomorphismand the unit element is the identity. It only fails to obey our convention in the sense that thisring is usually noncommutative. We only introduced it in order to be able state succinctly:

DEFINITION 3.1. An R-module is an abelian group M , equipped with a ring homomor-phism R→ End(M).

So any r ∈ R defines a homomorphism M → M ; we usually denote the image ofm ∈M simply by rm. If we write out the properties of anR-module structure in these terms,we get: r(m1−m2) = rm1− rm2, (r1− r2)m = r1m− r2m, 1.m = m, r1(r2m) = (r1r2)m.If R happens to be field, then we see that an R-module is the same thing as an R-vectorspace.

The notion of an R-module is quite ubiquitous if you think about it. A simple exampleis any ideal I ⊂ R. Any abelian group M is in a natural manner a Z-module. A R[x]-modulecan be understood as an real linear space V (an R-module) endowed with an endomorphism(the image of x in End(V )). A more involved example is the following: if X is a manifold,f is a C∞-function on X and ω a C∞ differential p-form on X, then fω is also a C∞

differential p-form on X. Thus the linear space of C∞-differential forms on X of a fixeddegree p is naturally a module over the ring of C∞-functions on X.

Here are a few companion notions, followed by a brief discussion.

3.2. In what follows is M an R-module. A map f : M → N from M to an R-module Nis called a R-homomorphism if it is a group homomorphism with the property that f(rm) =rf(m) for all r ∈ R and m ∈ M . If f is also bijective, then we call it an R-isomorphism; inthat case its inverse is also a homomorphism of R-modules.

For instance, given a ring homomorphism f : R → R′, then R′ becomes an R-moduleby rr′ := f(r)r′ and this makes f a homomorphism of R-modules.

A subset N ⊂ M is called an R-submodule of M if it is a subgroup and rn ∈ N for allr ∈ R and n ∈ N . Then the group quotient M/N is in a unique manner a R-module in sucha way that the quotient map M →M/N is a R-homomorphism: we let r(m+N) := rm+N

3. FINITENESS PROPERTIES AND THE HILBERT THEOREMS 13

for r ∈ R and m ∈ M . Notice that a R-submodule of R (here we regard R as a R-module)is the same thing as an ideal of R.

Given a subset S ⊂ M , then the set of elements m ∈ M that can be written as r1s1 +· · ·+ rksk with ri ∈ R and si ∈ S is a R-submodule of M . We call it the R-submodule of Mgenerated by S and we shall denote it by RS.

If there exists a finite set S ⊂ M such that M = RS, then we say that M is finitelygenerated as an R-module.

DEFINITION 3.3. We say that an R-module M is noetherian if the collection ofR-submodules of M satisfies the ascending chain condition: any ascending chainof R-submodules N1 ⊂ N2 ⊂ · · · becomes stationary.

It is clear that then every quotient module of a noetherian module is also noe-therian. The noetherian property of R as a ring (as previously defined) coincideswith this property of R as an R-module.

The following two propositions provide the passage from the noetherian prop-erty to finite generation:

PROPOSITION 3.4. AnR-moduleM is noetherian if and only if everyR-submoduleof M is finitely generated as an R-module.

PROOF. Suppose that M is a noetherian R-module and let N ⊂ M be a R-submodule. The collection of finitely generated R-submodules of M contained inN is nonempty. Hence it has a maximal element N0. If N0 = N , then N is finitelygenerated. If not, we run into a contradiction: just choose x ∈ N−N0 and considerN0 + Rx. This is a R-submodule of N . It is finitely generated (for N0 is), whichcontradicts the maximal character of N0.

Suppose now that every R-submodule of M is finitely generated. If N1 ⊂N2 ⊂ · · · is an ascending chain of R-modules, then the union N := ∪∞i=1Ni is aR-submodule. Let s1, . . . , sk be a finite set of generators of N . If sκ ∈ Niκ , andj := maxi1, . . . , ik, then it is clear that Nj = N . So the chain becomes stationaryas of index j.

PROPOSITION 3.5. Suppose that R is a noetherian ring. Then every finitely gen-erated R-module M is noetherian.

PROOF. By assumption M = RS for a finite set S ⊂ M . We prove the propo-sition by induction on the number of elements of S. If S = ∅, then M = 0 andthere is nothing to prove. Suppose now S 6= ∅ and choose s ∈ S, so that our induc-tion hypothesis applies to M ′ := RS′ with S′ = S − s: M ′ is noetherian. But sois M/M ′, for it is a quotient of the noetherian ring R via the surjective R-modulehomomorphism R→M/M ′, r 7→ rs+M ′.

Let now N1 ⊂ N2 ⊂ · · · be an ascending chain of R-submodules of M . ThenN1 ∩M ′ ⊂ N2 ∩M ′ ⊂ · · · becomes stationary, say as of index j1. Hence we onlyneed to be concerned with the stabilization of the submodules Nk/(Nk ∩M ′) ofM/M ′. These stabilize indeed (say as of index j2), since M/M ′ is noetherian. Sothe original chain N1 ⊂ N2 ⊂ · · · stabilizes as of index maxj1, j2.

We are now sufficiently prepared for the proofs of the Hilbert theorems. Theyare jewels of elegance and efficiency.

We will use the notion of initial coefficient of a polynomial, which we re-call. Given a ring R, then every nonzero f ∈ R[x] is uniquely written as rdxd +rd−1x

d−1 + · · · + r0 with rd 6= 0. We call rd ∈ R the initial coefficient of f and


denote it by in(f). For the zero polynomial, we simply define this to be 0 ∈ R.Notice that in(f) in(g) equals in(fg) when nonzero.

PROOF OF THEOREM 2.8. The assumption is here that R is a noetherian ring.In view of Proposition 3.4 we must show that every ideal I of R[x] is finitely gener-ated. Consider the subset in(I) := in(f) : f ∈ I of R. We first show that this isan ideal of R. If r ∈ R, f ∈ I, then r in(f) equals in(rf) when nonzero and sincerf ∈ I, it follows that r in(f) ∈ I. If f, g ∈ I with d := deg(f) − deg(g) ≥ 0, thenin(f)− in(g) = in(f − tdg). So in(I) is an ideal as asserted.

Since R is noetherian, in(I) is finitely generated: there exist f1, . . . , fk ∈ Isuch that in(I) = R in(f1) + · · · + R in(fk). Let di be the degree of fi, d0 :=maxd1, . . . , dk and R[x]<d0 the set of polynomials of degree < d0. So R[x]<d0 isthe R-submodule of R[x] generated by 1, x, . . . , xd0−1. We claim that

I = R[x]f1 + · · ·+R[x]fk + (I ∩R[x]<d0),

in other words, that every f ∈ I is modulo R[x]f1 + · · · + R[x]fk a polynomial ofdegree < d0. We prove this with induction on the degree d of f . Since for d < d0

there is nothing to prove, assume that d ≥ d0. We have in(f) = r1 in(f1) + · · · +rk in(fk) for certain r1, . . . rk ∈ R, where we may of course assume that every termri in(fi) is nonzero and hence equal to in(rifi). Then in(f) equals

∑i in(rifi) =

in(∑i rifix

d−di). So f −∑i rifix

d−di is an element of I of degree < d and hencelies in R[x]f1 + · · ·+R[x]fk + (I ∩R[x]<d0) by our induction hypothesis. Hence sodoes f .

Our claim implies the theorem: R[x]<d0 is a finitely generated R-module andso a noetherian R-module by Proposition 3.5. Hence the R-submodule I ∩R[x]<d0is a finitely generated R-module by Proposition 3.4. If fk+1, . . . , fk+l is a set ofR-generators of I ∩R[x]<d0 , then f1, . . . , fk+l is a set of R[x]-generators of I.

For the Nullstellensatz we need another finiteness result.

PROPOSITION 3.6 (Artin-Tate). Let R be a noetherian ring, B an R-algebra andA ⊂ B an R-subalgebra. Assume that B is finitely generated as an A-module. ThenA is finitely generated as an R-algebra if and only if B is so.

PROOF. By assumption there exist b1, . . . , bm ∈ B such that B =∑mi=1Abi.

If there exist a1, . . . , an ∈ A which generate A as an R-algebra (which meansthat A = R[a1, . . . , an]), then a1, . . . , an, b1, . . . , bm generate B as an R-algebra.

Suppose, conversely, that there exists a finite subset of B which generates Bas a R-algebra. By adding this subset to b1, . . . , bm, we may assume that b1, . . . , bmalso generate B as an R-algebra. Then every product bibj can be written as anA-linear combination of b1, . . . , bm:

bibj =

m∑k=1

akijbk, akij ∈ A.

Let A0 ⊂ A be the R-subalgebra of A generated by all the (finitely many) coeffi-cients akij . This is a noetherian ring by Corollary 2.9. It is clear that bibj ∈

∑k A0bk

and with induction it then follows that B = R[b1, . . . , bm] ⊂∑k A0bk. So B is

finitely generated as an A0-module. Since A is an A0-submodule of B, A is alsofinitely generated as an A0-module by Proposition 3.4. It follows that A is a finitelygenerated R-algebra.

4. THE AFFINE CATEGORY 15

This has a consequence for field extensions:

COROLLARY 3.7. A field extension L/K is finite if and only if L is finitely gener-ated as a K-algebra.

PROOF. It is clear that if L is a finite dimensional K-vector space, then L isfinitely generated as a K-algebra.

Suppose now b1, . . . , bm ∈ L generate L as a K-algebra. We must show thatevery bi is algebraic over K. Suppose that this is not the case. After renumberingwe can and will assume that (for some 1 ≤ r ≤ m) b1, . . . , br are algebraically inde-pendent over K and br+1, . . . , bm are algebraic over the quotient field K(b1, . . . , br)of K[b1, . . . , br]. So L is a finite extension of K(b1, . . . , br). We apply Proposition3.6 to R := K, A := K(b1, . . . , br) and B := L and find that K(b1, . . . , br) is asa K-algebra generated by a finite subset S ⊂ K(b1, . . . , br). If g is a common de-nominator for the elements of S, then clearly K(b1, . . . , br) = K[b1, . . . , br][1/g].Since K(b1, . . . , br) strictly contains K[b1, . . . , br], g must have positive degree. Inparticular, 1 + g 6= 0, so that 1/(1 + g) ∈ K(b1, . . . , br) can be written as f/gN , withf ∈ K[b1, . . . , br]. Here we may of course assume that f is not divisible by g inK[b1, . . . , br]. From the identity f(1+g) = gN we see that N ≥ 1 (for g has positivedegree). But then f = g(−f + gN−1) shows that f is divisible by g. We thus get acontradiction.

EXERCISE 16. Prove that a field which is finite generated as a ring (that is,isomorphic to a quotient of Z[x1, . . . , xn] for some n) is finite.

We deduce from the preceding corollary the Nullstellensatz.

PROOF OF THE NULLSTELLENSATZ 1.5. Let J ⊂ k[x1, . . . , xn] be an ideal. Wemust show that I(Z(J)) ⊂

√J . This amounts to: for every f ∈ k[x1, . . . , xn]−

√J

there exists a p ∈ Z(J) for which f(p) 6= 0. Consider k[x1, . . . , xn]/J and denote byf ∈ k[x1, . . . , xn]/J the image of f . Since f is not nilpotent, we have defined

A := (k[x1, . . . , xn]/J)[1/f ].

Notice that A is as a k-algebra generated by the images of x1, . . . , xn and 1/f(hence is finitely generated). Choose a maximal ideal m ⊂ A. Then the fieldA/m is a finitely generated k-algebra and so by Corollary 3.7 a finite extensionof k. But then it must be equal to k, since k is algebraically closed. Denote byφ : k[x1, . . . , xn]→ A→ A/m = k the corresponding surjection and put pi := φ(xi)and p := (p1, . . . , pn) ∈ An. So if we view xi as a function on An, then xi(p) = pi =φ(xi). The fact that φ is a homomorphism of k-algebras implies that it is given as‘evaluation in p’: any g ∈ k[x1, . . . , xn] takes in p the value φ(g). Since the kernelof φ contains J , every g ∈ J will be zero in p, in other words, p ∈ Z(J). On theother hand, f(p) = φ(f) is invertible, for it has the image of 1/f in A/m = k as itsinverse. So f(p) 6= 0.

4. The affine category

We begin with specifying the maps between closed subsets of affine spaces thatwe wish to consider.

DEFINITION 4.1. Let X ⊂ Am and Y ⊂ An be closed subsets. We say that amap f : X → Y is regular if the components f1, . . . , fn of f are regular functionson X (i.e., are given by the restrictions of polynomial functions to X).


Composition of a regular function on Y with f yields a regular function on X(for if we substitute in a polynomial of n variables g(y1, . . . , yn) for every variableyi a polynomial fi(x1, . . . , xm) of m variables, we get a polynomial of m variables).So f then induces a k-algebra homomorphism f∗ : A(Y ) → A(X). There is also aconverse:

PROPOSITION 4.2. Let be given closed subsets X ⊂ Am and Y ⊂ An and a k-algebra homomorphism φ : A(Y ) → A(X). Then there is a unique regular mapf : X → Y such that f∗ = φ.

PROOF. Let fi := φ(yi|Y ) ∈ A(X) (i = 1, . . . , n) and define f = (f1, . . . , fn) :X → An, so that f∗yi = fi = φ(yi|Y ). If j : Y ⊂ An denotes the inclusion, then wecan also write this as f∗yi = φj∗yi. In other words, the k-algebra homomorphismsf∗, φj∗ : k[y1, . . . , yn] → A(X) coincide on the generators yi. Hence they must beequal: f∗ = φj∗. It follows that f∗ is zero on I(Y ), which means that f takes itsvalues in Z(I(Y )) = Y , and that the resulting map A(Y )→ A(X) equals φ.

The same argument shows that if f : X → Y and g : Y → Z are regularmaps, then so is their composite gf : X → Z. So we have a category (with objectsthe closed subsets and regular maps as defined above). In particular, we have anotion of isomorphism: a regular map f : X → Y is an isomorphism if is hasa two-sided inverse g : Y → X which is also a regular map. This implies thatf∗ : A(Y ) → A(X) has a two-sided inverse g∗ : A(X) → A(Y ) which is also anisomorphism of k-algebras, and hence is an isomorphism of k-algebras. Proposition4.2 implies that conversely, an isomorphism of k-algebras A(Y ) → A(X) comesfrom a unique isomorphism X → Y .

We complete the picture by showing that any finitely generated reduced k-algebra A is isomorphic to some A(Y ); the preceding then shows that Y is uniqueup to isomorphism. Since A is finitely generated as a k-algebra, there exists a sur-jective k-algebra homomorphism φ : k[x1, . . . , xn] → A. If we put I := Ker(φ),then φ induces an isomorphism k[x1, . . . , xn]/I ∼= A. Put Y := Z(I) ⊂ An. SinceA is reduced, I is a radical ideal and hence equal to I(Y ) by the Nullstellensatz. Itfollows that φ factors through a k-algebra isomorphism A(Y ) ∼= A.

We may sum up this discussion in categorical language as follows.

PROPOSITION 4.3. The map which assigns to a closed subset of some An its coor-dinate ring defines an anti-equivalence between the category of closed subsets of affinespaces (and regular maps between them as defined above) and the category of reducedfinitely generated k-algebras (and k-algebra homomorphisms).

EXAMPLE 4.4. Consider the regular map f : A1 → A2, f(t) = (t2, t3). Theimage of this map is the hypersurface Z defined by x3 − y2 = 0. This is a home-omorphism on the image for it is a bijection of sets that have both the cofinitetopology. The inverse sends (0, 0) to 0 and is on Z − (0, 0) given by (x, y) 7→ y/x.In order to determine whether the inverse is regular, we consider f∗. We haveA(Z) = k[x, y]/(x3 − y2), A(A1) = k[t] and f∗ : k[x, y]/(x3 − y2)→ k[t] is given byx 7→ t2, y 7→ t3. This homomorphism is not surjective for its image misses t ∈ k[t].So f is not an isomorphism.

EXAMPLE 4.5. An affine-linear transformation of kn is of the form x ∈ kn 7→g(x) + a, where a ∈ kn and g ∈ GL(n, k) is a linear transformation. Its inverse isy 7→ g−1(y−a) = g−1(y)−g−1(a) and so of the same type. When we regard such an


affine linear transformation as a map from An to itself, then it is regular: its coordi-nates (g1, . . . , gn) are polynomials of degree one. So an affine-linear transformationis also an isomorphism of An onto itself. But if n ≥ 2, not every isomorphism ofAn onto itself need to be of this form. For example, (x, y) 7→ (x, y + x2) defines anautomorphism of A2 with inverse (x, y) 7→ (x, y − x2) (see also Exercise 18).

EXERCISE 17. Let C ⊂ A2 be the ‘circle’, defined by x2 + y2 = 1 and let p0 :=(−1, 0) ∈ C. For every p = (x, y) ∈ C − p0, the line through p0 and p has slopef(p) = y/(x+ 1). Denote by

√−1 ∈ k a root of the equation t2 + 1 = 0.

(a) Prove that when char(k) 6= 2, f defines an isomorphism5 onto A1 −±√−1.

(b) Consider the map g : C → A1, g(x, y) := x +√−1y. Prove that when

char(k) 6= 2, g defines an isomorphism of C onto A1 − 0.(c) Prove that when char(k) = 2, the defining polynomial x2 + y2 − 1 for C

is the square of a degree one polynomial so that C is a line.

EXERCISE 18. Let f1, . . . , fn ∈ k[x1, . . . , xn] be such that f1 = x1 and fi − xi ∈k[x1, . . . , xi−1] for i = 2, . . . , xn. Prove that f defines an isomorphism An → An.

4.6. QUADRATIC HYPERSURFACES IN CASE char(k) 6= 2. Let H ⊂ An be a hyper-surface defined by a polynomial of degree two:

f(x1, . . . , xn) =∑

1≤i≤j=n

aijxixj +

n∑i=1

aixi + a0.

By means of a linear transformation (this involves splitting off squares, hence re-quires the existence of 1/2 ∈ k), the quadratic form

∑1≤i≤j=n aijxixj can be

brought in diagonal form. This means that we can make all the coefficients aijwith i 6= j vanish. Another diagonal transformation (which replaces xi by

√aiixi

when aii 6= 0) takes every nonzero coefficient aii to 1 and then renumberingthe coordinates (which is also a linear transformation) brings f into the formf(x1, . . . , xn) =

∑ri=1 x

2i +

∑ni=1 aixi + a0 for some r ≥ 1. Splitting off squares

once more enables us to get rid of∑ri=1 aixi so that we get

f(x1, . . . , xn) =

r∑i=1

x2i +

n∑i=r+1

aixi + a0.

We now have the following cases.If the nonsquare part is identically zero, then we end up with the equation∑r

i=1 x2i = 0 for H.

If the linear part∑ni=r+1 aixi is nonzero (so that we must have r < n), then an

affine-linear transformation which does not affect x1, . . . , xr and takes∑ni=r+1 aixi+

a0 to −xn yields the equation xn =∑ri=1 x

2i . This is the graph of the function∑r

i=1 x2i on An−1 and so H is then isomorphic to An−1.

If the linear part∑ni=r+1 aixi is zero, but the constant term a0 is nonzero, then

we can make another diagonal transformation which replaces xi by√−a0xi) and

divide f by a0: then H gets the equation∑ri=1 x

2i = 1.

5We have not really defined what is an isomorphism between two nonclosed subsets of an affinespace. Interpret this here as: f∗ maps k[x, y][1/(x+1)]/(x2 + y2− 1) (the algebra of regular functionson C −p0) isomorphically onto k[t][1/(t2 +1)] (the algebra of regular functions on A1−±

√−1).

This will be justified by Proposition 4.10.


In particular, there are only a finite number of quadratic hypersurfaces up toisomorphism. This is also true in characteristic two, but the classification is moredelicate. (For instance, the hypersurfaces defined by x1x2 = 1 and x2

1 + x22 = 1 are

in this case not isomorphic.)

4.7. THE MAXIMAL IDEAL SPECTRUM. The previous discussion (and in particularProposition 4.3) leads us to associate to an arbitrary ring R a space with a topologyso that for R = A(Y ) we get a space homeomorphic to Y . Since the points of Ycorrespond to maximal ideals of A(Y ), we choose the underlying set of this space(which we shall denote by Specm(R)) to be the collection of maximal ideals of R.In order to avoid confusion about whether a maximal ideal is to be viewed as asubset of R or as a point of Specm(R) we agree that if m is a maximal ideal of R,then the corresponding element of Specm(R) is denoted xm and if x ∈ Specm(R),then the corresponding maximal ideal of R is denoted mx. We write κ(x) for theresidue field R/mx and ρx : R → κ(x) for the reduction modulo mx. We now havearranged that when R = A(Y ), x ∈ Specm(A(Y )) is the same thing as a point ofY . Since κ(x) is a finitely generated as a k-algebra, it is by Corollary 3.7 a finiteextension of k and hence equal to k (k is algebraically closed).

Every r ∈ R defines a ‘regular function’ fr on Specm(R) which takes in x ∈Specm(R) the value ρx(r) ∈ κ(x). So in general fr takes its values in a field thatmay depend on x, but for R of the form A(Y ) we know this field to be constantequal to k, so that fr : Specm(R)→ k. We denote by Z(r) ⊂ Specm(A) (or Z(fr))the zero set of this function. So Z(r) consists of the x ∈ Specm(R) with ρx(r) = 0,or equivalently, r ∈ mx. We write U(r) or (U(fr)) for its complement, the nonzeroset. We have U(rr′) = U(r) ∩ U(r′) and so the collection of U(r)r∈R is the basisof a topology on Specm(R). Note that a subset Specm(R) is closed precisely if itis an intersection of subsets of the form Z(r); this is equal to the common zero setof the set of functions defined by an ideal of R. The space Specm(R) is called themaximal ideal spectrum6 of R (but our notation for it is less standard). Notice that ifR = A(Y ), then the above discussion shows that Specm(R) can be identified withY as a topological space and that under this identification, A(Y ) becomes the ringof regular functions on Specm(R).

A ring homomorphism φ : S → R will in general not give rise to a mapSpecm(φ) : Specm(R) → Specm(S): if x ∈ Specm(R), then the composite homo-morphism S → R → κ(x) need not be onto and its kernel need not be a maximalideal (it will be a prime ideal, though). However, in case φ : S → R is a homomor-phism of k-algebras, then k ⊂ S maps onto κ(x) = k as the identity map so thatφ−1mx is a maximal ideal of S with residue field k. We then do get a map

Specm(φ) : Specm(R)→ Specm(S), xm 7→ xφ−1m.

For s ∈ S, the preimage of U(s) is U(φ(s)). This shows that Specm(φ) is continuous.We call the resulting pair (Specm(φ), φ) a morphism.

EXERCISE 19. Give an example of ring homomorphism φ : S → R and a maxi-mal ideal m ⊂ R, such that φ−1m is not a maximal ideal of S. (Hint: take a look atExercise 13.)

6I. Gelfand was presumably the first to consider this in the context of functional analysis: hecharacterized the Banach algebras that appear as the algebras of continuous C-valued functions oncompact Hausdorff spaces and so it might be appropriate to call this the Gelfand spectrum.


EXERCISE 20. Prove that if R is a finitely generated k-algebra, then the map r ∈R 7→ fr is a k-algebra homomorphism from R to the algebra of k-valued functionson Specm(R) with kernel

√(0). Show that for every subset X ⊂ Specm(R), the set

I(X) of r ∈ R with fr|X = 0 is a radical ideal of R.

EXERCISE 21. Prove that Specm(R) is irreducible when R is a finitely generatedalgebra over a field and nonzero7.

DEFINITION 4.8 (Preliminary definition). An affine k-variety8 is a topologicalspace Y endowed with a finitely generated k-algebra A(Y ) of regular functions onY which identifies Y with the maximal ideal spectrum of A(Y ).

In more elementary terms, there should exist a homeomorphism h of Y onto aclosed subset of some An such that A(Y ) = h∗k[x1, . . . , xn]. We shall often denotean affine variety simply by the underlying topological space, e.g., by Y , where it isthen understood that A(Y ) is part of the data. Likewise a morphism will simply bedenoted by f : X → Y ; it is then understood that composition with f takes A(Y )to A(X) and defines f∗ : A(Y )→ A(X).

With this terminology Proposition 4.3 takes now a rather trivial form:

4.9. The functor which assigns to an affine k-variety its k-algebra of regularfunctions identifies the category of affine k-varieties with the dual of the category ofreduced finitely generated k-algebras, the inverse being given by the functor Specm.

From our previous discussion it is clear that if Y is an affine variety, then everyclosed subset Y ′ ⊂ Y determines an affine variety Y ′ with A(Y ′) := A(Y )/I(Y ′).Here I(Y ′) is of the course the ideal of regular functions on Y that vanish on Y ′.

Our next aim is to give any basic open subset U of Y the structure of an affinevariety.

PROPOSITION 4.10. Let Y be an affine variety. Then every basic open subsetU ⊂ Y is in a natural manner an affine variety so that an inclusion U ⊂ U ′ of anytwo such is a morphism and the algebra of regular functions on U = U(g) isA(Y )[1/g](assuming U 6= ∅, so that g is not nilpotent).

PROOF. We observe that A(Y )[1/g] is defined as a k-algebra and as such it isfinitely generated (just add to a generating set for A(Y ) the generator 1/g) andis without zero divisors. So we have an affine variety Specm(A(Y )[1/g]). Anymaximal ideal of A(Y )[1/g] intersects A(Y ) in a maximal ideal of A(Y ) which doesnot contain g. Conversely, if m is a maximal ideal of A(Y ), then mA(Y )[1/g] isa maximal ideal of A(Y )[1/g] unless g ∈ m. So the injection A(Y ) → A(Y )[1/g]induces an injection of Specm(A(Y )[1/g]) in Y with image U(g) = U . This allowsus to regard U as an affine variety whose coordinate ring is A(Y )[1/g].

Assume that we have an inclusion U(g′) ⊂ U(g). Then Z(g′) ⊃ Z(g) and soby the Nullstellensatz, g′ ∈

√(g). This implies that (g′)n ∈ (g) for some positive

integer n. So we have a natural homomorphism A(Y )[1/g] → A(Y )[1/(g′)n] =A(Y )[1/g′]. This also shows that A(Y )[1/g] only depends on U , for if U(g′) = U(g),we also a homomorphism in the opposite direction which serves as inverse so thatA(Y )[1/g] = A(Y )[1/g′].

7This corrects an earlier formulation, with thanks to my students.8Since we fixed k, we will often drop k and speak of an affine variety.


Let f : X → Y be a morphism of affine varieties. Since f is continuous, a fiberf−1(y), or more generally, the preimage f−1Y ′ of a closed subset Y ′ ⊂ Y , will beclosed in X. It is the zero set of the ideal in A(X) generated by f∗I(Y ′). But thisideal need not be a radical ideal. Here is a simple example:

EXAMPLE 4.11. Let f : X = A1 → A1 = Y be defined by f(x) = x2. Thenf∗ : k[y] → k[x] is given by f∗y = x2. If we assume k not to be of characteristic 2,and we take a ∈ Y − 0, then the fiber f−1(a) consists of two distinct points andis define by the ideal generated by f∗(y − a) = x2 − a. This a radical ideal and thefiber has coordinate ring k[x]/(x2 − a). However, the fiber over 0 ∈ Y = A1 is thesingleton 0 ⊂ X = A1 and the ideal generated by f∗y = x2 is not a radical ideal.This example indicates that there might good reason to accept nilpotent elementsin the coordinate ring of f−1(0) by endowing f−1(0) with the ring of functionsA(f−1(0)) := k[x]/(x2). Since this is a k-vector space of dimension 2, we thusretain the information that two points have come together and the fiber should bethought of as a point with multiplicity 2.

Example 4.4 shows that a homeomorphism of affine varieties need not be anisomorphism. Here is another class of examples.

EXAMPLE 4.12 (THE FROBENIUS MORPHISM). Assume that k has positive char-acteristic p and consider the morphism Fp : A1 → A1, x 7→ xp. If we rememberthat A1 can be identified with k, then we observe that under this identification, Fpis a field automorphism: Fp(a − b) = (a − b)p = ap − bp = Fp(a) − Fp(b) and ofcourse Fp(ab) = (ab)p = Fp(a)Fp(b) and Fp is surjective (every element of k has apth root since k is algebraically closed) and injective. But the endomorphism F ∗p ofk[x] induced by Fp sends x to xp and has therefore image k[xp]. Clearly, F ∗p is notsurjective.

The fixed point set of Fp (so the set of a ∈ A1 with ap = a) is via the iden-tification of A1 with k just the prime subfield Fp ⊂ k and we therefore denote itby A1(Fp) ⊂ A1. Likewise, the fixed point set A1(Fpr ) of F rp is the subfield of kwith pr elements. Since the algebraic closure Fp of Fp in k is the union of the finitesubfields of k, the affine line over Fp equals ∪r≥1A1(Fpr ). This generalizes in astraightforward manner to higher dimensions: by letting Fp act coordinatewise onAn, we get a morphism An → An (which we still denote by Fp) which is also abijection. The fixed point of F rp is An(Fpr ) and An(Fp) = ∪r≥1An(Fpr ).

EXERCISE 22. Assume that k has positive characteristic p. Let q = pr be a powerof p with r > 0 and denote by Fq ⊂ k the subfield of a ∈ k satisfying aq = a. Wewrite F for F rp : An → An. Suppose Y ⊂ An is the common zero set of polynomialswith coefficients in Fq ⊂ k.

(a) Prove that f ∈ k[x1, . . . , xn] has its coefficients in Fq if and only if F ∗f =fq.

(b) Prove that an affine-linear transformation of An with coefficients in Fqcommutes with F .

(c) Prove that F restricts to a bijection FY : Y → Y and that the fixed pointset of FmY is Y (Fqm) := Y ∩ An(Fqm).

(d) Suppose that k is an algebraic closure of Fp. Prove that every closedsubset of An is defined over a finite subfield of k.

5. THE PRODUCT 21

REMARK 4.13. After this exercise we cannot resist to mention the Weil zetafunction. This function and its relatives—among them the Riemann zeta function—codify arithmetic properties of algebro-geometric objects in a very intricate manner.In the situation of Exercise 22, we can use the numbers |Y (Fqm)| (= the number offixed points of Fm in Y ) to define a generating series

∑m≥1 |Y (Fqm)|tm. It appears

to be more convenient to work with the Weil zeta function:

ZY (t) := exp( ∞∑m=1

|Y (Fqm)| tm

m

),

which has the property that t ddt logZY yields the generating series above. Thisseries has remarkable properties. For instance, a deep theorem due to BernardDwork (1960) asserts that it represents a rational function of t. Another deeptheorem, due to Pierre Deligne (1974), states that the roots of the numerator anddenominator have for absolute value a nonpositive half-integral power of q and thatmoreover, these powers have an interpretation in terms of an ‘algebraic topologyfor algebraic geometry’. All of this was predicted by Andre Weil in 1949. (This canbe put in a broader context by making the change of variable t = q−s. Indeed,now numerator and denominator have their zeroes when the real part of s is anonnegative half-integer and this makes Deligne’s result reminiscent of the famousconjectured property of the Riemann zeta function.)

EXERCISE 23. Compute the Weil zeta function of affine n-space relative to thefield of q elements.

5. The product

Let m and n be nonnegative integers. If f ∈ k[x1, . . . , xm] and g ∈ k[y1, . . . , yn],then we have a regular function f ∗ g on Am+n defined by

f ∗ g(x1, . . . , xm, y1, . . . , yn) := f(x1, . . . , xm)g(y1, . . . , yn).

It is clear that U(f ∗ g) = U(f) × U(g), which shows that the Zariski topology onAm+n refines the product topology on Am × An. Equivalently, if X ⊂ Am andY ⊂ An are closed, then X × Y is closed in Am+n. We give X × Y the topologyit inherits from Am+n (which is usually finer than the product topology). For thecoordinate rings we have defined a map:

A(X)×A(Y )→ A(X × Y ), (f, g) 7→ f ∗ g

which is evidently k-bilinear (i.e., k-linear in either variable). We want to provethat the ideal I(X × Y ) defining X × Y in Am+n is generated by I(X) and I(Y )(viewed as subsets of k[x1, . . . , xm, y1, . . . yn]) and that X × Y is irreducible whenX and Y are. This requires that we translate the formation of the product into al-gebra. This centers around the notion of the tensor product, the definition of whichwe recall. (Although we here only need tensor products over k, we shall define thisnotion for modules over a ring, as this is its natural habitat. This is the setting thatis needed later anyhow.)

If R is a ring and M and N are R-modules, then we can form their tensor product overR, M ⊗R N : as an abelian group M ⊗R N is generated by the expressions a ⊗R b, a ∈ M ,b ∈ N and subject to the conditions (ra)⊗R b = a⊗R (rb), (a+ a′)⊗R b = a⊗R b+ a′ ⊗R band a ⊗R (b + b′) = a ⊗R b + a ⊗R b′. So a general element of M ⊗R N can be written


like this:∑Ni=1 ai ⊗R bi, with ai ∈ M and bi ∈ N . We make M ⊗R N an R-module if we

stipulate that r(a⊗R b) := (ra)⊗R b (which is then also equal to r⊗R (rb)). Notice that themap

⊗R : M ×N →M ⊗R N, (a, b) 7→ a⊗R b,is R-bilinear (if we fix one of the variables, then it becomes an R-linear map in the othervariable).

In case R = k we shall often omit the suffix k in ⊗k.

EXERCISE 24. Prove that ⊗R is universal for this property in the sense that every R-bilinear mapM×N → P ofR-modules is the composite of⊗R and a uniqueR-homomorphismM ⊗R N → P . In other words, the map

HomR(M ⊗R N,P )→ BilR(M ×N,P )), f 7→ f ⊗Ris an isomorphism of R-modules.

EXERCISE 25. Let m and n be nonnegative integers. Prove that Z/(n)⊗Z Z/(m) can beidentified with Z/(m,n).

If A is an R-algebra and N is an R-module, then A ⊗R N acquires the structure of anA-module which is characterized by

a.(a′ ⊗R b) := (aa′)⊗R b.For instance, if N is an R-vector space, then C⊗RN is a complex vector space, the complexi-fication of N . If A and B are R-algebras, then A⊗RB acquires the structure of an R-algebracharacterized by

(a⊗R b).(a′ ⊗R b′) := (aa′)⊗R (bb′).

Notice that A→ A⊗R B, a 7→ a⊗R 1 and B → A⊗R B, b 7→ 1⊗R b are R-algebra homo-morphisms. For example, A⊗RR[x] = A[x] as A-algebras (and hence A⊗RR[x1, . . . , xn] =A[x1, . . . , xn] with induction).

EXERCISE 26. Prove that C⊗R C is as a C-algebra isomorphic to C⊕C with componen-twise multiplication.

PROPOSITION 5.1. For closed subsets X ⊂ Am and Y ⊂ An the bilinear mapA(X)×A(Y )→ A(X×Y ), (f, g) 7→ f∗g induces an isomorphism µ : A(X)⊗A(Y )→A(X × Y ) of k-algebras (so that in particular A(X)⊗A(Y ) is reduced).

If X and Y are irreducible, then so is X × Y , or equivalently, if A(X) and A(Y )are domains, then so is A(X)⊗A(Y ).

PROOF. Since the obvious map

k[x1, . . . , xm]⊗ k[y1, . . . , yn]→ k[x1, . . . , xm, y1, . . . , yn]

is an isomorphism, it follows that µ is onto. In order to prove that µ is injective, letlet first observe that every u ∈ A(X)⊗ A(Y ) can be written u =

∑Ni=1 fi ⊗ gi with

g1, . . . , gN are k-linearly independent. Given p ∈ X, then the restriction of µ(u) =∑Ni=1 fi ∗ gi to p × Y ∼= Y is the regular function up :=

∑Ni=1 fi(p)gi ∈ A(Y ).

Since the gi’s are linearly independent, we have up = 0 if and only if fi(p) = 0 forall i. In particular, the subset X(u) ⊂ X of p ∈ X for which up = 0, is equal to∩Ni=1Z(fi) and hence closed.

If µ(u) = 0, then up = 0 for all p ∈ X and hence fi = 0 for all i. So u = 0. Thisproves that µ is injective.

Suppose now X and Y irreducible and suppose that u is zero divisor: uv = 0for some v ∈ A(X×Y ) with u, v 6= 0. Since the restriction of uv = 0 to p×Y ∼= Yis upvp and A(Y ) is a domain, it follows that up = 0 or vp = 0. So X = X(u)∪X(v).

6. FUNCTION FIELDS AND RATIONAL MAPS 23

Since X is irreducible we have X = X(u) or X(v). This means that u = 0 or v = 0,which contradicts our assumption.

EXERCISE 27. Let X and Y be closed subsets of affine spaces. Prove that eachirreducible component of X × Y is the product of an irreducible component of Xand one of Y .

It is clear that the projections πX : X × Y → X and πY : X × Y → Y areregular. We have observed that X × Y has not a topological product in general.Still it is the ‘right’ product in the sense of category theory: it has the followinguniversal property, almost seems too obvious to mention: if Z is a closed subset ofsome affine space, then any pair of regular maps f : Z → X, g : Z → Y defines aregular map Z → X × Y characterized by the property that its composite with πXresp. πY yields f resp. g (this is of course (f, g)).

6. Function fields and rational maps

In this section we interpret the formation of the fraction ring of an algebra ofregular functions.

Let Y be an affine variety. Recall that for f, g ∈ A(Y ), a fraction f/g ∈Frac(A(Y )) is defined if g not a zerodivisor. We need the following lemma.

LEMMA 6.1. Let Y be an affine variety and let g ∈ A(Y ). Then g is not a zerodivisor if and only if g is nonzero on every irreducible component of Y . This is alsoequivalent to U(g) being dense in Y .

PROOF. Suppose g is zero on some irreducible component C of Y and supposeC 6= Y . If Y ′ denotes the union of the irreducible components of Y distinct form C,then either Y ′ = ∅ (and so g = 0 in A(Y )) or Y ′ 6= ∅ and so there exists a nonzerog′ ∈ IY (Y ′). Then gg′ vanishes on C ∪ Y ′ = Y and so gg′ = 0. It follows that g isthen a zero divisor.

Conversely, assume that g is a zerodivisor of A(Y ). So there exists a nonzerog′ ∈ A(Y ) with gg′ = 0. Let C be an irreducible component of Y on which g′ isnonzero. Then we must have g|C = 0.

The proof of the last clause is left to you.

This lemma tells us that this means that the affine variety U(g) is open-densein Y . Clearly, f/g defines a regular function U(g)→ k. Another such fraction f ′/g′

yields a regular function on an open-dense U(g′) and their product f/g.f ′/g′ =ff ′/gg′ resp. difference f/g − f ′/g′ = (fg′ − f ′g)/gg′ in Frac(A(Y )) defines aregular function on U(gg′) which is there just the product resp. difference of theassociated regular functions. In particular, if f ′/g′ = f/g in A(Y ), then the twoassociated regular functions coincide on U(gg′). There is a priori no best wayto represent a given element of Frac(A(Y )) as a fraction (as there is in a UFD),and so we must be content with the observation that an element of Frac(A(Y ))defines a regular function on some (unspecified) open dense subset of Y , with theunderstanding that two such functions are regarded as equal if they coincide on anopen-dense subset contained in a common domain of definition. When an elementof Frac(A(Y )) is thus understood, then we call it a rational function on Y . Thisis the algebro-geometric analogue of a meromorphic function in complex functiontheory. When Y is irreducible, then Frac(A(Y )) is a field, called the function fieldof Y , and will be denoted k(Y ).


We shall now give a geometric interpretation of finitely generated field exten-sions of k and the k-linear field homomorphisms between them. We begin with anotion that generalizes that of a rational function.

DEFINITION 6.2. Let X and Y be affine varieties. A rational map from X to Y isgiven by a pair (U,F ), where U is an affine open-dense subset of X and F : U → Yis a morphism, with the understanding that a pair (V,G) defines the same rationalmap if F and G coincide on U ∩V . We denote a rational map like this f : X 99K Y .

We say that the rational map is dominant if for a representative pair (U,F ),F (U) is dense in Y . (This is then also so for any other representative pair. Why?)

PROPOSITION 6.3. Any finitely generated field extension of k is k-isomorphic tothe function field of an irreducible affine variety. If X and Y are irreducible and affine,then a dominant rational map f : X 99K Y determines a k-linear field embeddingf∗ : k(Y ) → k(X) and every k-linear field embedding k(Y ) → k(X) is of this form(for a unique f).

PROOF. Let K/k be a finitely generated field extension of k: there exist ele-ments a1, . . . , an ∈ K such that every element of K can be written as a fractionof polynomials in a1, . . . , an. So if R denotes the k-subalgebra R of K generatedby a1, . . . , an, (a domain since K is a field), then K is the field of fractions of R.Since R is the coordinate ring of a closed irreducible subset X ⊂ An (defined bythe kernel of the obvious ring homomorphism k[x1, . . . , xn] → R), it follows thatK can be identified with k(X).

Suppose we are given an open-dense affine subvariety U ⊂ X and a morphismF : U → Y with F (U) dense in Y . Now F ∗ : A(Y ) → A(U) will be injective, forif g ∈ A(Y ) is in the kernel: F ∗(g) = 0, then F (U) ⊂ Z(g′). Since F (U) is densein Y , this implies that Z(g) = Y , in other words, that g = 0. Hence the compositemap A(Y )→ A(U)→ k(U) = k(X) is an injective homomorphism from a domainto a field and therefore extends to a field embedding k(Y )→ k(X).

It remains to show that every k-linear field homomorphism Φ : k(Y )→ k(X) isso obtained. For this, choose generators b1, . . . , bm of A(Y ). Then Φ(b1), . . . ,Φ(bm)are rational functions on X and so are regular on an open-dense affine subsetU ⊂ X. Since b1, . . . , bm generate A(Y ) as a k-algebra, it follows that Φ maps A(Y )to A(U) ⊂ k(X). This is a k-algebra homomorphism and so defines a morphismF : U → Y such that F ∗ = Φ|A(Y ). The image of F will be dense by the argumentabove: if the image of F is contained in a closed subset of Y distinct from Y , thenits maps to some Z(g) with g ∈ A(Y )−0. But this implies that Φ(g) = F ∗(g) = 0,which cannot happen, since Φ is injective. It is clear that Φ is the extension of F ∗

to the function fields.As to the uniqueness: if (V,G) is another solution, then choose a nonempty

affine open-dense subset W ⊂ U ∩ V so that F and G both define morphismsW → Y . The maps A(Y )→ A(W ) they define are equal, for are given by Φ. HenceF andG coincide onW . SinceW is dense in U∩V , they also coincide on U∩V .

EXERCISE 28. Let f ∈ k[x1, . . . , xn+1] be irreducible of positive degree. Its zeroset X ⊂ An+1 is then closed and irreducible. Denote by d the degree of f in xn+1.

(a) Prove that the projection π : X → An induces an injective k-algebrahomomorphism π∗ : k[x1, . . . , xn] → A(X) = k[x1, . . . , xn]/(f) if andonly if d > 0.


(b) Prove that for d > 0, π is dominant and that the resulting field homomor-phism k(x1, . . . , xn)→ k(X) is a finite extension of degree d.

COROLLARY 6.4. There is a category with objects the irreducible affine varietiesand morphisms the rational dominant maps. Assigning to an irreducible affine va-riety its function field makes this category anti-equivalent to the category of finitelygenerated field extensions of the base field k.

PROPOSITION-DEFINITION 6.5. A rational map f : X 99K Y is an isomorphismin the above category (that is, induces a k-linear isomorphism of function fields) if andonly if there exists a representative pair (U,F ) of f such that F maps U isomorphicallyonto an open subset of Y . If these two equivalent conditions are satisfied, then f iscalled a birational map. If a birational map X 99K Y merely exists (in other words,if there exists a k-linear field isomorphism between k(X) and k(Y )), then we say thatX and Y are birationally equivalent.

PROOF. If f identifies a nonempty open subset of X with one of Y , then f∗ :k(Y )→ k(X) is clearly a k-algebra isomorphism.

Suppose now we have a k-linear isomorphism k(Y ) ∼= k(X). Represent thisisomorphism and its inverse by (U,F ) and (V,G) respectively. Then F−1V is affineand open-dense in U and GF : F−1V → X is defined. Since GF induces theidentity on k(X), GF is the identity on a nonempty open subset of F−1V andhence on all of F−1V . This implies that F maps F−1V injectively to G−1U . Forthe same reason, G maps G−1U injectively to F−1V . So F defines an isomorphismbetween the open subsets F−1V ⊂ X and G−1U ⊂ Y .

EXERCISE 29. We here assume k not of characteristic 2. Let for X ⊂ A2 bedefined by x2

1 +x22 = 1. Prove that X is birationally equivalent to the affine line A1.

(Hint: take a look at Exercise 17.) More generally, prove that the quadric in An+1

defined by x21 + x2

2 + · · ·x2n+1 = 1 is birationally equivalent to An. What about the

quadric cone in An+1 defined by x21 + x2

2 + · · ·x2n+1 = 0 (n ≥ 2)?

In case k(X)/k(Y ) is a finite extension, one may wonder what the geometricmeaning is of the degree d of that extension, perhaps hoping that this is just thenumber of elements of a general fiber of the associated rational map X 99K Y . Wewill see that this often true (namely when the characteristic of k is zero, or moregenerally, when this characteristic does not divide d), but not always, witness thefollowing example.

EXAMPLE 6.6. Suppose k has characteristic p > 0. We take X = A1 = Y and letf be the Frobenius map: f(a) = ap. Then f is homeomorphism, but f∗ : A(Y ) →A(X) is given by y 7→ xp and so induces the field extension k(y) = k(xp) ⊂ k(x),which is of degree p. From the perspective of Y , we have enlarged its algebra ofregular functions by introducing a formal pth root of its coordinate (which yieldsanother copy of A1, namely X). From the perspective of X, A(Y ) is just f∗A(X).

This is in fact the basic example of a purely inseparable field extension, i.e., afield extension L/K with the property that every element of L has a minimal poly-nomial in K[T ] that has at most one root in L. Such a polynomial must be of theform T p

r − c, with c ∈ K not a p-th power of an element of K when r > 0, wherep is the characteristic of K (for p = 0 we necessarily have L = K). So if L 6= K,then the Frobenius map Fp : a ∈ K 7→ ap ∈ K is not surjective. Purely inseparable


extensions are not detected by Galois theory, for they have trivial Galois group: af-ter all, there is only one root to move around.

For any algebraic extension L/K, the elements of L that are separable over K makeup an intermediate extension Lsep/K that is (of course) separable and is such that L/Lsep

is purely inseparable. When L is an algebraic closure of K, then Lsep is called a separablealgebraic closure of K: it is a separable algebraic extension of K which is maximal for thatproperty. Then L is as an extension of Lsep obtained by successive adjunction of p-powerroots of elements of Lsep: it is an extension minimal for the property of making the Frobeniusmap a ∈ L 7→ ap ∈ L surjective. In case L/K is a finite normal extension (i.e., a finiteextension with the property that every f ∈ K[x] that is the minimal polynomial of someelement of L has all roots in L), then the fixed point set of the Galois group of L/K is asubextension Linsep/L that is purely inseparable, whereas L/Linsep separable. In that casethe natural map Lsep ⊗K Linsep → L is an isomorphism of K-algebras.

EXERCISE 30. Let f : X → Y be a dominant morphism of irreducible affinevarieties which induces a purely inseparable field extension k(Y )/k(X). Prove thatthere is an open-dense subset U ⊂ Y such that f restricts to a homeomorphismf−1U → U . (Hint: show first that it suffices to treat the case when k(X) is obtainedfrom k(Y ) by adjoining the pth root of an element f ∈ k(Y ). Then observe thatif f is regular on the affine open-dense U ⊂ Y , then Y contains as an open densesubset a copy of the locus of (x, t) ∈ U × A1 satisfying tp = f .)

If K/k is a finitely generated field extension and generated by b1, . . . , bm, say,then we may after renumbering assume that for some r ≤ m, b1, . . . , br are alge-braically independent (so that the k-linear field homomorphism k(x1, . . . , xr)→ Kwhich sends xi to bi is injective) and br+1, . . . , bm are algebraic over k(b1, . . . , br).This number r is invariant of K/k, called its transcendence degree. So if K = k(X),then this defines a dominant rational map X 99K Ar. We now recall the theorem ofthe primitive element:

THEOREM 6.7. Every separable field extension L/K of finite degree is generatedby a single element (which we then call a primitive element for L/K).

It follows that if br+1, . . . , bm are separable over k(b1, . . . , br), then we canfind a b ∈ K that is algebraic over k(b1, . . . , br) such that K is obtained fromk(b1, . . . , br) by adjoining b. This b is a root of an irreducible polynomial F ∈k(b1, . . . , br)[T ]. If we clear denominators, we may assume that the coefficients ofF lie in k[b1, . . . , br][T ] = k[b1, . . . , br, T ]. Then we can take for X the hypersurfacein Ar+1 defined by F , when viewed as an element of k[x1, . . . , xr, xr+1].

In particular, every irreducible affine variety Y is birationally equivalent to ahypersurface in Ar+1, where r is the transcendence degree of k(Y ). This suggeststhat this transcendence degree might be a good way to define the dimension of Y .We shall return to this.

Much of the algebraic geometry in the 19th century and early 20th centurywas of a birational nature: birationally equivalent varieties were regarded as notreally different. This sounds rather drastic, but it turns out that many interestingproperties of varieties are an invariant of their birational equivalence class.


Here is an observation which not only illustrates how affine varieties over al-gebraically nonclosed fields can arise when dealing with affine k-varieties, but onethat is also a tell-tale sign that we ought to enlarge the maximal ideal spectrum.Let f : X → Y be a dominant morphism of irreducible affine varieties. This impliesthat f∗ : A(Y )→ A(X) is injective and that f(X) contains an open-dense subset ofY . Then we may ask whether there exists something like a general fiber: is therean open-dense subset V ⊂ Y such that the fibers f−1(y), y ∈ V all “look the same”?The question is too vague for a clear answer and for most naive ways of makingthis precise, the answer will be no. But there is a cheap way out by simply refusingto specify one such V and allowing it to be arbitrarily small: we would like to takea limit over all open-dense subsets of Y . This we cannot do (yet) topologically, butwe can do this algebraically by making all the nonzero elements of A(Y ) in A(X)invertible: we take

(A(Y )− 0)−1A(X) = k(Y )⊗A(Y ) A(X)

(this equality follows from the fact that A(X) = A(Y ) ⊗A(Y ) A(X)). This isa reduced finitely generated k(Y )-algebra and we may regard its maximal idealspectrum Specm(k(Y ) ⊗A(Y ) A(X)) as an affine variety over the (algebraicallynonclosed) field k(Y ): now every regular function on X which comes from Y istreated as a scalar (and will be invertible when nonzero). If we decide to add toY ∼= Specm(k(Y )) a point η(Y ) with residue field k(Y ), then we can simply saythat the generic fiber of f is the fiber Xη(Y ) over η(Y )9. But once we decide todo this for Y , we should of course do this for every irreducible affine variety. Inparticular, we must then add for every irreducible subset Z of Y (or equivalently,for every prime ideal p ⊂ A(Y )) a point to Specm(k(Y )) with residue field k(Z)(= Frac(A(Y )/p)). We do this below in the same kind of generality as the maximalideal spectrum, namely for an arbitrary ring.

6.8. THE PRIME IDEAL SPECTRUM. Let R be a ring. Its spectrum Spec(R) isa topological space whose underlying set is the set of its prime ideals, where weemploy the following notation: if p is a prime ideal of R, then we denote thecorresponding element of Spec(R) by xp and if x ∈ Spec(R), we denote by px ⊂ Rthe corresponding prime ideal. If r ∈ R, then we write U(r) for the set of x ∈Spec(R) with r /∈ px. We have U(r) ∩ U(r′) = U(rr′) and so this collection ofsubsets is a basis for a to topology on Spec(R), called (from now on) the Zariskitopology. It is also characterized by the property that the closed sets are thosedefined by an ideal of R: if we put Z(r) := Spec(R)−U(r), then any closed subsetof Spec(R) is of the form Z(S) = ∩r∈SZ(r) for some subset S ⊂ R (and vice versa).Notice that this just the set of x ∈ Spec(R) with px ⊃ S and hence only depends onthe ideal I(S) generated by S: Z(S) = Z(I(S)).

Since the zero ring (for which we have 0 = 1) has no prime ideals, its spectrumis the empty set.

As the following exercise shows, points of Spec(R) need not be closed:

9It is often preferable to work over an algebraically closed field. We can remedy this simply bychoosing an algebraic closure L of k(Y ). The maximal ideal spectrum of L ⊗A(Y ) A(X) is then anaffine L-variety, and yields a notion of a general fiber that is even closer to our geometric intuition.


EXERCISE 31. Let x, y ∈ Spec(R). Prove that x lies in the closure of y if andonly if px ⊃ py. Conclude that x is closed in Spec(R) if and only if px is a maximalideal of R.

We can think (at least for now) of r ∈ R a defining a ‘regular function’ onSpec(R), but its value field varies with the point: to x ∈ Spec(R) we associate thefield κ(x) := Frac(R/px), called the residue field of x. Then r ∈ R determines a‘regular function’ on Spec(R): it assigns to x the image of r in κ(x) (denoted r(x)).(We will come up with something better when we discuss sheaves.)

One immediate advantage of the prime ideal spectrum over the maximal idealspectrum is that any ring homomorphism φ : R′ → R determines a map

Spec(φ) : Spec(R)→ Spec(R′), xp 7→ xp′ ,

where p′ := φ−1p = ker(R′ → R → R/p). This is a prime ideal of R′ indeed, forφ now induces an embedding of R′/p′ in R/p and so R′/p′ has no zero divisors(but if p were a maximal ideal, then we would not be able to conclude that φ−1pis maximal). The embedding R′/p′ → R/p of domains of course extends to anembedding of their fields of fractions, κ(xp′) → κ(xp). For r′ ∈ R′, the ‘regularfunction’ it defines on Spec(R′) as above precomposed with φ yields the ‘regularfunction’ on Spec(R) defined by φ(r′) and so Spec(φ)−1(U(r′)) = U(φ(r′)). HenceSpec(φ) is continuous.

A generic point of Spec(R) is a point that does not lie in the closure of anotherpoint. In other words, it is associated to a minimal prime ideal of R, or equivalently,to an irreducible component Spec(R). IfR is a domain, there is only one such point,namely x(0) with residue field Frac(R).

Notice that for φ as above, the fiber of Spec(φ) over x′ = xp′ ∈ Spec(R′) is(as a subspace of Spec(R)) the set of x ∈ Spec(R) with for which p′ = φ−1px, orequivalently, for which p ∩ φ(R′) = φ(p′). In Exercise 33 you will show that suchprime ideals correspond bijectively to prime ideals of the κ(x′)-algebra κ(x′)⊗R′ R.Hence it makes sense to regard Spec(κ(x′) ⊗R′ R) as the fiber over x′. (Here withthe given of the κ(x′)-algebra κ(x′) ⊗R′ R); we will later employ definitions forwhich this is automatic.) If R′ is a domain, then the fiber over the generic point,Spec(Frac(R′)⊗R′ R), is called the generic fiber of Spec(φ).

EXAMPLE 6.9. The spectrum of Z has a point xp for every prime number p(with residue field κ(xp) = Fp), and one for the zero ideal, yielding the genericpoint x0 (with residue filed κ(x0) = Q). For every ring R we have a natural ringhomomorphism Z → R. This defines a map Spec(R) → Spec(Z). In case R isnot the zero ring (so that Spec(R) is nonempty), the fiber over xp is Spec(Fp ⊗R) = Spec(R/pR) (which is empty if p is invertible in R) and the fiber over x0 isSpec(Q⊗R).

EXERCISE 32. Determine the points and their residue fields of Spec(k[X]) andSpec(R[X]). Also try your hand at Spec(Z[X]).

EXERCISE 33. Let φ : R′ → R be a ring homomorphism and p′ ⊂ R′ a primeideal. Prove that for every prime ideal p ⊂ R with φ−1p = p′, the image ofFrac(R′/p′) ⊗R′ p in Frac(R′/p′) ⊗R′ R is a prime ideal of latter. Prove that thisdefines a bijection between the prime ideals p ⊂ R with φ−1p = p′ and the primeideals of Frac(R′/p′)⊗R′ R.

7. FINITE MORPHISMS 29

Now let us return to the case of an affine variety: we take R = A, with A a re-duced finitely generated k-algebra and we write X for Specm(A). Any x ∈ Spec(A)corresponds to a prime ideal in A. But this in turn corresponds to a closed ir-reducible subset Zx ⊂ X so that A/px = A(Zx). The residue field κ(x) is thenFrac(A/px) = k(Zx). So when x is a closed point (so that px is a maximal ideal ofA), then Zx is a singleton and we have κ(x) = k. It is clear how any f ∈ A willdefine a function of Spec(A): for x ∈ Spec(A), f(x) is the restriction of f to Zx(viewed as a rational function on Zx, which just happens to be regular).

We can now bring the definition of an affine variety in an almost final form.

DEFINITION 6.10 (Almost final form of the definition). An affine k-variety is theprime ideal spectrum of a k-algebra of a reduced finitely generated k-algebra.

This definition does not seem to amount to much. The reason for making itis to some extent psychological: we reversed the arrows of our category, so thatwe can draw on our geometric intuition (and employ associated terminology). Thefinal definition has to wait until we discuss sheaves.

7. Finite morphisms

Let A be a ring and B an A-algebra.

PROPOSITION-DEFINITION 7.1. We say that b ∈ B is integrally dependent on Aif one the following equivalent properties is satisfied.

(i) b is a root of a monic polynomial xn + a1xn−1 + · · ·+ an ∈ A[x],

(ii) A[b] is as an A-module finitely generated,(iii) A[b] is contained in a A-subalgebra C ⊂ B which is finitely generated as an

A-module.In particular, if B is finite over A (which is short for: B is a finitely generated

A-module), then B is integral over A (which is short for: every element of B is integralover A).

PROOF. (i)⇒ (ii). If b is a root of xn + a1xn−1 + · · ·+ an ∈ A[x], then clearly

A[b] is generated as a A-module by 1, b, b2, . . . , bn−1.(ii)⇒ (iii) is obvious.(iii) ⇒ (i). Suppose that C is as in (iii). Choose an epimorphism π : An → C

of A-modules and denote the standard basis of An by (e1, . . . , en). We may (andwill) assume that π(e1) = 1. By assumption, bπ(ei) =

∑nj=1 xijπ(ej) for certain

xij ∈ A. Put σ := (bδij − xij)i,j ∈ End(A[b]n). Then πσ = 0.If σ′ ∈ End(A[b]n) is the matrix of cofactors of σ, then σσ′ = det(σ)1n by

Cramer’s rule. We find that det(σ) = det(σ)π(e1) = π(det(σ)e1) = π(σσ′)(e1) =(πσ)σ′(e1) = 0. Clearly, det(σ) is a monic polynomial in b.

The integral closure of A in B is the set elements of B that are integrally depen-dent on A and will be denoted A

B.

COROLLARY 7.2. The integral closure of A in B is an A-subalgebra of B.

PROOF. It is enough to prove the following: if b, b′ ∈ B are integrally depen-dent over A, then every element of A[b, b′] is algebraically dependent over A. Sinceb′ is integrally dependent over A, it is so over A[b]. Hence A[b][b′] = A[b, b′] is a


finitely generated A[b]-module. Since A[b] is a finitely generated A-module, it fol-lows that A[b, b′] is a finitely generated A-module. Hence by (iii) every element ofA[b, b′] is integrally dependent over A.

EXERCISE 34. Prove that ‘being integral over’ is transitive: if B is an A-algebraintegral over A, then any B-algebra that is integral over B is as an A-algebra inte-gral over A.

We say that the ring homomorphism A → B is an integral extension if it isinjective (so that A may be regarded as a subring of B) and B is integral over A.

PROPOSITION 7.3. Let A ⊂ B be an integral extension and let p ⊂ A be a primeideal of A. Then the going up property holds: p is of the form q ∩ A, where q isa prime ideal of B. If also is given is a prime ideal q′ of B with the property thatq′ ∩ A ⊂ p, then we can take q ⊃ q′. Moreover the incomparability property holds:two distinct prime ideals of B having the same intersection with A cannot obey aninclusion relation (equivalently, if q′ ∩A = p, then q = q′ is the only solution).

If B is a domain, then Frac(B) is an algebraic field extension of Frac(A), whichis finite whenever B is finite over A.

PROOF. Consider the localizations Ap resp. ApB obtained by making the ele-ments of A− p invertible. Then Ap is a local ring with maximal ideal p := pAp andApB is an integral extension ofAp. If we find a prime ideal q ofApB with q∩Ap = p,then q := q ∩ B is a prime ideal of B the property that q ∩ A = q ∩ A = p ∩ A = p.Hence there is no loss in generality in assuming that A is a local ring and p is itsunique maximal ideal mA.

We claim that mAB 6= B. Otherwise 1 ∈ B can be written as an mA-linearcombination of a finite set elements of B. Denote by B′ the A-algebra generatedby this finite set. Then B′ is an integral extension of A and finitely generated as anA-algebra. This implies that B′ is a finitely generated A-module. Since 1 ∈ mAB

′

we have B′ = mAB′, and it then follows from Nakayama’s lemma (recalled below)

that B′ = 0. This is clearly a contradiction.Now that we know that mAB 6= B, we can take for q any maximal ideal of B

which contains the ideal mAB: then qB ∩ A is a maximal ideal of A, hence equalsmA.

For the refinement and the incomparability property we can, simply by passingto the integral extension A/(q′ ∩ A) ⊂ B/q′, assume that q′ = 0. This reducesthe refinement to the case already treated, while the incomparability property thenamounts showing that q 6= (0) implies q ∩ A 6= (0). If b is a nonzero element of q,then it is a root of a monic polynomial: bn+a1b

n−1 + · · ·+an−1b+an = 0 (ai ∈ A),where we can of course assume that an 6= 0 (otherwise divide by b). We then findthat 0 6= an ∈ Bb ∩A ⊂ q ∩A.

Since B is a union of finitely generated A-modules, the last clause follows ifwe prove that Frac(B) = Frac(A)B. Let b ∈ B. It is a root of a monic polynomialxn + a1x

n−1 + · · · + an = 0 (ai ∈ A) with an 6= 0 and so 1/b = −1/an.(bn−1 +

a1bn−2 + · · ·+ an−1) ∈ Frac(A)B.

REMARK 7.4. Note that the going-up property implies that for any strictly as-cending sequence p0 ( p1 ( · · · ( pn of prime ideals in A is the intersection withA of a (necessarily strictly) ascending sequence of prime ideals q0 ⊂ q1 ⊂ · · · ⊂ qnin B. Conversely, the incomparability property implies that if we intersect a strictly


ascending sequence of prime ideals q0 ( q1 ( · · · ( qn in B with A, we get astrictly ascending sequence of prime ideals of A.

For completeness we state and prove:

LEMMA 7.5 (Nakayama’s lemma). Let R be a local ring with maximal ideal mand M a finitely generated R-module. Then a finite subset S ⊂ M generates Mas a R-module if (and only if) the image of S in M/mM generates the latter as aR/m-vector space. In particular (take S = ∅), mM = M implies M = 0.

PROOF. In fact, we first reduce to the case when S = ∅ by passing to M/RS.Our assumptions then say that M = mM . We must show that M = 0. Let π :Rn →M be an epimorphism of R-modules and denote the standard basis of Rn by(e1, . . . , en). By assumption there exist xij ∈ m such that π(ei) =

∑sj=1 xijπ(ej).

So if σ := (δij − xij)i,j ∈ End(Rn), then πσ = 0. Notice that det(σ) ∈ 1 + m. Since1 + m consists of invertible elements, Cramer’s rule shows that σ is invertible. Soπ = π(σσ−1) = (πσ)σ−1 = 0 and hence M = 0.

EXERCISE 35. Prove that an integral extension A ⊂ B defines surjective mor-phisms Spec(B)→ Spec(A) and Specm(B)→ Specm(A).

DEFINITION 7.6. We say that a morphism of affine varieties f : X → Y is finiteif the k-algebra homomorphism f∗ : A(Y ) → A(X) is finite, i.e., makes A(X) afinitely generated A(Y )-module.

EXERCISE 36. Prove that if Y is an affine variety, then the disjoint union of itsirreducible components is finite over Y .

In the corollary below we use our recent definition 6.10 of an affine variety.

COROLLARY 7.7. Let f : X → Y be a finite morphism of affine varieties suchthat f∗ : A(Y ) → A(X) is injective. Then f is surjective, in particular, every closedirreducible subset Y ′ ⊂ Y is the image of a closed irreducible subset X ′ ⊂ X. Iffurthermore is given a closed irreducible subset X ′′ ⊂ X with f(X ′′) ⊃ Y ′, then wemay choose X ′ ⊂ X ′′.

If in addition X is irreducible, then so is Y and f∗ : k(Y ) → k(X) is a finitealgebraic extension.

PROOF. Apply Proposition 7.3 to f∗.

EXERCISE 37. Prove that a finite morphism f : X → Y of affine varieties isclosed and that every fiber f−1(y) is finite (possibly empty).

THEOREM 7.8 (Noether normalization). LetK be a field. Every finitely generatedK-algebraB is a finite extension of a polynomial algebra overK: there exist an integerr ≥ 0 and an injection K[x1, . . . , xr] → B such that B is finite over K[x1, . . . , xr]. Infact, if we are given an epimorphism φ : K[x1, . . . , xm]→ B of K-algebras, then φ isan isomorphism or we can take in this statement r < m.

The proof will be with induction onm and the induction step is worth recordingseparately.

LEMMA 7.9. Let φ : K[x1, . . . , xm] → B be an epimorphism of K-algebras,which is not an isomorphism. Then there exists a K-algebra homomorphism φ′ :K[x1, . . . , xm−1]→ B such that φ(xm) is integral over the image of φ′.


PROOF. We define for any positive integer s a K-algebra automorphism σs ofK[x1, . . . , xm] by σs(xm) = xm and σs(xi) := xi + xs

m−i

m for i < m (its inversefixes xm and sends xi to xi − xs

m−i

m for i < m). So if I = (i1, . . . , im) ∈ Zm≥0, andxI := xi11 · · ·ximm , then

σs(xI) = (x1 + xs

m−1

m )i1 · · · (xm−1 + xsm)im−1ximm .

When viewed as an element of K[x1, . . . , xm−1][xm], this is a monic polynomial inxm of degree pI(s) := i1s

m−1 + i2sm−2 + · · ·+ im. Now give Zm≥0 the lexicographic

ordering: this means that I > J implies pI(s) > pJ(s) for s large enough. Choosea nonzero f ∈ ker(φ). Then σs(f) ∈ ker(φσ−1

s ). If I ∈ Zm≥0 is the largest multi-exponent of a monomial that appears in f with nonzero coefficient, then for s largeenough, σs(f) is a constant times a monic polynomial in xm of degree pI(s) withcoefficients in K[x1, . . . , xm−1] and hence φσ−1

s (xm) = φ(xm) becomes integralover φσ−1

s K[x1, . . . , xm−1]. So φ′ := φσ−1s |K[x1, . . . , xm−1] is then as desired.

PROOF OF NOETHER NORMALIZATION. Let φ : K[x1, . . . , xm] → B be an epi-morphism of K-algebras which is not an isomorphism (otherwise we are done).We prove the refined version of the theorem with induction on m. According toLemma 7.9, there exists a K-algebra homomorphism φ′ : K[x1, . . . , xm−1] → Bsuch that φ(xm) is integral over B′ := φ′(K[x1, . . . , xm−1]). By induction, B′ is afinite extension of some polynomial algebra K[x1, . . . , xr] with r ≤ m − 1. Henceso is B.

REMARK 7.10. If A is a domain, then according to Proposition 7.3, Frac(A) willbe a finite extension of K(x1, . . . , xr) and so r must be the transcendence degree ofFrac(A)/K. In particular, r is an invariant of A.

COROLLARY 7.11. For every affine variety Y there exists an integer r ≥ 0 and afinite surjective morphism f : Y → Ar.

This corollary gives us a better grasp on the geometry of Y , especially when Yis irreducible, for it shows that Y can be spread over Ar in a finite-to-one mannerover affine r-space. The last assertion of Proposition 7.3 has a converse, also dueto Noether, which we state without proof.

*THEOREM 7.12 (Emmy Noether). Let K be a field and A be a finitely generatedK-domain. Then for any finite field extension L/Frac(A), the integral closure A

Lof

A in L is finite over A.

If we take L = Frac(A), then AL

is called the normalization of A and if Aequals its normalization, then we say that A is normal. We carry this terminologyto the geometric setting by saying that an irreducible affine variety X is normalwhen A(X) is. Any nonempty open subset of An is normal. More generally:

LEMMA 7.13. Any unique factorization domain A is normal.

PROOF. Suppose b ∈ Frac(A) is integral over A: bd+a1bd+· · ·+ad with ai ∈ A.

Write b = r/s with r, s relatively prime. Then rd + a1rsd−1 + · · ·+ ads

d and we seethat rd is divisible by s. This can only happen when s is a unit, so that b ∈ A.


Proposition 7.12 has a remarkable geometric interpretation: let be given anirreducible affine variety Y and a finite field extension L/k(Y ). Then Proposi-tion 7.12 asserts that A

Lis a finitely generated A(Y )-module. It is also an do-

main (because it is contained in a field) and so it defines an irreducible varietyYL := Spec(A

L). Since A ⊂ AL is an integral extension, we have a finite surjective

morphism YL → Y . Notice that this morphism induces the given field extensionL/k(Y ). This shows that every finite field extension of k(Y ) is canonically realizedby a finite morphism of irreducible affine varieties!

If L is a separable algebraic closure of k(Y ), then this does not apply, forL/k(Y ) will not be finite, unless Y is a singleton. But L can be written as a mono-tone union of finite (separable) field extensions: L = ∪∞i=1Li with Li ⊂ Li+1 andLi+1/Li finite. This yields a sequence of finite surjective morphisms

Y ← YL1← YL2

← YL3← · · ·

of which the projective limit can be understood as a “pro-affine variety” (a pointof this limit is given by a sequence (yi ∈ YLi)∞i=1 such that yi+1 maps to yi for all

i). Its algebra of regular functions is ∪∞i=1A(Y )Li

= A(Y )L

(which is usually not afinitely generated k-algebra) and its function field is L.

Of special interest is the case of a finite Galois extension. We begin with the cor-responding result from commutative algebra which has also important applicationsin algebraic number theory:

PROPOSITION 7.14. Let A be a normal domain and let L/Frac(A) be a finitenormal extension with Galois group G. Then G leaves invariant the integral closureAL

of A in L, and for every prime ideal p ⊂ A, G acts transitively on the set of primeideals q ⊂ AL that lie over p (i.e., with q ∩A = p).

PROOF. That any g ∈ G leaves invariant AL

is clear, for g fixes the coefficientsof an equation of integral dependence over A.

Let q and q′ lie over p. We first show that q′ ⊂ ∪g∈Ggq. If b ∈ q′, then the normof b over Frac(A), which is defined as the product

∏g∈G gb, lies in Frac(A), and

hence inAL∩Frac(A). SinceA is assumed to be normal, we haveA

L∩Frac(A) = A.Hence

∏g∈G gb ∈ A ∩ q′ = p ⊂ q. So we must have gb ∈ q for some g ∈ G, or

equivalently, b ∈ g−1q.From q′ ⊂ ∪g∈Ggq and the fact that each gq is a prime ideal, it follows from the

prime avoidance lemma below that q′ ⊂ gq for some g ∈ G. Since both q′ and gqlie over p, the incomparability property implies that q′ = gq.

LEMMA 7.15 (The prime avoidance lemma). Let R be a ring, q1, . . . , qn primeideals in R and I ⊂ R an ideal contained in ∪ni=1qi. Then I ⊂ qi for some i.

PROOF. We prove this induction on n. The case n = 1 being trivial, we mayassume that n > 1 and that for every i = 1, . . . , n, I is not contained in ∪j 6=iqj .Choose ai ∈ I \ ∪j 6=iqj . So then ai ∈ qi. Consider a := a1a2 · · · an−1 + an. Thena ∈ I and hence a ∈ qi for some i. If i < n, then an = a−a1a2 · · · an−1 ∈ qi and weget a contradiction. If i = n, then a1a2 · · · an−1 = a − an ∈ qn and hence aj ∈ qnfor some j ≤ n− 1. This is also a contradiction.

We transcribe this to algebraic geometry:


COROLLARY 7.16. Let Y be a normal variety and L/k(Y ) be a normal Galoisextension with Galois group G. Then G acts on YL in such a manner that it commuteswith f : YL → Y : for all g ∈ G we have fg = f . If Y ′ ⊂ Y is closed an irreducible,then G acts transitively on the irreducible components of f−1Y ′. In other words, Gacts transitively on the fibers of f (including the fibers over nonclosed points).

This also leads for normal domains to a supplement of the going up property:

COROLLARY 7.17 (Going down property). Suppose A is a normal domain andB ⊃ A an integral extension without zero divisors. Given prime ideals p in A and q′

in B such that q′ ∩A ⊃ p, then there exists a prime ideal q in B with q ∩A ⊂ p.

PROOF. Choose an algebraic closure Frac(B) of Frac(B) and let L be the sub-field of Frac(B) generated by all the Frac(A)-embeddings of Frac(B) in Frac(B) (sothis is obtained by adjoining the roots of the irreducible polynomials in Frac(A)[x]having a root in Frac(B)). Galois theory tells us that this is a finite normal ex-tension of Frac(B) (with Galois group G, say), which brings us in the situation ofProposition 7.14 above. Since L contains Frac(B) , A

Lcontains B. Observe that

AL

is finite over B and (hence) over A.Put p′ := q′ ∩ A so that p′ ⊃ p. According to Proposition 7.3 we can find in A

L

nested prime ideals p′ ⊃ p which meet A in p′ ⊃ p. Similarly, there exists a primeideal q′ in A

Lwhich meets B in q′. Since q′ and p′ both meet A in p′, there exists

according to Proposition 7.14 a g ∈ G which takes p′ to q′. Then upon replacing p′

by g′p′, we may assume that p′ = q′. Then q := p ∩ B is as desired: it meets A in pand is contained in p′ ∩B = q′ ∩B = q.

8. Dimension

One way to define the dimension of a topological space X is as follows: theempty set has dimension −1 and X has dimension ≤ n if it admits a basis of opensubsets such that the boundary of every basis element has dimension ≤ n− 1. Thisis close in spirit to the definition that we shall use here (which is however adaptedto the Zariski topology; as you will find in Exercise 38, it is useless for Hausdorffspaces).

DEFINITION 8.1. Let X be a nonempty topological space. We say that the Krulldimension of X is at least d if there exists an irreducible chain of length d in X, thatis, a strictly descending chain of closed irreducible subsets Y 0 ) Y 1 ) · · · ) Y d

of X. The Krull dimension of X is the supremum of the d for which an irreduciblechain of length d exists and we then write dimX = d. We stipulate that the Krulldimension of the empty set is −1.

LEMMA 8.2. For a subset Z of a topological space X we have dimZ ≤ dimX.

PROOF. If Y ⊂ Z is irreducible, then so is its closure Y in X. So if we have anirreducible chain of length d in Z, then the closures of the members of this chainyield an irreducible chain of length d in X. This proves that dimZ ≤ dimX.

EXERCISE 38. What is the Krull dimension of a nonempty Hausdorff space?

EXERCISE 39. Let U be an open subset of the space X. Prove that for anirreducible chain Y • in X of length d with U ∩ Y d 6= ∅, U ∩ Y • is an irreducible

8. DIMENSION 35

chain of length d in U . Conclude that if U is an open covering of X, then dimX =supU∈U dimU .

EXERCISE 40. Suppose that X is a noetherian space. Prove that the dimensionof X is the maximum of the dimensions of its irreducible components. Prove alsothat if all the singletons (= one element subsets) in X are closed, then dim(X) = 0if and only if X is finite.

It is straightforward to transcribe this notion of dimension into algebra:

DEFINITION 8.3. Let R be a ring. We say that the Krull dimension of R is at leastd if there exists an prime chain of length d in R, that is, a strictly ascending chain p•of prime ideals p0 ( p1 ( · · · ( pd in R. The Krull dimension of R is the supremumof the d for which this is the case and we then write dimR = d. We stipulate thatthe trivial ring R = 0 (which has no prime ideals) has Krull dimension −1.

The prime ideals of a ring R are the preimages of the prime ideals of Rred :=

R/√

(o) and so these rings have the same dimension.It is clear that for a closed subset X ⊂ An, dimA(X) = dimX. Similarly, we

have that for a finitely generated k-algebra A, dimA = dim Spec(A).Remark 7.4 shows immediately:

LEMMA 8.4. The Krull dimension is invariant under integral extension: if B isintegral over A, then A and B have the same Krull dimension.

The notion of Krull dimension was easily defined, but can be difficult to usein concrete cases. How can we be certain that given prime chain has maximalpossible length? It is not even clear how we can tell whether the Krull dimensionof a given ring is finite. We will settle this for a finitely generated k-domain (i.e., ofthe form A(Y ) with Y an irreducible affine variety), by showing that this is just thetranscendence degree over k of its field of fractions.

THEOREM 8.5. Let K be a field. For a finitely generated K-domain A, the Krulldimension of A equals the transcendence degree of Frac(A)/K. Moreover, every max-imal prime chain in A (i.e., one that cannot be extended to a longer prime chain) haslength dimA.

PROOF. We prove both assertions with induction on the transcendence degreeof Frac(A). By Noether normalization there exists an integer r ≥ 0 and a K-linearembedding of K[x1, . . . , xr] in A such that A is finite over K[x1, . . . , xr]. ThenFrac(A) is a finite extension of K(x1, . . . , xr) and so the transcendence degree ofFrac(A)/K is r.

Let (0) = q0 ( q1 ( · · · ( qm be a prime chain of A of length m. By theincomparability property, p• := q• ∩ Sr is then a prime chain in Sr, also of lengthm. Choose an irreducible f ∈ p1. After renumbering the coordinates, we mayassume that f does not lie in K[x1, . . . , xr−1] and so f =

∑Ni=0 aN−ix

ir with ai ∈

K[x1, . . . , xr−1], a0 6= 0 and N > 1. Then the image of xr in Frac(Sr/(f)) is aroot of the monic polynomial xNr +

∑N−1i=0 (aN−i/a0)xir ∈ K(x1, . . . , xr−1)[xr] so

that Frac(Sr/(f)) is a finite extension of K(x1, . . . , xr−1). Hence Frac(Sr/(f) hastranscendence degree r − 1 over K. By our induction induction hypothesis, theKrull dimension of Sr/(f) equals r−1. The image of the subchain p1 ( · · · ( pm inSd/(f) will still be strictly ascending (for it will be so in Sr/p1) and so m−1 ≤ r−1.Hence m ≤ r.


On the other hand, r is attained as the length of a prime chain in A, for theprime chain of length r in Sr given by (0) ( (x1) ( (x1, x2) ( · · · ( (x1, x2, . . . , xr),lifts to one in A, by Remark 7.4.

The last assertion relies on the going down theorem. If we assume that q• isa maximal prime chain in A, then the going down property Corollary ?? (whichapplies here since Sr is normal) implies that p• is one in Sr. Since our irreduciblef generates a prime ideal, it must then generate p1. The image of p1 ( · · · ( pm inSr/p1 = Sr/(f) is then a maximal prime chain in Sr/(f) and hence of length r − 1by induction hypothesis. This shows that m = r.

If S ⊂ R is a multiplicative system, then the preimage of a prime ideal q ofS−1R under the ring homomorphism R→ S−1R is a prime ideal q of R which doesnot meet S and we have q = S−1q. This sets up a bijection between the primeideals of S−1R and those of R not meeting S. If we take S = R − p, with p aprime ideal, then this implies that dimRp is the supremum of the prime chains inR which end with p. Since the prime chains in R which start with p correspond toprime chains in R/p, it follows that dimRp+dimR/p is the supremum of the primechains in R which contain p. According to Theorem 8.5, the latter is for a domainR finitely generated over a field always equal to the dimension of R.

COROLLARY 8.6. Let X be an irreducible affine variety. Then every irreduciblechain in X is a subchain of one of length dimX. Moreover, every nonempty openaffine U ⊂ X has the same dimension as X.

EXERCISE 41. Prove that a hypersurface in An has dimension n− 1.

EXERCISE 42. Let X be an irreducible affine variety and Y ⊂ X an irreducibleclosed subvariety. Prove that dimX − dimY is equal to the Krull dimension ofA(X)I(Y ).

EXERCISE 43. Prove that when X and Y are irreducible affine varieties, thendim(X×Y ) = dimX+dimY . (Hint: Embed each factor as a closed subset of someaffine space. You may also want to use the fact that the equality to be proven holdsin case X = Am and Y = An.)

9. Nonsingular points

In this section we focus on the local properties of an affine varietyX = Spec(A)(so A = A(X) is here a reduced finitely generated k-algebra) at a point o (whichmost of the time will be a closed point of X) and so a central role will be playedby the local algebra Apo := (A− po)

−1A whose maximal ideal is (A− po)−1po. We

will write OX,o for this local algebra and mX,o for its maximal ideal.

If k = C and X ⊂ Cn is a closed subset of dimension d, then we hope thatthere is a nonempty open subset of X where X is ‘smooth’, i.e., where X looks likea complex submanifold of complex dimension d. Our goal is to define smoothnessin algebraic terms (so that it make sense for our field k) and then to show that theset of smooth points of a variety is open and dense in that variety.

Our point of departure is the implicit function theorem. One version states thatif U ⊂ Rn is an open neighborhood of o ∈ Rn and fi : U → R, i = 1, . . . n − d aresuch that fi(o) = 0 and the total differentials at o, df1(o), . . . , dfn−d(o) are linearlyindependent in o (this is equivalent to: the Jacobian matrix of (f1, . . . , fn−d) at o is

9. NONSINGULAR POINTS 37

of rank n−d), then the common zero set of f1, . . . , fn−d is a submanifold of dimen-sion d at owhose tangent space there is the common zero set of df1(o), . . . , dfn−d(o).In fact, this solution set is there the graph of a map: we can express n − d of thecoordinates as smooth functions in the d remaining ones. Conversely, any subman-ifold of Rn at o of dimension d is locally thus obtained.

We begin with the observation that for any ring R partial differentiation ofa polynomial f ∈ R[x1, . . . , xn] is well-defined and produces another polynomial.The same goes for a fraction φ = f/g in R[x1, . . . , xn][g−1]: a partial derivative ofφ is a rational function (in this case with denominator g2). We then define the totaldifferential of a rational function φ ∈ R[x1, . . . , xn] as usual:

dφ :=

n∑i=1

∂φ

∂xi(x)dxi,

where for now, we do not worry about interpreting the symbols dxi: we think of dφsimply as a regular map from an open subset of An to a k-vector space of dimensionn with basis dx1, . . . , dxn, leaving its intrinsic characterization for later. However,caution is called for when R is a field of positive characteristic:

EXERCISE 44. Prove that f ∈ k[x] has zero derivative, if and only if f is constantor (when char(k) = p > 0) a pth power of some g ∈ k[x].

Generalize this to: given f ∈ k[x1, . . . , xn], then df = 0 if and only if f isconstant or (when char(k) = p > 0) a pth power of some g ∈ k[x1, . . . , xn].

We should also be aware of the failure of the inverse function theorem:

EXAMPLE 9.1. Let C ⊂ A2 be the curve defined by y2 = x3 + x. By anyreasonable definition of smoothness we should view the origin (0, 0) as a smoothpoint of C. Indeed, the projection f : C → A1, (x, y) 7→ y, would be a local-analytic isomorphism in case k = C. But the map is not locally invertible withinour category: the inverse requires us to find a rational function x = u(y) whichsolves the equation y2 = x3 + x and it is easy to verify that none exists. (We cansolve for x formally: x = u(y) = y2 − y6 + 3y10 + · · · , where it is important to notethat the coefficients are all integers so that this works for every characteristic.) Infact, the situation is worse: no affine neighborhood U of (0, 0) in C is isomorphicto an open subset V of A1. The reason is that this would imply that k(C) = k(U) isisomorphic to k(V ) = k(x) and one can show that this is not so.

Somewhat related to this is an issue illustrated by the following example.

EXAMPLE 9.2. Consider the curve C ′ ⊂ A2 defined by xy = x3 + y3 and leto = (0, 0). The polynomial x3 + y3 − xy is irreducible in k[x, y], so that A(C ′) iswithout zero divisors. Hence OC′,o ⊂ k(C ′) is also without zero divisors. But C ′

seems to have two branches at o which apparently can only be recognized formally:one such branch is given by y = u(x) = x2 + x5 + 3x8 + · · · and the other byinterchanging the roles of x and y: x = v(y) = y2 + y5 + 3y8 + · · · . If we useξ := x− v(y) and η := y − u(x) as new formal coordinates, then C ′ is simply givenat 0 by the reducible equation ξη = 0.

These examples make it clear that for a local understanding of a variety X at o,the local ringOX,o still carries too much global information, and that one way to get


rid of this overload might be by passing formal power series. This is accomplishedby passage to what is known as

9.3. FORMAL COMPLETION. Let R be a ring and I ( R a proper ideal. Weendow every R-module M with a topology, the I-adic topology of which a basisis the collection additive translates of the submodules InM , i.e., the collection ofsubsets a+ InM , a ∈M , n ≥ 0. This is a topology indeed: given two basic subsetsa + InM , a′ + In

′M , then for any element c in their intersection, the basic subset

c + Imaxn,n′M is also in their intersection. The fact that our basis is translationinvariant implies that with this topology, M is a topological abelian group ((a, b) ∈M ×M 7→ a− b ∈M is continuous). If we endow R also with the I-adic topologyand R ×M with the product topology, then the map (r, a) ∈ R ×M → ra ∈ Mwhich gives the action by R is also continuous.

For the topology on M to be Hausdorff, it suffices that the intersection of allneighborhoods of 0 is reduced to 0: ∩n≥0I

nM = 0. It is then even metriz-able: if φ : Z+ → (0,∞) is any strictly monotonously decreasing function withlimn→∞ φ(n) = 0 (one often takes u−n, where u > 1), then a metric ρ is defined by

ρ(a, a′) := infφ(n) : a− a′ ∈ InM.

This metric is nonarchimedean in the sense that ρ(a, a′′) ≤ maxρ(a, a′), ρ(a′, a′′).and a sequence (an ∈ M)∞n=0 is then a Cauchy sequence if and only if for everyinteger n ≥ 0 all but finitely many terms lie in the same coset of InM in M ;in other words, there exists an index in ≥ 0 such that aj − ain ∈ InM for allj ≥ in. This makes it clear that the notion of Cauchy sequence is independentof the choice of φ. We also note that the Cauchy sequence defines a sequence ofcosets (αn ∈ M/InM)n≥0 with the property that αn is the reduction of αn+1 ∈M/In+1. Recall that a metric space is said to be complete if every Cauchy sequencein that space converges. A standard construction produces a completion of everymetric space M : its points are represented by Cauchy sequences in M , with theunderstanding that two such sequences represent the same point if the distancebetween the two nth terms goes to zero as n → ∞. In the present situation thisyields what is called the I-adic completion, MI . From the above discussion we seethat an element of MI is uniquely given by a sequence (αn ∈ M/InM)n≥0 whoseterms are compatible in the sense that αn is the reduction of αn+1 for all n. It isan R-module for componentwise addition and multiplication so that the obviousmap M → MI , a 7→ (a+ In)n≥0 is a R-module homomorphism. (This makes MI asubmodule of the R-module

∏∞n=0M/InM .) It is easy to verify that MI is complete

for the I-adic topology and that the homomorphism M → MI is continuous andhas dense image.

Notice that RI is in fact a subring of∏∞n=0R/I

n in a way that makes R → RIa ring homomorphism. Moreover, MI is a RI -module (use that M/InM is a R/In-module for every n) and an R-homomorphism φ : M → N of R-modules extendsuniquely to an RI -homomorphism φI : MI → NI (use that φ sends InM to InNand hence induces a homomorphism M/In → N/In).

EXAMPLE 9.4. Take the ring k[x1, . . . , xn]. Its completion relative the maximalideal (x1, . . . , xn) is just the ring of formal power series k[[x1, . . . , xn]]. We get thesame result if we do this for the localization OAn,o of k[x1, . . . , xn] at (x1, . . . , xn).


EXAMPLE 9.5. Take the ring Z. Its completion with respect to the ideal (p),p prime, yields the ring of p-adic integers Z(p): an element of Z(p) is given by asequence (ai ∈ Z/(pi))∞i=1 with the property that ai is the image of ai+1 underthe reduction Z/(pi+1) → Z/(pi). We get the same result if we do this for thelocalization Z(p) = a/b : a ∈ Z, b ∈ Z− (p). If n is an integer ≥ 2, then it followsfrom the Chinese remainder theorem that Z(n) =

∏p|n Z(p).

EXERCISE 45. Let I and J be ideals of a ring R and m a positive integer withJm ⊂ I. Prove that the J -adic topology is finer than the I-adic topology and thatthere is a natural continuous ring homomorphism RJ → RI . Conclude that whenR is noetherian, RI can be identified with R√I .

*LEMMA 9.6 (Artin-Rees). Let R be a noetherian ring, I ⊂ R an ideal, M afinitely generatedR-module andM ′ ⊂M aR-submodule. Then there exists an integern0 ≥ 0 such that for all n ≥ 0:

M ′ ∩ In+n0M = In(M ′ ∩ In0M).

The proof (which is ingeneous, but not difficult) can be found in any book oncommutative algebra (e.g., Atiyah-Macdonald). The Artin-Rees lemma implies thatfor every n ≥ 0 there exists a n′ ≥ 0 such that M ′ ∩ In′M ⊂ InM ′ (we can taken′ = n+n0). This implies that the inclusion M ′ ⊂M is not merely continuous, butthat the I-adic topology on M ′ is induced by that of M . We will use the Artin-Reeslemma via this property only.

COROLLARY 9.7. In the situation of Lemma 9.6, the homomorphism M ′I → MI

induced by the inclusion M ′ ⊂ M is a closed embedding with the property that thepreimage of M is M ′: if the image of a ∈ M under M → MI lies in the imageof M ′I → MI , then a ∈ M ′. Moreover, MI/M

′I can be identified with the I-adic

completion of M/M ′.

One might phrase this by saying that when R is noetherian, I-adic completionis an exact functor on the category of finitely generated R-modules.

PROOF OF COROLLARY 9.7. We first part of the corollary amounts to the fol-lowing property: if a sequence (ai ∈M ′)∞i=0 in M ′ is a Cauchy sequence in M , thenit is already one in M ′ and when its limit lies in the image of M , then this limitlies in fact in the image of M ′ (an element of the kernel of M ′I → MI is given bya sequence (ai ∈ M ′)∞i=0 in M ′ which converges in M to 0 and so this implies thatM ′I → MI is injective).

Suppose therefore given a sequence (ai ∈ M ′)∞i=0 which is a Cauchy sequencein M . Then for every n there exists an in such that ai − ain ∈ InM for i ≥ in. So ifwe take i ≥ in+n0

, then ai−ain ∈M ′∩ In+n0M = In(M ′∩ In0M) ⊂ InM ′. Hence(ai ∈ M ′)∞i=0 is a Cauchy sequence in M ′. If (ai ∈ M ′)∞i=0 converges to the imageof a ∈ M , then for every n there exists an i′n such that ai − a ∈ InM for i ≥ i′n.So if we take i ≥ i′n+n0

, then by the preceding argument, ai − a ∈ InM ′. Hencea ∈ ai + InM ′ ⊂M ′ and (ai ∈M ′)∞i=0 converges to the image of a in M ′.

As to the last statement, we observe that RI -homomorphism MI → M/M ′Iinduced by the surjection M → M/M ′ is also surjective: any b ∈ M/M ′I is rep-resentable by a Cauchy sequence (bi ∈ M/M ′)∞i=0 with bm − bn ∈ In(M/M ′) for


m ≥ n. Since In(M/M ′) ∼= InM/(InM ∩M ′), we can represent bn+1− bn by somedn+1 ∈ InM . Let d0 ∈M lift b0. Then the series d0 + d1 + · · ·+ dn + · · · convergesin M and maps to b.

The kernel of MI → M/M ′I clearly contains M ′ and since a kernel is closed,it will also contain its closure M ′I in MI . In order to see that it is not bigger,let a ∈ ker(MI → M/M ′I) be arbitrary. If we represent a by a Cauchy sequence(ai ∈M)∞i=0, then for every n there exists an in such that for i ≥ in, ai ∈M ′+InM .Write ai accordingly: ai = a′i + di with a′i ∈ M ′ and di ∈ InM . Since (di ∈ M)∞i=0

converges to 0 ∈ M , (a′i ∈ M ′)∞i=0 is also a Cauchy sequence converging to a. Aswe saw above, this is then a Cauchy sequence in M ′ with the same limit and soa ∈ M ′I .

Let R be a noetherian local ring with maximal ideal m and residue field κ. Weuse simply for completion with respect to m.

Then m is a finitely generated R-module. Since the ring R acts on m/m2 viaR/m = κ, m/m2 is a finite dimensional vector space over κ.

DEFINITION 9.8. The Zariski cotangent space T ∗(R) of R is the κ-vector spacem/m2 and its κ-dual, T (R) := Homκ(m/m2, κ) (which is also equal to HomR(m, κ)),is called the Zariski tangent space T (R) of R. The embedding dimension embdim(R)is the dimension of m/m2 as a vector space over κ.

If X is an affine variety and o ∈ X, then the Zariski cotangent space T ∗oX resp.the Zariski tangent space ToX resp. the embedding dimension embdimoX of X ato are by definition that of OX,o.

For instance, the embedding dimension of An at any closed point o ∈ An isn. This follows from the fact that the map do : f ∈ mAn,o 7→ df(o) ∈ kn definesan isomorphism of k-vector spaces mAn,o/m

2An,o

∼= kn. We note in passing that wehere have a way of understanding the total differential at o in more intrinsic termsas the map do : OAn,o → mAn,o/m

2An,o which assigns to f ∈ OAn,o the image of

f−f(o) ∈ mAn,o in mAn,o/m2An,o. Thus, a differential of f at o can be understood a k-

linear function df(o) : ToAn → k and (do(xi) = dxi(o))ni=1 is a basis of mAn,o/m

2An,o.

Notice that the embedding dimension and the Zariski tangent space of a localring only depend on its formal completion.

EXERCISE 46. Let (R′,m) and (R,m′) be local rings with the same residuefield10 κ. Prove that a nonzero ring homomorphism φ : R′ → R induces an κ-linear map of Zariski tangent spaces T (φ) : T (R)→ T (R′)

An application of Nakayama’s lemma to M = m yields:

COROLLARY 9.9. The embedding dimension of a noetherian local ring R is thesmallest number of generators of its maximal ideal. The embedding dimension is zeroif and only if R is a field.

DEFINITION 9.10. A local ring R with maximal ideal m is said to be regular if itis noetherian and its Krull dimension equals its embedding dimension.

10By this we mean that φ induces an isomorphism R′/m′ ∼= R/m and that we identify the tworesidue fields via this isomorphism.


A point o of an affine variety X is called regular if its local ring OX,o is andsingular when it is not. (We write Xreg resp. Xsing for the subsets of X defined bythis property.) An affine variety without singular points is said to be nonsingular.

We will show that for a closed point o ∈ X this is indeed an intrinsic character-ization of ‘being like a submanifold’.

We mention without proof that an arbitrary point of X is a regular point of X preciselywhen its closure contains a closed point with that property. In other words, if that point isgiven by the closed irreducible subset Z ⊂ X, then it is regular if and only if Z contains aclosed point that is also a regular point of X.

We begin with a formal version of the implicit function theorem.

LEMMA 9.11. Let o ∈ An be a closed point and let f1, . . . , fn−d ∈ mAn,o be suchthat their images in mAn,o/m

2An,o are linearly independent. Then f1, . . . , fn−d gen-

erate a prime ideal p in the local ring OAn,o and the formal completion of OAn,o/pwith respect to its maximal ideal is isomorphic to k[[x1, . . . , xd]] as a complete lo-cal k-algebra. Moreover, there exists an affine neighborhood U of o in An on whichf1, . . . , fn−d are regular and generate a prime ideal in A(U) with the property thatthe corresponding irreducible affine variety X ⊂ U has o as a regular point.

PROOF. Let us abbreviate OAn,o by O and its maximal ideal by m. Extendf1, . . . , fn−d to a system of regular functions f1, . . . , fn ∈ m such that their imagesin m/m2 are linearly independent. (After an affine linear transformation, we maythen just as well assume that o is the origin of An and that fi ≡ xi (mod m2).)Hence the monomials of degree N in f1, · · · , fn map to a k-basis of mN/mN+1.This implies that the monomials of degree ≤ N in f1, · · · , fn make up a k-basis ofO/mN+1. It follows that the map

yi ∈ k[[y1, . . . , yn]] 7→ fi ∈ O

determines an isomorphism k[[y1, . . . , yn]] ∼= O of complete local rings (a ring iso-morphism that is also a homeomorphism). Its inverse defines a topological embed-ding of O in k[[y1, . . . , yn]] (sending fi to yi). The ideal generated by (y1, . . . , yn−d)in k[[y1, . . . , yn]] is the closure of the image of p. It is clearly a prime ideal. Accord-ing to Corollary 9.7 the preimage of that prime ideal in O is p (hence p is a primeideal) and the embedding

O/p → k[[y1, . . . , yn]]/(y1, . . . , yn−d) = k[[yn−d+1, . . . , yn]]

realizes the m-adic completion of O/p.Let U ⊂ An be an affine neighborhood of o on which the f1, . . . , fn−d are

regular and denote by P ⊂ A(U) the preimage of p under the localization homor-morphism A(U) → O. This is a prime ideal which contains f1, . . . , fn−d and hasthe property that it generates the same ideal in O as (f1, . . . , fn−d). Choose a finiteset of generators φ1, . . . , φr of P and write φi =

∑n−dj=1 uijfj with uij ∈ O. By pass-

ing to a smaller U , we may then assume that each uij is regular on U so that P isin fact equal to the ideal (f1, . . . , fn−d) in A(U) generated by f1, . . . , fn−d. Hence(f1, . . . , fn−d) defines an irreducible affine variety X ⊂ U which has o is a regularpoint of dimension d.


THEOREM 9.12. Let X ⊂ An be closed and let o ∈ X be a closed point of X. Thenthe local ring OX,o is regular of dimension d if and only there exist regular functionsf1, . . . , fn−d on an affine neighborhood U of o in An such that these functions generateIU (X ∩ U) and df1, . . . , dfn−d are linearly independent in every point of U . In thatcase, X ∩ U is regular and for every closed point q ∈ X ∩ U the Zariski tangent spaceTqX is the kernel of the linear map (df1(q), . . . , dfn−d(q)) : TqAn → kn−d.

PROOF. Suppose that OX,o is regular of dimension d. Let IX,o ⊂ OAn,o be theideal of regular functions at o vanishing on a neighborhood of o in X, in otherwords, the kernel of OAn,o → OX,o. The latter is a surjective homomorphism oflocal rings and so the preimage of mX,o resp. m2

X,o is IX,o + mAn,o = mAn,o resp.IX,o + m2

An,o. It follows that the quotient mAn,o/(IX,o + m2An,o)

∼= mX,o/m2X,o has

dimension d. So the image (IX,o+m2An,o)/m

2An,o of IX,o in the n-dimensional vector

space mAn,o/m2An,o must have dimension n − d. Choose f1, . . . , fn−d ∈ IX,o such

that df1(o), . . . , dfn−d(o) are linearly independent. We show among other thingsthat these functions generate IX,o (this will in fact be the key step).

According to Lemma 9.11, the ideal pi ⊂ OAn,o generated by f1, . . . , fi is primeand so we have a prime chain

(0) ⊆ p0 ( p1 ( · · · ( pn−d ⊆ IX,o.

As OX,o is of dimension d, there also exists a prime chain of length d containingIX,o:

IX,o ⊆ q0 ( q1 ( · · · ( qd ⊆ OAn,o.

Since OAn,o has dimension n, these two prime chains cannot make up a primesequence of length n+ 1 and so pn−d = IX,o = q0.

In particular, f1, . . . , fn−d generate IX,o. Let U ′ be an affine neighborhoodof o in An on which the fi’s are regular and has the property that X ∩ U ′ is thecommon zero set of f1, . . . , fn−d. Since the differentials df1(o), . . . , dfn−d(o) arelinearly independent, there exist n − d indices 1 ≤ ν1 < ν2 < · · · < νn−d ≤ n suchthat δ := det((∂fi/∂xνj )i,j) ∈ A(U ′) is nonzero in o. Then U := U(δ) ⊂ U ′ has allthe asserted properties (with the last property following from Lemma 9.11).

The converse says that if U , o and f1, . . . , fn−d are as in the theorem, thenthe functions f1, . . . , fn−d generate a prime ideal Io in OAn,o such that OAn,o/Io isregular. This follows Lemma 9.11.

The last assertion is clear from the preceding.

PROPOSITION 9.13. The regular closed points of an affine variety X form anopen-dense subset Xreg of the set of closed points of X.

For the proof we will assume Proposition 9.14 below, which we will leave un-proved for now. (Note that it tells us only something new in case k has positivecharacteristic.)

*PROPOSITION 9.14. Every finitely generated field extension L/k (so with k as inthese notes, i.e., algebraically closed) is separably generated by which we mean thatthere exists an intermediate extension k ⊂ K ⊂ L such that K/k is purely transcen-dental (i.e., of the form k(x1, . . . , xr)) and L/K is a finite separable extension.

According to 6.7 and the subsequent discussion, this implies that every irre-ducible affine variety is birationally equivalent to a hypersurface.


PROOF OF PROPOSITION 9.13. Without loss of generality we may assume thatX is irreducible. Since we already know thatXreg is open, it remains to see that it isnonempty. It thus becomes an issue which only depends on k(X). Hence it sufficesto treat the case of a hypersurface in An so that I(X) is generated by an irreduciblepolynomial f ∈ k[x1, . . . , n]. In view of Lemma Lemma 9.11 it then suffices toshow that df is not identically zero on X. Suppose otherwise, i.e., that each partialderivative ∂f/∂xi vanishes on X. Then each ∂f/∂xi must be multiple of f andsince the degree of ∂f/∂xi is less than that of f , this implies that it is identicallyzero. But then we know from Exercise 44 that the characteristic p of k is positiveand that f is of the form gp. This contradicts the fact that f is irreducible.

EXERCISE 47. Let X be a nonsingular variety. Prove that X is connected if andonly if it is irreducible.

REMARK 9.15. This enables us to find for an affine variety X of dimension da descending chain of closed subsets X = Xd ⊃ Xd−1 ⊃ · · · ⊃ X0 such thatdimXi ≤ i and all the (finitely many) connected components of Xi − Xi−1 arenonsingular subvarieties of dimension i (such a chain is called a stratification ofX).We let Xd−1 be the union of Xsing and the irreducible components of dimension< d and if (with downward induction) Xi has been defined, then we take for Xi−1

the union of the singular locus Xireg and the irreducible components of dimension

≤ i − 1. Then dimXi−1 ≤ i − 1 and every connected component of Xi − Xi−1

is an nonempty open subset of some Xireg and hence a nonsingular subvariety of

dimension i.9.16. DIFFERENTIALS AND DERIVATIONS. The differential that we defined earlier has an

intrinsic, coordinate free description that turns out to be quite useful. Let us begin with theobservation that the formation of the total differential of a polynomial, φ ∈ k[x1, . . . , xn] 7→dφ :=

∑ni=1

∂φ∂xi

(x)dxi is a k-linear map which satisfies the Leibniz rule: d(φψ) = φdψ+ψdφ.This property is formalized with the following definition. Let us fix a ring R (the base ring)and an R-algebra A.

DEFINITION 9.17. Let M be a A-module. An R-derivation of A with values in M is a mapD : A → M is an R-module homomorphism which satisfies the Leibniz rule: D(a1a2) =a1D(a2) + a2D(a2) for all a1, a2 ∈ A.

The last condition usually prevents D from being an A-module homomorphism. Let usnote that (by taking a1 = a2 = 1) that we must have D(1) = 0. Then it also follows thatfor every r ∈ R, D(r) = rD(1) = 0. Note also that if a happens to be invertible in A, then0 = D(1) = D(a/a) = 1/aD(a) + aD(1/a) so that D(1/a) = −D(a)/a2.

Given a1, . . . , an ∈ A, then the values of D on a1, . . . , an determine its values onthe subalgebra A′ by the ai’s, for if φ : R[x1, . . . , xn] → A denotes the corresponding R-homomorphism and f ∈ R[x1, . . . , xn], then

Dφ(f) =

n∑i=1

φ( ∂f∂xi

)Dai.

If we combine this with the formula for D(1/a), we see that this even determines D onthe largest subalgebra A′ of A generated by the ai’s and the invertible elements in A′. Inparticular, if we are given a field extension L/K, then a K-derivation of L with valuesin some L-vector space is determined by its values on a set of generators of L as a fieldextension of K.

Observe that the set of R-derivations of A in M form an R-module: if D1 and D2 areR-derivations of A with values in M , and a1, a2 ∈ A, then a1D1 + a2D2 is also one. Wedenote this module by DerR(A,M).


EXERCISE 48. Prove that if D1, D2 ∈ DerR(A,A), then [D1, D2] := D1D2 − D2D1 ∈DerR(A,A). What do we get for R = k and A = k[x1, . . . , xn]?

It is immediate from the definition that for every homomorphism of A-modules φ :M → N , the composition of a D as above with φ is an R-derivation of A with values in N .We can now construct a universal R-derivation of A, d : A → ΩA/R (where ΩA/R must ofcourse be an R-module) with the property that every D as above is obtained by composingd with a unique homomorphism of A-modules D : ΩA/R → N . The construction that isforced upon us starts with the free A-module A(A) which has A itself as a generating set—let us denote the generator associated to a ∈ A by d(a)—which we then divide out by theA-submodule of A(A) generated by the expressions d(ra)− rd(a), d(a1 +a2)− d(a1)− d(a2)

and d(a1a2) − a1d(a2) − a2d(a2), with r ∈ R and a, a1, a2 ∈ A. The quotient A-moduleis denoted ΩA/R and the composite of d with the quotient map by d : A → ΩA/R. Thelatter is an R-derivation of A by construction. Given an R-derivation D : A → M , then themap which assigns to d(a) the value Da extends (obviously) as a A-module homomorphismA(A) → M . It has the above submodule in its kernel and hence determines an A-modulehomomorphism of D : ΩA/R → M . This has clearly the property that D = Dd. We callΩA/R the module of Kahler differentials. We will see that the map d : A → ΩA/R can bethought of as an algebraic version of the formation of the (total) differential.

The universal derivation of a finitely generated R-algebra is obtained as follows. Firstwe do the case of a polynomial algebra P := R[x1, . . . , xn]. Assigning to f ∈ P the rowvector (∂f/∂x1, . . . , ∂f/∂xn) is an R-derivation on P with values in Pn. By the universalproperty of ΩP/R this must be given by a unique P -homomorphism ΩP/R → Pn. For anyR-derivation D : P →M we have Df =

∑ni=1(∂f/∂xi)Dxi and the Dxi can be prescribed

arbitrarily. So if we take M = Pn as above, we see that ΩP/R → Pn is an isomorphism.Thus the universal R-derivation d : P → ΩP/R may be regarded as the intrinsic way offorming the total differential.

Next consider a quotient A := P/I of P , where I ⊂ P is an ideal. If M is an A-moduleand D′ : A → M is an R-derivation, then its composite with the projection π : P → A,D = D′π : P → M , is an R-derivation of P with the property that Df = 0 for every f ∈ I.Conversely, every R-derivation D : P → M in an A-module M with this property factorsthrough an an R-derivation D′ : A → M . We observe here that the map f ∈ I 7→ D(f)factors through I/I2, for if f, g ∈ I, then D(fg) = fDg + gDf ∈ IM = 0. Since I/I2 isan A-module, we thus obtain a short exact sequence of A-modules

I/I2 → ΩP/R/IΩP/R → ΩA/R → 0.

If we are given a generating set f1, . . . , fm for I, then ΩA/R can be identified with thequotient of An by the A-submodule generated by the row vectors (∂fi/∂x1, . . . , ∂fi/∂xn),i = 1, . . . ,m. Note that if R is a noetherian ring, then, as we know, so is A and it also followsthat ΩA/R is a noetherian R-module.

This applies to the case when R = k and A = A(X) for some affine variety X. We thenwrite Ω(X) for ΩA(X)/k.

EXERCISE 49. Prove that ΩA/R behaves well with localization: if S ⊂ A is a multi-plicative subset, then every R-derivation with values is some A-module M extends natu-rally to R-derivation of S−1A-with values in S−1M . Prove that we have a natural mapS−1ΩA/R → ΩS−1A/R and that this map is a A-homomorphism.

For an affine variety X and x ∈ X, we write ΩX,x for ΩOX,x/k. The preceding exerciseimplies that ΩX,x is the localization of Ω(X) at x: ΩX,x = (A(X)− px)−1Ω(X).

10. LOCAL NATURE OF A REGULAR FUNCTION AND THE NOTION OF A VARIETY 45

EXERCISE 50. Let X be an affine variety and let o ∈ X be a closed point. Show that theZariski tangent space ofX at o can be understood (and indeed, be defined) as the space of k-derivations of OX,o with values in k, where we regard k as a OX,o-module via OX,o/mX,o ∼=k. Prove that this identifies its dual, the Zariski cotangent space, with ΩX,o/mX,oΩX,o.

EXERCISE 51. Show (perhaps with the help of the preceding exercises) that if o ∈ An isa closed point, then ΩOAn,o/k is the free OAn,o-module generated by dx1, . . . , dxn and thatif X is an affine variety in An that has o a regular point, then ΩX,o is a free OX,o-module ofrank dimOX,x.

We must be careful with this construction when dealing with formal power series rings.For instance, as we have seen, the completion of the local k-algebra OA1,0 with respect to itsmaximal ideal is k[[x]] and the embedding of OA1,0 → k[[x]] is given by Taylor expansion.A k-derivation D of k[[x]] with values in some k[[x]]-module is not determined by Dx. Allit determines are the values on the k-subalgebra OA1,0. This issue disappears however if werequire that D commutes with infinite summation when the series converges relative to the(x)-adic metric, for then D(

∑∞r=0 crx

r) will be equal to∑∞r=0 crrx

r−1Dx.

EXERCISE 52. Prove that the composite d : A → ΩA/R → (ΩA/R)I factors through aderivation dI : AI → (ΩA/R)I and prove that it is universal among all the R-derivations ofA in R-modules that are complete for the I-adic topology (i.e., for which M = MI).

So if o is a regular closed point of an affine variety X, then ΩX,o is a free OX,o-moduleof rank dimOX,x.

10. Local nature of a regular function and the notion of a variety

In any topology or analysis course you learn that the notion of continuity islocal: there exists a notion of continuity at a point so that a function is continuousif it is so at every point of its domain. We shall see that in algebraic geometry thenotion of a regular function also has a local nature.

DEFINITION 10.1. Let X be an affine variety, U ⊂ X an open subset and φ :U → k a function. We say that φ is regular at x ∈ U if it is so on an affineneighborhood of x in X. In other words: there should exist gx ∈ A(X) − px andfx ∈ A(X) such that φ|U(gx) ∩ U is representable as the fraction fx/gx. We saythat φ is locally regular if it is so in every point of U and we denote the k-algebra ofsuch functions by O(U).

We will see that for U affine, locally regular implies regular. After that, wewill abandon the adjective locally and just speak of regular functions, whether thedomain is affine or not. Both the notion of ‘locally regular’ and the proof that foraffine varieties this is the same notion as regular have a counterpart in the contextof ring spectra and we will in fact prove an intermediate result in that setting.

PROPOSITION 10.2. The natural map A(X) → O(X) is an isomorphism of k-algebras. A map φ : X → Y between affine varieties is a morphism if and only if φ iscontinuous and for any f ∈ O(V ) (with V ⊂ Y open) we have f∗φ = φf ∈ O(f−1V ).

REMARK 10.3. Before we begin the proof we make the following observation.Suppose (ri ∈ R)i∈I is a collection of elements such that Spec(R) = ∪i∈IU(ri).This means that the ideal generated by the ri is not contained in any maximalideal and hence must be all of R. In particular, this ideal contains 1 so that1 = a1ri1 + · · · + anrin for certain i1, . . . , in ∈ I and a1, . . . , an ∈ R. It followsthat Spec(R) = ∪nν=1U(riν ). This shows that Spec(R) is quasi-compact: any open


covering of Spec(R) admits a finite subcovering. In particular, any affine varietyhas this property.

PROOF OF PROPOSITION 10.2. It is clear that the map is injective: if f ∈ A(X)is in the kernel, then f must be identically zero as a function and hence f = 0.

For surjectivity, let φ : X → k be regular in every point of X. We must showthat φ is representable by some f ∈ A(X). By definition there exists for everyx ∈ X, gx ∈ A(X) − px and fx ∈ A(X) such that φ|U(gx) is representable as thefraction fx/gx. Since X is quasicompact, the covering U(gx)x∈X of X possessesa finite subcovering U(gxi)Ni=1. Let us now write fi for fxi and gi for gxi . Thenfi/gi and fj/gj define the same regular function on U(gi) ∩ U(gj) = U(gigj) andso gifj − gjfi is annihilated by (gigj)

mij for some mij ≥ 0. Let m := maxmijso that gm+1

i gmj fj = gmi gm+1j fi for all i, j. Since φ|U(gi) is also represented by

(figmi )/gm+1

i we may then without loss of generality assume that m = 0, so thatgifj = gjfi for all i, j.

Since the U(gi) cover X, the ideal generated by g1, . . . , gN must be all of A(X)

and hence contain 1. So 1 =∑Ni=1 aigi for certain ai ∈ A(X). Now consider

f :=∑Ni=1 aifi ∈ A(X). We have for every j,

fgj =

N∑i=1

aifigj =

N∑i=1

aigifj = fj

and so f represents φ on Uj . It follows that φ is represented by f .The last statement is left as an exercise.

Let us denote by OX the collection of the k-algebras O(U), where U runs overall open subsets ofX. The preceding proposition says thatOX is a sheaf of k-valuedfunctions on X, by which we mean the following:

DEFINITION 10.4. Let X be a topological space and K a ring. A sheaf O ofK-valued functions11 on X assigns to every open subset U of X a K-subalgebraO(U) of the K-algebra of K-valued functions on the set of closed points in U withthe property that

(i) for every inclusion U ⊂ U ′ of open subsets of X, ‘restriction to U ’ mapsO(U ′) in O(U) and

(ii) given a collection (Ui)i∈I of open subsets of X and f : ∪i∈IUi → K, thenf ∈ O(∪iUi) if and only if f |Ui ∈ O(Ui) for all i.

If (X,OX) and (Y,OY ) are topological spaces endowed with a sheaf of K-valued functions, then a continuous map f : X → Y is called a morphism if forevery open V ⊂ Y , composition with f takes OY (V ) to OX(f−1V ).

With the notion of a morphism, we have a category of topological spaces en-dowed with a sheaf of K-valued functions. In particular, we have the notion ofisomorphism: this is a homeomorphism f : X → Y which for every open V ⊂ Ymaps OY (V ) onto OX(f−1V ).

Note that a sheaf O of K-valued functions on X restricts to a sheaf O|U forevery open U ⊂ X.

11We give the general definition of a sheaf later. This will do for now. A defect of this definition isthat a sheaf of K-valued functions on X need not restrict to one on an arbitrary subspace of X. Anotheris the special role attributed to the closed points.


The definition above is a way of expressing that we are dealing with a localproperty—just as we have a sheaf of continuous R-valued functions on a topologicalspace, a sheaf of differentiable R-valued functions on a manifold and a sheaf ofholomorphic C-valued functions on a complex manifold. In fact for the followingdefinition, which includes our final definition of an affine variety, we take our cuefrom the definition of a manifold.

DEFINITION 10.5. An affine k-variety is a topological space X endowed with asheaf OX of k-valued functions which is isomorphic to a pair as above12.

A k-prevariety is a topological space X endowed with a sheaf OX of k-valuedfunctions such that X can be covered by finitely many open subsets U such that(U,OX |U) is an affine k-variety.

A quasi-affine k-variety is a k-prevariety which is isomorphic to an open subsetof an affine variety.

Given k-prevarieties (X,OX) and (Y,OY ), then a morphism f : (X,OX) →(Y,OY ) is simply a morphism in the category of spaces endowed with a sheaf OXof k-valued functions, so f is continuous and for every open V ⊂ Y , compositionwith f takes OY (V ) to OX(f−1V ).

We often designate a prevariety and its underlying topological space by thesame symbol, a habit which rarely leads to confusion. The composite of two mor-phisms is evidently a morphism so that we are dealing here with a category. Theprefix ‘pre’ in prevariety refers to the fact that we have not imposed a separationrequirement which takes the place of the Hausdorff property that one normallyimposes on a manifold (see Example 10.8 below).

Let X be a prevariety. By assumption X is covered by finitely many affine opensubvarieties U1, . . . , UN . Suppose κi is an isomorphism of Ui onto an affine varietyXi that sits as a closed subset in some affine space. Then Xi,j := κi(Ui ∩ Uj) isan open subset of Xi and κi,j := κjκ

−1i is an isomorphism of Xi,j onto Xj,i ⊂ Xj .

We can recover X from the disjoint union of the X1, . . . , XN by means of a gluingprocess: if we use κi,j to identify Xi,j with Xj,i for all i, j we get back X. Thecollection (Ui, κi)Ni=1 is called an affine atlas for X.

EXERCISE 53. Let (X,OX) be a prevariety.(i) Prove that X is a noetherian space.

(ii) Prove that X contains an open-dense subset which is affine.(iii) Let Y ⊂ X be locally closed (i.e., the intersection of a closed subset with

an open subset). Prove that Y is in natural manner a prevariety in such amanner that the inclusion Y ⊂ X is a morphism of prevarieties.

Much of what we did for affine varieties extends in a straightforward mannerto this more general context. For instance, a rational function f : X 99K k is definedas before: it is represented by a regular function on a subset ofX that is open-densein its set of closed points and two such represent the same rational function if theycoincide on a nonempty open-dense subset in their common domain of definition.

If X is irreducible, then the rational functions on X form a field k(X), thefunction field of X. If U ⊂ X is an open nonempty affine subset, then k(X) =k(U) = Frac(O(U)) (but we will see that it is not true in general that k(X) =

12Note that X can then be identified with the spectrum of OX(X)


Frac(O(X))). In particular, U1, . . . , UN all have the same dimension trdegk k(X).According to Exercise 39, this is then also the (Krull) dimension of X.

Similarly, if X and Y are prevarieties, then a rational map f : X 99K Y isrepresented by morphism from a nonempty open-dense subset of X to Y with theunderstanding that two such defined the same map if they coincide on a nonemptyopen-dense subset. If some representative morphism has dense image in Y , thenf is said to be dominant and then f induces a field extension f∗ : k(Y ) → k(X).Conversely, an embedding of fields k(Y ) → k(X) determines a dominant rationalmapX 99K Y . If U ⊂ X is open and nonempty, then k(U) = k(X) and the inclusionis a birational equivalence.

The notion of a regular point is local and so automatically carries over to thisgeneral setting. It is then also clear that the regular points of a prevariety X definean open-dense subset Xreg of X.

10.6. THE PRODUCT OF TWO PREVARIETIES. Our discussion of the product ofclosed subsets of affine spaces dictates how we should define the product of twoprevarieties X and Y : if (p, q) ∈ X × Y , then let p ∈ U ⊂ X and q ∈ V ⊂ Ybe affine open neighborhoods of the components. We require that the topology onU × V be the Zariski topology so that a basis of neighborhoods of (p, q) consists ofthe loci (U × V )(h) where a h ∈ O(U)⊗O(V ) with h(p, q) 6= 0 is nonzero. We alsorequire that ring of sections of OX×Y over such a basic neighborhood (U × V )(h)be O(U)⊗O(V )[1/h].

EXERCISE 54. Prove that this product has the usual categorical characteriza-tion: the two projections X × Y → X and X × Y → Y are morphisms and if Zis a prevariety, then a pair of maps (f : Z → X, g : Z → Y ) defines a morphism(f, g) : Z → X × Y if and only both f and g are morphisms.

The Hausdorff property is not of a local nature: a non-Hausdorff space can verywell be locally Hausdorff. The standard example is the space X obtained from twocopies of R by identifying the complement of 0 in either copy by means of theidentity map. Then X is locally like R, but the images of the two origins cannot beseparated. A topological spaceX is Hausdorff precisely when the diagonal ofX×Xis a closed subset relative to the product topology. As we know, the Zariski topologyis almost never Hausdorff. But on the other hand, the selfproduct of the underlyingspace has not the product topology either and so requiring that the diagonal isclosed is not totally unreasonable a priori. In fact, imposing this condition turnsout to be the appropriate way of avoiding the pathologies that can result from anunfortunate choice of gluing data.

DEFINITION 10.7. A k-prevariety X is called a k-variety if the diagonal is closedin X ×X (where the latter has the Zariski topology as defined above).

The diagonal in An ⊂ An × An is closed, so An is a variety. This implies thatthe same is true for any quasi-affine subset of An. Hence a quasi-affine prevarietyis in fact a variety.

EXAMPLE 10.8. The simplest example of a prevariety that is not a variety is theobvious generalization of the space described above: let X be obtained from twocopies A1

+ and A1− of A1 by identifying A1

+ − 0 with A1− − 0 by means of the

identity map. If o± ∈ X denotes the image of origin of A1±, then (o+, o−) ∈ X ×X

lies in the closure of the diagonal, but is not contained in the diagonal.


A subset of a variety X is called a subvariety if it is open in a closed subset ofX. The proof of the following assertion is left as an exercise (see also Exercise 53).

PROPOSITION 10.9. A subvariety is in a natural manner a variety such that theinclusion becomes a morphism. The product of two varieties is a variety.

EXAMPLE 10.10. We can take this further: if U ⊂ Am is open and f : U → An isa morphism, then consider the graph of f , U := (x, y) ∈ Am+n : x ∈ U, y = f(x).It is easy to see that U is a subvariety of Am+n. The map x ∈ U 7→ (x, f(x)) ∈ Uand the projection U → U are regular and each others inverse. So they define anisomorphism U → U . Notice that via this isomorphism f appears as a projectionmapping: (x, y) ∈ U 7→ y ∈ V .

EXERCISE 55. The goal of this exercise is to show that U := A2−(0, 0) is notaffine. Let Ux ⊂ U resp. Uy ⊂ U be the complement of the x-axis resp. y-axis sothat U = Ux ∪ Uy.

(a) Prove that Ux is affine and that O(Ux) = k[x, y][1/x].(b) Prove that every regular function on U extends to A2 so that O(U) =

k[x, y].(c) Show that U is not affine.

CHAPTER 2

Projective varieties

1. Projective spaces

Two distinct lines in the plane intersect in a single point or are parallel. Inthe last case one would like to say that the lines intersect at infinity so that thestatement becomes simply: two distinct lines in a plane meet in a single point.There are many more examples of geometric configurations for which the specialcases disappear by the simple remedy of adding points at infinity. A satisfactoryapproach to this which makes no a priori distinction between ordinary points andpoints at infinity involves the notion of a projective space. The following notion,perhaps a bit abstract, is quite useful. It will soon become more concrete.

DEFINITION 1.1. A projective space of dimension n over k is a set P endowedwith an extra structure that can be given by a pair (V, `), where V is k-vectorspace of dimension n + 1 and ` is a bijection between P and the collection of 1-dimensional linear subspaces of V : p ∈ P 7→ `p ⊂ V . It is here understood thatanother such pair (V ′, `′) defines the same structure if and only if there exists ak-linear isomorphism φ : V → V ′ such that for every p ∈ P , φ sends `p to `′p. (Weare in fact saying that thus is defined an equivalence relation on the collection ofsuch pairs and that a projective structure is given by an equivalence class.)

In particular, if V is a finite dimensional k-vector space, then the collection ofits 1-dimensional linear subspaces is in a natural manner a projective space; wedenote it by P(V ). We often write Pn or Pnk for P(kn+1) and call it simply projectiven-space (over k).

EXERCISE 56. Prove that the linear isomorphism φ in Definition 1.1 is uniqueup to scalar multiplication.

DEFINITION 1.2. Given a projective space P of dimension n over k, then asubset L of P is said to be linear subspace of dimension d if, for some (or any) pair(V, `) as above, there exists a linear subspace VL ⊂ V of dimension d+ 1 such that`(L) is the collection of 1-dimensional linear subspaces of VL.

A map f : P → P ′ between two projective spaces over k is said to be linearmorphism if for corresponding structural data (V, `) and (V ′, `′) for P resp. P ′ thereexists a linear injection F : V → V ′ such that F sends `p to `′f(p) for all p ∈ P .

So a linear subspace has itself the structure of a projective space and its inclu-sion in the ambient projective space is a linear morphism. Conversely, the image ofa linear morphism is linear subspace.

A linear subspace of dimension one resp. two is often called a line resp. a plane.A linear subspace of codimension one (of dimension one less than the ambientprojective space) is called a hyperplane. It is now clear that two distinct lines in a

51

52 2. PROJECTIVE VARIETIES

plane intersect in a single point: this simply translates the fact that the intersectionof two distinct linear subspaces of dimension two in a three dimensional vectorspace is of dimension one.

Let P be a projective space of dimension n. We can of course describe itsstructure by a pair (V, `) for which V = kn+1. This gives rise to a ‘coordinate systemon P ’ as follows: if we denote the coordinates of kn+1 by (X0, . . . , Xn), then everypoint p ∈ P(V ) is representable as a ratio [p0 : · · · : pn] of n + 1 elements of k thatare not all zero: choose a generator p ∈ `p and let pi = Xi(p). Any other generatoris of the form λp with λ ∈ k − 0 and indeed, [λp0 : · · · : λpn] = [p0 : · · · : pn].This is why [X0 : · · · : Xn] is called a homogeneous coordinate system on P(V ) eventhough an individual Xi is not a function on P(V ) (but the ratios Xi/Xj are, albeitthat for i 6= j they are not everywhere defined).

1.3. LINEAR CHARTS AND STANDARD ATLAS. Let U ⊂ P be a hyperplane com-plement in P , i.e., the complement of a hyperplane H ⊂ P . We show that U isin a natural manner an affine space. To see this, let the structure on P be givenby the pair (V, `). Then the hyperplane H corresponds to a hyperplane VH ⊂ V .A coset A of VH in V is an affine hyperplane in the classical sense of the word.Given a coset A of VH distinct from VH , then assigning to v ∈ A the 1-dimensionallinear subspace spanned by v defines a bijection A ∼= P − H whose inverse weredenote by κA : P − H ∼= A. This puts on U a structure of an affine space. Thisis independent of the choice of A: any another choice A′ is obtained from A bymultiplication by some nonzero scalar u ∈ k and hence κA′ is the composite of κAand the map u· : A ∼= A′; since the latter is an isomorphism of affine spaces, κA′defines the same affine structure on P − H as κA. For a linear subspace L ⊂ Pthe bijection U ∼= A restricts to one between L ∩ U and the affine-linear subspaceVL ∩ A. So if L ∩ U 6= ∅, then L ∩ U is an affine subspace of U . Concretely, ifwe choose a homogeneous coordinate system (X0, . . . , Xn) such that H is given byX0 = 0, then (x1 := X1/X0, . . . , xn := Xn/X0) maps A bijectively onto An andany other choice will differ from this one by an affine-linear transformation of An.We call such an isomorphism from a hyperplane complement in P onto An a linearchart for P .

Thus the homogeneous coordinate system [X0, . . . , Xn] defines a chart for everyi = 0, . . . , n: if Ui ⊂ P is the hyperplane complement defined by Xi 6= 0, then

κi : Ui ∼= An, [X0 : · · · : Xn] 7→ (X0/Xi, . . . , Xi/Xi, . . . , Xn/Xi),

is a chart with inverse (x1, . . . , xn) 7→ [x1 : · · · : xi : 1 : xi+1 : · · · : xn]. Noticethat the Ui’s cover P . We call a collection of linear charts (Ui, κi)

ni=0 thus obtained

from a homogeneous coordinate system a standard atlas for P . Let us determinethe coordinate change for a pair of charts, say for κnκ−1

0 . The image of U0 ∩ Ununder κ0 resp. κn is the open subset U(xn) resp. U(x1) of An and

κnκ−10 : U(xn)→ U(x1), (x1, x2, . . . , xn) 7→ (1/xn, x1/xn, . . . , xn−1/xn).

Notice that this defines an isomorphism of affine varieties (the inverse is κ0κ−1n ).

Thus P becomes the set of closed points of a prevariety. In particular, P ac-quires a (Zariski) topology for which U ⊂ P is open if and only if κ(U ∩Ui) is openin the set of closed points of An. We now change the interpretation of P a bit asto by adding a point for every subset of P that is closed and irreducible for thistopology (and was not already in P ). Then we have a prevariety (P,OP ) with the

2. THE ZARISKI TOPOLOGY ON A PROJECTIVE SPACE 53

property that U ⊂ P is open if and only if κ(U ∩ Ui) is open in An and f ∈ OP (U)if and only if fκ−1

i ∈ Oκi(U) for all i. We shall see shortly that this structure isindependent of the coordinate system (X0, . . . , Xn) and that P is in fact a k-variety.

We could also proceed in the opposite direction and start with an affine spaceand realize it as the hyperplane complement of a projective space. Changing ourpoint of view accordingly, we might say that a projective space arises as a ‘comple-tion with points at infinity’ of a given affine space. For instance, An embeds in Pnby (x1, . . . , xn) 7→ [1 : x1 : · · · : xn] with image the hyperplane complement definedby X0 6= 0.

2. The Zariski topology on a projective space

We begin with giving a simpler, more classical characterization of the Zariskitopology on a projective space.

Let P be a projective space of dimension n over k and let [X0 : · · · : Xn] be a ho-mogeneous coordinate system for P . Suppose F ∈ k[X0, . . . , Xn] is homogeneousof degree d so that F (tX0, . . . , tXn) = tdF (X0, . . . , Xn) for λ ∈ k. The property ofthis being zero only depends on [X0 : · · · : Xn] and hence the zero set of F definesa subset ZF ⊂ P (even though F is not a function on P ). We denote its nonzeroset by UF := P − ZF .

PROPOSITION 2.1. The collection UF , where F runs over the homogeneous poly-nomials in k[X0, . . . , Xn], is a basis for the Zariski topology on P . This topology isindependent of the choice of our homogeneous coordinate system [X0, . . . , Xn] and ev-ery linear chart is a homeomorphism that identifies the sheaf of regular functions onits domain with OAn .

PROOF. That the collection UF F is a basis of a topology follows from the ob-vious equality UF ∩UF ′ = UFF ′ . The independence of the coordinate choice resultsfrom the observation that under a linear substitution a homogeneous polynomialtransforms into a homogeneous polynomial (of the same degree).

Let now U = P − H be a hyperplane complement. Choose a homogeneouscoordinate system [Y0, . . . , Yn] such that U is defined by Y0 6= 0 and denote byj : An ∼= U , (y1, . . . , yn) 7→ [1 : y1 : · · · : yn] the inverse. If F ∈ k[Y0, . . . , Yn] ishomogeneous, then j−1UF = U(f), where f(y1, . . . , yn) := F (1, y1, . . . , yn). So theinclusion j is continuous. Conversely, if f ∈ k[y1, . . . , yn] is nonzero of degree d,then its ‘homogenization’ F (Y0, . . . , Yn) := Y d0 f(Y1/Y0, . . . , Yn/Y0) is homogeneousof degree d and U(f) = j−1UF . So j is also open.

By letting (U, j−1) run over the charts (Uν , κν), we see that have thus recoveredthe Zariski topology. The last statement follows from the fact that each κνj is anisomorphism between two open subsets of An.

EXERCISE 57. Let 0 6= F ∈ k[X0, . . . , Xn] be homogeneous of degree d.(a) Prove that every function UF → k of the form G/F r with r ≥ 0 and G

homogeneous of the same degree as F r is regular.(b) Prove that conversely every regular function on UF is of this form.

We now discuss the projective analogue of the (affine) I ↔ Z correspondenceand at the same time clarify what a nonclosed point is. Let us denote for an inte-ger d ≥ 0 by k[X0, . . . , Xn]d the set of F ∈ k[X0, . . . , Xn] that are homogeneousof degree d. It is clear that k[X0, . . . , Xn] is the direct sum of the k[X0, . . . , Xn]d,


d = 0, 1, . . . and that k[X0, . . . , Xn]d · k[X0, . . . , Xn]e = k[X0, . . . , Xn]d+e. A goodway to understand the decomposition of k[X0, . . . , Xn] into its homogeneous sum-mands is in terms of the action in An+1 defined by scalar multiplication: if F ∈k[X0, . . . , Xn]d, then for every t ∈ k, we have F (tX0, . . . , tXn) = tdF (X0, . . . , Xn).So this decomposition is in fact the eigenspace decomposition for this action ofthe multiplicative group k× on k[X0, . . . , Xn]. (Indeed, this simple decomposi-tion should be understood as the algebraic expression of that action.) This impliesamong other things that for any F ∈ k[X0, . . . , Xn]d the zero set Z(F ) ⊂ An+1 =kn+1 is invariant under scalar multiplication. This is still true for an intersectionof such zero sets, in other words, if I ⊂ k[X0, . . . , Xn] is an ideal generated byhomogeneous polynomials of positive degree, then Z(I) ⊂ An+1 is invariant underscalar multiplication. Such a closed subset is called an affine cone; the origin iscalled the vertex of that cone. Observe that since we insisted that the degrees of thehomogeneous generators of I be positive, we have I ⊂ (X0, . . . , Xn), and so Z(I)will always contain the vertex 0.

If Y ⊂ Pn is closed, Y can be written as an intersection ∩αZFα , where eachFα is homogeneous. The common zero set of the Fα in An+1, ∩αZ(Fα), is thecone that as a set is just the union of the 1-dimensional linear subspaces of kn+1

parameterized by Y . We denote this affine cone by Cone(Y ).Given a subset Y ⊂ Pn, then for d ≥ 1, we denote by IY,d the set of F ∈

k[X0, . . . , Xn]d with Y ⊂ ZF (we put IY,0 = 0). This is a vector space and we haveIY,d · k[X0, . . . , Xn]e ⊂ IY,d+e. So the direct sum of the IY,d, d = 1, 2, . . . , denotedby IY ⊂ k[X0, . . . , Xn], is an ideal of k[X0, . . . , Xn]. Notice that with this definition,I∅ is the ideal generated by X0, . . . , Xn.

The k-algebra k[X0, . . . , Xn] is graded and the ideal IY is homogeneous in thesense below.

DEFINITION 2.2. A (nonnegatively) graded ring is a ring R whose underlyingadditive group comes with a direct sum decomposition R• = ⊕∞k=0Rd such that thering product maps Rd × Re in Rd+e, or equivalently, is such that

∑∞d=0Rdt

d is asubring of R[t]. An ideal I of such a ring is said to be homogeneous if it is the directsum of its homogeneous parts Id := I ∩Rd.

If R• is a graded ring, then clearly R0 is a subring of R. If I• is a homogeneousideal, then R/I = ⊕∞d=0Rd/Id is again a graded ring. Of special interest is thehomogeneous ideal R+ := ⊕∞k=1Rd, for which we have R/R+ = R0. We will bemostly concerned with the case when R0 = k, so that R+ is then a maximal ideal(and the only one that is homogeneous).

LEMMA 2.3. If I, J are homogeneous ideals of a graded ring R•, then so are I∩J ,IJ , I + J and

√I. Moreover, a minimal prime ideal of R• is a graded ideal.

PROOF. The proofs of the first statement are not difficult. We omit them.As to the last, it suffices to show that if p ⊂ R• is a prime ideal, then the direct

sum of its homogeneous parts, p• := ⊕n(p ∩ Rn) is also a prime ideal. Indeed,suppose r, s ∈ R are such that rs ∈ p• and neither r nor s lies in p•. Write r and sas a sum of its homogenous parts and let rk ∈ Rk resp. sl ∈ Rl be the lowest degreepart or r resp. s that is not in p. Then it is easily seen that rksl ∈ p and so rk ∈ p orsl ∈ p and we get a contradiction.

LEMMA 2.4. If X ⊂ An+1 is an affine cone (with vertex 0), then I(X) is ahomogeneous ideal.

2. THE ZARISKI TOPOLOGY ON A PROJECTIVE SPACE 55

PROOF. Let F ∈ I(X). We must show that each homogeneous componentof F lies in I(X). As X is invariant under scalar multiplication, the polynomialF (tX0, . . . , tXn) =

∑d≥1 t

dFd(X0, . . . , Xn) (as an element of k[x1, . . . , xn][t]) van-ishes on X×A1 in An+1×A1. Clearly the zero set of I(X)[t] is X×A1 ⊂ An+1×A1

and since the quotient is k[x1, . . . , xn][t]/I(X)[t] = A(X)[t] is reduced, we haveI(X × A1) = I(X)[t]. It follows that Fd ∈ I(X) for all d.

The ideal I(X) is of course a radical ideal and so we find:

COROLLARY 2.5. The maps Y 7→ Cone(Y ) and Y 7→ IY set up bijections between(i) the collection of closed subsets of Pn, (ii) the collection of affine cones in An+1

and (iii) the collection of homogeneous radical ideals of k[X0, . . . , Xn] contained in(X0, . . . , Xn).

Note that here Y = ∅ ↔ 0 ⊂ An+1 ↔ the homogeneous ideal (X0, . . . , Xn).The bijections above restrict to bijections between (i) the collection of irreduciblesubsets of Pn (which we can also think of as Pn, its nonclosed points included), (ii)the collection of irreducible affine cones in An+1 strictly containing the vertex 0 and(iii) the collection of homogeneous prime ideals of k[X0, . . . , Xn] strictly containedin (X0, . . . , Xn).

DEFINITION 2.6. The homogeneous coordinate ring of a closed subset Y of Pn isthe graded ring S(Y )• := k[X0, . . . , Xn]/IY , S(Y )d = k[X0, . . . , Xn]d/IY,d.

Notice that if we ignore the grading of S(Y )•, then we just get the coordinatering of the affine cone over Y , A(Cone(Y )).

EXERCISE 58. Let R• be a graded ring.(b) Prove that if I is a prime ideal in the homogeneous sense: if rs ∈ I for

some r ∈ Rk, s ∈ Rl implies r ∈ I or s ∈ I, then I is a prime ideal.(c) Prove that the intersection of all homogeneous prime ideals of R• is its

ideal of nilpotents.

EXERCISE 59. Prove that a closed subset Y ⊂ Pn is irreducible if and only if IYis a prime ideal.

Since k[X0, . . . , Xn] is a noetherian ring, any ascending chain of homogeneousideals in this ring stabilizes. This implies that any projective space is noetherian(and according to Exercise 53 Pb, as any prevariety, is noetherian). In particular,every subset of a projective space has a finite number of irreducible componentswhose union is all of that subset.

EXERCISE 60. Let S• be a graded k-algebra that is reduced, finitely generatedand has S0 = k.

(a) Prove that S• is as a graded k-algebra isomorphic to the homogeneouscoordinate ring of a closed subset Y .

(b) Prove that under such an isomorphism, the homogeneous radical idealscontained in the maximal ideal S+ := ⊕d≥1Sd correspond to closed sub-sets of Y under an inclusion reversing bijection: homogeneous idealsstrictly contained in S+ and maximal for that property correspond topoints of Y .

(c) Suppose S• a domain. Show that a fraction F/G that is homogeneous ofdegree zero (F ∈ Sd and 0 6= G ∈ Sd for some d) defines a function onUg.


EXERCISE 61. Let Y be an affine variety.(a) Show that a homogeneous element of the graded ring A(Y )[X0, . . . , Xn]

defines a closed subset of Y × Pn as its zero set.(b) Prove that every closed subset of Y ×Pn is an intersection of finitely many

zero set of homogeneous elements of A(Y )[X0, . . . , Xn].(c) Prove that we have a bijective correspondence between closed subsets of

Y × Pn and homogeneous radical ideals in A(Y )[X0, . . . , Xn].

3. The Segre embeddings

First we show how a product of projective spaces can be realized as a closedsubset of a projective space. This will imply among other things that a projectivespace is a variety. Consider the projective spaces Pm and Pn with their homoge-neous coordinate systems [X0 : · · · : Xm] and [Y0 : · · · : Yn]. We also consider aprojective space whose homogeneous coordinate system is the set of matrix coeffi-cients of an (m+ 1)× (n+ 1)-matrix [Z00 : · · ·Zij : · · · : Zmn]; this is just Pmn+m+n

with an unusual indexing of its homogeneous coordinates.

PROPOSITION 3.1 (The Segre embedding). The map f : Pm × Pn → Pmn+m+n

defined by Zij = XiYj , i = 0, . . . ,m; j = 0, . . . , n is an isomorphism onto a closedsubset of Pmn+m+n. If m = n, then the diagonal of Pm × Pm is the preimage of thelinear subspace of Pm2+2m defined by Zij = Zji and hence is closed in Pm × Pm.

PROOF. In order to prove that f defines an isomorphism onto a closed subsetof Pmn+m+n, it is enough to show that for every chart domain Uij of the standardatlas of Pmn+m+n, f−1Uij is open in Pm × Pn and is mapped by f isomorphicallyonto a closed subset of Uij . For this purpose we may (simply by renumbering)assume that i = j = 0. So then U00 ⊂ Pmn+m+n is defined by Z00 6= 0 andis parametrized by the coordinates zij := Zij/Z00, (i, j) 6= (0, 0). It is clear thatf−1U00 is defined by X0Y0 6= 0. This is just UX0

× UY0and hence parametrized by

x1 := X1/X0, . . . , xm := Xm/X0 and y1 := Y1/Y0, . . . , yn := Yn/Y0. In terms ofthese coordinates, f : f−1U00 → U00 is given by zij = xiyj , where (i, j) 6= (0, 0)and where we should read 1 for x0 and y0 (so that zi0 = xi and z0j = yj). It is nowclear that the image is in fact the graph of the morphism Am × An → Amn withcoordinates xiyj , 1 ≤ i ≤ m, 1 ≤ j ≤ n. This graph is closed in Am×An×Amn andisomorphic to Am × An. So f indeed restricts to an isomorphism of f−1U00 onto aclosed subset of U00.

In case m = n, we must also show that the condition XiYj = XjYi for 0 ≤ i <j ≤ m implies that [X0 : · · · : Xm] = [Y0 : · · · : Ym], assuming that not all Xi resp.Yj are zero. Suppose Xi 6= 0 and put t := Yi/Xi ∈ k×. Then XiYj = XjYi impliesYj = tXj for all j. So t 6= 0 and [Y0 : · · · : Ym] = [X0 : · · · : Xm].

COROLLARY 3.2. A projective space over k is a variety.

PROOF. Proposition 3.1 shows that the diagonal of Pm × Pm is closed.

DEFINITION 3.3. A variety is said to be projective if it is isomorphic to a closedirreducible subset of some projective space. A variety is called quasi-projective if isisomorphic to an open subset of some projective variety.

COROLLARY 3.4. Every irreducible closed (resp. locally closed) subset of Pn is aprojective (resp. quasi-projective) variety. The collection of projective (resp. quasi-projective) varieties is closed under a product.

4. PROJECTIONS 57

PROOF. The first statement follows from 10.9 and the second from Proposition3.1.

EXERCISE 62. (a) Prove that the image of the Segre embedding is thecommon zero set of the homogenenous polynomials ZijZkl − ZilZkj .

(b) Show that for every (p, q) ∈ Pm×Pn the image of p×Pn and Pm×qin Pmn+m+n is a linear subspace.

(c) Prove that the map Pn → P(n2+3n)/2 defined by Zij = XiXj , 0 ≤ i ≤ j ≤n is an isomorphism on a closed subset defined by quadratic equations.Find these equations for n = 2.

(d) As a special case we find that the quadric hypersurface in P3 defined byZ0Z1 − Z2Z3 = 0 is isomorphic to P1 × P1. Identify in this case the twosystems of lines on this quadric.

EXERCISE 63 (Intrinsic Segre embedding). Let V and W be finite dimensionalk-vector spaces. Describe the Segre embedding for P(V ) × P(W ) intrinsically as amorphism P(V )× P(W )→ P(V ⊗W ).

4. Projections

Let V be a finite dimensional k-vector space. If W ⊂ V is a linear subspace,then we can form the quotient vector space V/W and we have linear surjectionπ : V → V/W . We cannot projectivize this as a morphism from P(V ) to P(V/W ),but we shall see that it defines at least a morphism Pπ : P(V ) − P(W ) → P(V/W )(which then determines a rational map P(V ) 99K P(V/W ) in case W 6= V ). Asa map, Pπ is defined as follows: for p ∈ P(V ) − P(W ), `p is a one dimensionalsubspace of V not contained in W so that π(`p) is a one dimensional subspace ofV/W ; we then put Pπ(p) := [π(`p)]. To see that this is morphism, let n := dimP(V ),m := dimP(V/W ) (so that dimV = n + 1 and dimW = n − m) and choose acoordinate system (X0, . . . , Xn) for V such that W is given by X0 = · · · = Xm = 0.Then (X0, . . . , Xm) serves as a coordinate system for V/W and π : V → V/W issimply given by (X0, . . . , Xn) → (X0, . . . , Xm). In projective coordinates this is ofcourse given by [X0 : · · · : Xn] → [X0 : · · · : Xm] and so is indeed a morphismwhere it makes sense, namely where X0, . . . , Xm not all vanish, that is, on thecomplement of P(W ).

DEFINITION-LEMMA 4.1. The blowup of P(V ) along P(W ), denoted BlP(W ) P(V ),is the closure of the graph of Pπ in P(V )× P(V/W ) (hence is a projective variety). Itenjoys the following properties:

The projection on the first factor, p1 : BlP(W ) P(V ) → P(V ) is an isomorphismover P(V )− P(W ) whereas the preimage over P(W ) is P(W )× P(V/W ).

The projection to the second factor p2 : BlP(W ) P(V ) → P(V/W ) is a locallytrivial bundle of projective spaces of dimension dimW , to be precise, if U ⊂ P(V/W )is a hyperplane complement (hence an affine space), then there exists an isomorphismp−1

2 U ∼= Pn−m × U whose second component is given by p2.

PROOF. We use a homogeneous coordinate system [X0 : · · · : Xn] for P(V )as above (so that P(W ) is given by X0 = · · · = Xm = 0). If we denote thecorresponding coordinate system for P(V/W ) by [Y0 : · · · : Ym], then the graph ofPπ in P(V ) × P(V/W ) is given by the pairs ([X0 : · · · : Xn], [Y0 : · · · : Ym]) with[X0 : · · · : Xm] proportional to [Y0 : · · · : Ym] and (X0, . . . , Xm) 6= (0, . . . , 0). The


proportionality property is equivalent to: XiYj = XjYi for all 0 ≤ i < j ≤ m. LetZ be the subset of ⊂ P(V ) × P(V/W ) defined by these equations: XiYj = XjYi,0 ≤ i < j ≤ m. In terms of the Segre embedding of Pn × Pm → Pnm+n+m thisamounts to Zij = Zji in this range and so Z is a closed subset of P(V )× P(V/W ).The equations defining Z are trivially satisfied when (X0, . . . , Xm) = (0, . . . , 0) andso Z contains P(W ) × P(V/W ). We have just shown that the graph of Pπ equalsthe complement of P(W )× P(V/W ) in Z.

Let us now see what the projection Z → P(V/W ) is like over the open subsetUY0 of P(V/W ) defined by Y0 6= 0. We parametrize UY0 by Am via (y1, . . . , ym) ∈Am 7→ [1 : y1 : · · · : ym]. Then its preimage ZUY0 := Z ∩ (P(V ) × UY0

) in Z isparametrized by Pn−m × UY0

by means of the morphism

([X0 : Xm+1 : · · · : Xn], [1 : y1 : · · · : yn]) ∈ Pn−m × UY0 7→([X0 : X0y1 : · · · : X0ym : Xm+1 : · · · : Xn], [1 : y1 : · · · : yn]) ∈ ZUY0 .

This is an isomorphism (the inverse is obvious) which commutes with the projectionon UY0

. Notice that it makes ZUY0 ∩(P(V )×P(V/W )) correspond to the locus whereX0 = 0, which is the product of a hyperplane in Pn−m times UY0 . In particular,ZUY0 ∩ (P(V )× P(V/W )) lies in the closure of the graph of Pπ. This remains true,of course, if we replace UY0

by UYi , i = 1, . . . ,m, or by any other hyperplanecomplement in P(V/W ). It follows that Z is the closure of the graph of Pπ and thatZ enjoys the stated properties.

It is worthwhile to describe this map in purely projective terms. This starts withthe observation that the π-preimage of a one-dimensional linear subspace of V/Wis a linear subspace of V which contains W as a hyperplane and that every suchsubspace so arises. So we can think of P(V/W ) as parameterizing the projectivelinear subspaces of P(V ) that contain P(W ) as a projective hyperplane.

EXERCISE 64. Prove that the projective structure on this collection of linearsubspaces is intrinsic: prove that if P is a projective space and Q ⊂ P is a projectivelinear subspace of codimension m > 0, then the collection of projective linearsubspaces Q ⊂ P which contain Q as a projective hyperplane is in a natural mannera projective space (denoted here by PQ) of dimension m−1. Show that the blowupBlQ P of P along Q can be identified with the set of pairs (p, [Q]) ∈ P × PQ withp ∈ Q.

5. Elimination theory and closed projections

Within a category of reasonable topological spaces (say, the locally compactHausdorff spaces), the compact ones can be characterized as follows: K is compactif and only if the projection K × X → X is closed for every space X in that cate-gory. In this sense the following theorem states a kind of compactness property forprojective varieties.

THEOREM 5.1. If U is a variety, then the projection πU : Pn × U → U isclosed. Moreover, if Z ⊂ Pn × U is closed and irreducible, then there exists anopen-dense subset Y ′ ⊂ πU (Z) such that for every closed point y ∈ Y ′ we havedimZ = dimπU (Z) + dimZy, where Zy ⊂ Pn denotes the fiber πU |Z over y.

We derive this theorem from the main theorem of elimination theory, which wewill state and prove first.

5. ELIMINATION THEORY AND CLOSED PROJECTIONS 59

Given an integer d ≥ 0, let us write Vd for k[X0, X1]d, the k-vector space ofhomogeneous polynomials in k[X0, X1] of degree d. The monomials (Xd−i

0 Xi1)di=0

form a basis, in particular, dimVd = d+ 1. Given F ∈ Vm and G ∈ Vn, then

uF,G : Vn−1 ⊕ Vm−1 → Vn+m−1, (A,B) 7→ AF +BG

is a linear map between two k-vector spaces of the same dimension m + n. Theresultant R(F,G) of F and G is defined as the determinant of this linear map withrespect to the monomial bases of the summands of Vn−1 ⊕ Vm−1 and of Vn+m−1.So R(F,G) = 0 if and only if uF,G fails to be injective.

Notice that R(F,G) is a polynomial in the coefficients of F and G:if F =

∑mi=0 aiX

m−i0 Xi

1 and G =∑nj=0 biX

n−i0 Xi

1, then

R(F,G) = det

a0 a1 · · · am 0 · · · · · · 00 a0 a1 · · · am · · · · · · 00 0 a0 · · · 0

· · ·0 0 · · · 0 a0 a1 · · · amb0 b1 · · · · · · bn 0 · · · 00 b0 b1 · · · · · · bn 0 · · · 00 0 b0 · · · 0

· · ·0 · · · 0 b0 b1 · · · · · · bn

So the resultant defines an element of A(Vm × Vn) = A(Vm)⊗A(Vn).

LEMMA 5.2. R(F,G) = 0 if and only if F and G have a common zero in P1.

PROOF. If R(F,G) = 0, then uF,G is not injective, so that there exist a nonzero(A,B) ∈ Vn−1 ⊕ Vm−1 with AF + BG = 0. Suppose that B 6= 0. It is clear thatF divides BG. Since deg(B) = m − 1 < m = degF , it follows that F and G musthave a common factor.

If conversely F and G have a common zero in P1, then they must have acommon linear factor L, say: F = LF1, G = LG1 and we see that (G1,−F1) ∈Vn−1 ⊕ Vm−1 is nonzero and in the kernel of uF,G.

PROOF OF THEOREM 5.1. Let Z ⊂ Pn × U . It is clear that πU (Z) is closedin U if for every open affine subset U ′ ⊂ U , πU (Z) ∩ U ′ is closed in U ′. SinceπU (Z) ∩ U ′ = πU ′(Z ∩ (Pn × U ′)) we may (and will) assume that is U affine.

We first treat the case n = 1 and then proceed with induction on n.Let Z ⊂ P1 × U be closed. Denote by IZ,• the homogeneous ideal in the

graded algebra A(U)[X0, X1] of functions vanishing on Z. Then Z is the commonzero set of the members of IZ,• (see Exercise 61). For every homogeneous pairF,G ∈ ∪mA(U)[X0, X1]m, we can form the resultant R(F,G) ∈ A(U). We claimthat πU (Z) is the common zero set Z(R) ⊂ U of the set of resultants R(F,G) ofpairs of homogeneous forms F,G taken in ∪mIZ,m, hence is closed in U .

Suppose that y ∈ πU (Z) so that (y, p) ∈ Z for some p ∈ P1. Then p is acommon zero of each pair Fy, Gy, where F,G ∈ ∪mIZ,m and the subscript y refersto substituting y for the first argument. So R(F,G) vanishes in y. It follows thaty ∈ Z(R).

Next we show that if y /∈ πU (Z), then y /∈ Z(R). Since y×P1 is not containedin Z, there exists an integer m > 0 and a F ∈ IZ,m with Fy 6= 0. Denote by


p1, . . . , pr ∈ P1 the distinct zeroes of Fy. We show that there exists a G ∈ IZ,n forsome n such that Gy does not vanish in any pi. This suffices, for this means thatR(Fy, Gy) 6= 0 and so y /∈ Z(R). To this end, we note that for any given 1 ≤ i ≤ r,Z⋃∪j 6=i(y, pj) is closed in U × P1, so that there will exist a G(i) ∈ ∪mIZ,m with

G(i)y zero in all the pj with j 6= i, but nonzero in pi. Upon replacing each G(i) by

some positive power of it , we may assume that G(1), . . . , G(r) all have the samedegree n, say. Then G := G(1) + · · ·+G(r) ∈ IZ,n and Gy(pi) = G(i)(pi) 6= 0.

In order to prove the dimension assertion, we assume Z irreducible so thatY := πU (Z) is irreducible and closed. We show that either Z = P1 × Y (so thatdimZ = dimY +1) or that there exists an open-dense Y ′ ⊂ Y such that ZY ′ := Z∩(P1 × Y ′)→ Y ′ is a finite morphism (so that dimY = dimY ′ = dimZY ′ = dimZ).

Consider the ideal in A(U) generated by the coefficients of all the members ofIZ ⊂ A(U)[X0, X1] and denote by Y1 ⊂ U its zero set. This is closed subset ofY consisting of the y ∈ U with the property that P1 × y ⊂ Z. If Z1 = Y , thenZ = P1×Y . Assume therefore that Z1 is a proper closed subset of Y . Choose p ∈ P1

such that p×Y is not contained in Z. By adapting the coordinates in P1, we mayassume that p =∞. Let Y2 := πU (∞×U ∩Z). This is also a proper closed subsetof Y and so Y contains an an affine open-dense subset Y ′ ⊂ Y − Y1 − Y2 such thatZY ′ is contained in A1 × Y ′ and has finite fibers over Y ′. Then ZY ′ can be given asa subset of A1×Y ′ by a polynomial in A(Y ′)[t]. After possible shrinking Y ′, we canassume that the top coefficient of this polynomial is a unit in A(Y ′). Then ZY ′ willbe given by a monic polynomial and hence ZY ′ → Y ′ is a finite morphism.

Now assume n ≥ 2. Let q = [0 : · · · : 0 : 1] and consider the blowup Pn :=

Blq Pn → Pn. Recall that an element of Pn is the set of pairs in ([X0 : · · · :

Xn], [Y0 : · · · : Yn−1]) in Pn × Pn−1 with [X0 : · · · : Xn−1] = [Y0 : · · · : Yn−1]. Wehave seen that over the open subset UYi ⊂ Pn−1 defined by Yi 6= 0, the projectionPn → Pn−1 is isomorphic to the projection P1 × UYi → UYi . Hence the projectionπ1 : Pn × U → Pn−1 × U is over UYi × U like P1 × UYi × U → UYi × U . So thisprojection is closed over UYi × U . It follows that the projection π1 is closed. Thepreimage Z of Z under the projection Pn × U → Pn × U is closed and by whatwe just proved, π1(Z) is closed in Pn−1 × U . By induction, the image of the latterunder the projection π2 : Pn−1 × U → U is closed. But this is just πU (Z).

For the dimension assertion, we may without loss of generality assume that Zis not contained in q × U (otherwise choose a different q) and that πU (Z) = U

(otherwise replace U by π(Z)). We then replace Z by the closure of Z ∩ (Pn −q)×U in Pn×U so that Z is irreducible, is of the same dimension as Z and mapsalso onto U . It therefore suffices to prove the dimension assertion for the projectionπ : Z → U . We put W := π1(Z) ⊂ Pn−1 × U .

The case n = 1 treated above shows that for the projection π1 : Pn → Pn−1×Ueither Z = π−1

1 W (so that Z is a P1-bundle over W and dim Z = dimW + 1), ordim Z = dimW and Z is finite over W − V ⊂ W where V ⊂ W is a proper closedsubset. In that last case, we note that for every closed irreducible subset C ⊂ Wwhich is not contained in V , we have dim(π−1

1 C ∩ Z) = dimC: this is clear foran irreducible component of π−1

1 C ∩ Z which meets π−1(W − V ) and any otherirreducible component will be contained in a P1-bundle over C ∩ V and hence ofdimension 1 + dim(C ∩ V ) ≤ dimC.

5. ELIMINATION THEORY AND CLOSED PROJECTIONS 61

If π2 : Pn−1 × U → U is the projection, then (by induction) there exists anopen-dense subset U ′ ⊂ U such that the closed fibers of W → U resp. V → U allhave dimension dimW −dimU resp. have dimension smaller than dimW −dimU .So all the closed fibers of Z → U over U ′ have dimension dim Z − dimU .

We mention a few corollaries.

COROLLARY 5.3. Let X be a projective variety. Then any morphism from X to avariety is closed (and hence has closed image).

PROOF. Assume that X is closed in Pn. A morphism f : X → Y to a varietyY can be factored as the obvious isomorphism of X onto the graph Γf of f , theinclusion of this graph in X×Y (which is evidently closed), the inclusion of X×Yin Pn×Y (which is closed sinceX ⊂ Pn is closed) and the projection onto Y (whichis closed by Theorem 5.1). So f is closed.

It is an elementary result from complex function theory (based on Liouville’sthoerem) that a holomorphic function on the Riemann sphere is constant. Thisimplies the corresponding assertion for holomorphic functions on complex projec-tive n-space PnC (to see that a holomorphic function on PnC takes the same value onany two distinct points, simply apply the previous remark to its restriction to thecomplex projective line passing through them, viewed as a copy of the Riemannsphere). The following corollary is an algebraic version of this fact.

COROLLARY 5.4. Let X be a projective variety. Then any morphism from X to aquasi-affine variety is constant. In particular, any regular function on X is constant.

PROOF. If f : X → Y is a morphism to a quasi-affine variety Y , then its com-posite with an embedding of Y in some affine space An is given by n regular func-tions on X. So it suffices to prove the special case when Y = A1. By the previouscorollary this image is closed in A1. But if we think of f as taking its values in P1

(via the embedding y ∈ A1 7→ [1 : y] ∈ P1), then we see that f(X) is also closedin P1. So f(X) cannot be all of A1. Since X is irreducible, so is the image and itfollows that f(X) is a singleton. In other words, f is constant.

PROPOSITION 5.5. Let Z ⊂ Pn an irreducible and closed subset. If Q ⊂ Pn is alinear subspace with the property that Q ∩ Z = ∅, then dimZ + dimQ < n and theprojection π : Pn → PnQ maps Z onto an irreducible and closed subset π(Z) with theproperty that every fiber of Z → π(Z) is finite.

PROOF. Every fiber of Z → π(Z) is the intersection of Z with a linear subspaceQ ⊂ Pn which contains Q a hyperplane. It is closed in Q and hence this fiber isprojective (or strictly speaking, every irreducible component of that fiber is). On theother hand it is contained in the hyperplane complement Q−Q, which is affine. Sothe fiber is also affine. Then Corollary 5.4 implies that each irreducible componentof this fiber is a singleton. Hence the fiber is finite.

It follows from Theorem 5.1 that dimZ = dimπ(Z) ≤ dimPnQ−dimQ− 1.

EXERCISE 65. Let P be a projective space of dimension n.(a) The dual P of P is by definition the collection of hyperplanes in P . Prove

that P has a natural structure of a projective space.(b) Identify the double dual of P with P itself.


(c) The incidence locus I ⊂ P × P is the set of pairs (p, q) ∈ P × P with theproperty that p lies in the hyperplane Hq defined by q. Prove that I is anonsingular variety of dimension 2n− 1.

(d) Show that we can find homogeneous coordinates [Z0 : · · · : Zn] for P and[W0 : · · · : Wn] for P such that I is given by

∑ni=0 ZiWi = 0.

EXERCISE 66. Let F ∈ k[X0, . . . , Xn]d define a nonsingular hypersurface H inPn. Prove that the map H → Pn which assigns to p ∈ H the projectived tangentspace of H at p is given by [ ∂F∂Z0

: · · · : ∂F∂Zn

]. Prove that the image of this map isclosed in Pn (this image is called the dual of H). What can you say in case d = 2?

6. Constructible sets

We now ask about the image of an arbitrary morphism of varieties. This neednot be a variety as the following simple example shows.

EXAMPLE 6.1. Consider the morphism f : A2 → A2, (x1, x2) 7→ (x1, x1x2). Apoint (y1, y2) ∈ A2 is in the image of f if and only if the equation y2 = y1x2 has asolution in x2. This is the case precisely when y1 6= 0 or when y1 = y2 = 0. So theimage of f is the union of the open subset y1 6= 0 and the singleton (0, 0). This isnot a locally closed subset, but the union of two such.

DEFINITION 6.2. A subset of variety is called constructible if it can be written asthe union of finitely many (locally closed) subvarieties.

THEOREM 6.3. Let f : X → Y be a morphism of varieties. Then f takesconstructible subsets of X to constructible subsets of Y . In particular, f(X) is con-structible.

We first show that the theorem follows from

PROPOSITION 6.4. Let X be an irreducible variety and let f : X → Y be amorphism of varieties. Then f(X) contains a nonempty open subset of its closure (inother words, f(X) contains a locally closed subvariety of Y which is dense in f(X)).

PROOF THAT PROPOSITION 6.4 IMPLIES THEOREM 6.3. A constructible subset isa finite union of irreducible subvarieties and so it is clearly enough to prove that theimage of each of these is constructible. In other words, it suffices to show that theimage of a morphism f : X → Y of varieties with X irreducible is constructible.We prove this with induction on the dimension of X. According to Proposition6.4 the closure of f(X) contains a nonempty open subset U such that f(X) ⊃ U .It is clear that f−1U is a nonempty open subset of X. If Z := X − f−1U , thenf(X) = U ∪ f(Z). The irreducible components of Z have smaller dimension thanX and so f(Z) is constructible by induction.

PROOF OF PROPOSITION 6.4. Since X is covered by finitely many of its openaffine subsets, we may without loss of generality assume that X is affine. So wecan assume that X is closed in some An. Then the graph of f identifies X with aclosed subset of An × Y and so we see that we only need to prove that for everyclosed subset X ⊂ An × Y , πY (X) contains a nonempty open subset of πY (X).Since we can factor πY in an obvious manner into successive line projections (byforgetting the last coordinate)

An × Y → An−1 × Y → · · · → A1 × Y → Y,

7. THE VERONESE EMBEDDING 63

it suffices to do the case n = 1. Upon replacing Y by πY (X), we are now left toshow that if X ⊂ A1 × Y is closed and such that πY (X) is dense in Y , then πY (X)contains a nonempty open subset of Y .

Since X is irreducible, so is its image πY (X) and hence also Y . Upon replacingY be a nonempty affine open subset and X by the preimage of that open subset,we may also assume that Y is affine.

When X = A1 × Y , there is nothing to show. Otherwise X is contained in ahypersurface Z(f) ⊂ A1 × Y with equation f ∈ A(Y )[x], f(x, y) =

∑Ni=0 ai(y)xi,

with ai ∈ A(Y ) and aN not identically zero. Next, by replacing Y by the nonzeroset U(aN ) and X by its preimage, we may assume that aN is nowhere zero. Weprove that then πY (X) = Y . Since is aN now invertible in A(Y ), we may assumethat f is a monic polynomial. Then Z(f) → Y is a finite morphism and henceclosed by Exercise 37. Since X is a closed subset of Z(f), it follows that πY (X) isclosed in Y . Since πY (X) is also dense in Y , it follows that πY (X) ⊃ Y .

7. The Veronese embedding

Let be given a positive integer d. We index the monomials in Z0, . . . , Zn thatare homogenous of degree d by their exponents: these are the sequences of non-negative integers d = (d0, . . . , dn) with sum d. They are

(n+dd

)in number. We use

this index set to label the homogeneous coordinates Zd of P(n+dd )−1.

PROPOSITION 7.1 (The Veronese embedding). The map fd : Pn → P(n+dd )−1

defined by Zd = Xd00 · · ·Xdn

n is an isomorphism onto a closed subset of the targetprojective space.

PROOF. In order to prove that fd is an isomorphism onto a closed subset, it isenough to show that for every chart domain Ud of the standard atlas of the targetspace, its preimage f−1

d Ud is open in Pn and is mapped by fd isomorphically onto aclosed subset of Ud. This preimage is defined by Xd0

0 · · ·Xdnn 6= 0. Let us renumber

the coordinates such that d0, . . . , dr are positive and dr+1 = · · · = dn = 0. Thenf−1d Ud = U0 ∩ · · · ∩Ur ⊂ U0. So if we use the standard coordinates (x1, . . . , xn) onU0, then f−1

d Ud is defined by x1 · · ·xr 6= 0.The coordinates on Ud are the functions Ze/Zd with e 6= d. Let us write

ze−d := ze0−d0,e1−d1,...,en−dn for this function. This notation is chosen as to makethe expression fd in terms of these coordinates simply:

fd : U0(x1 · · ·xr) = f−1d Ud → Ud, ze−d = xe1−d11 · · ·xen−dnn (e 6= d).

Since the Laurent monomial xe1−d11 · · ·xen−dnn has degree d0 − e0, the componentsof the above morphism are all the nonconstant Laurent monomials xk11 · · ·xknn withki ≥ −di and of total degree k1 + · · · + kn ≤ d0. In particular, each xi appearsas a component and so does the Laurent monomial x−d11 · · ·x−drr . The graph ofthe regular function xd11 · · ·xdrr on U0(x1 · · ·xr) identifies U0(x1 · · ·xr) with theclosed hypersurface Z = Z(yxd11 · · ·xdrr − 1) in An+1. All other components offd|U0(x1 · · ·xr) : U0(x1 · · ·xr) → Ud are products of x1, . . . , xn, x

−d11 · · ·x−drr and

so its image is the graph of a morphism Z → A(n+dd )−n−2. So we get a closed

embedding.

The following proposition is remarkable for its repercussions in intersectiontheory.


PROPOSITION 7.2. Let H ⊂ Pn be a hypersurface. Then Pn −H is affine and forevery closed irreducible subset Z ⊂ Pn of positive dimension, Z ∩H is nonempty andof dimension ≥ dim(Z)− 1, with equality holding if Z is not contained in H.

For the proof we need the following complement to Proposition 5.5.

PROPOSITION 7.3. For every proper closed subset Z ( Pn there exists a linearsubspace in P of dimension n− dim(Z)− 1 which misses Z.

PROOF. We may (and will) assume that Z is irreducible. Let i be a nonnegativeinteger < n−dim(Z). We prove with induction on i that Z misses a linear subspaceof dimension i. If i = 0, then we must have dim(Z) < n and so any singleton inPn − Z is as required. When i > 0, there exists by induction hypothesis a linearsubspace Q ⊂ Pn of dimension (i − 1) which does not meet Z. Then projectionfrom Q defines a morphism π : Z → PnQ from Z to a projective space of dimensionn − i. Any fiber of this map is an intersection of Z with a linear subspace in Pn ofdimension i passing through Q. We claim that at least one of them is empty. For ifnot, then π is surjective and hence dominant. In particular, dim(Z) ≥ dim(PnQ) =

n− i, which evidently contradicts our assumption that i < n− dim(Z).

PROOF OF PROPOSITION 7.2. The hypersurface H is given by a homogeneouspolynomial of degree d, say by

∑d cdX

d00 · · ·Xdn

n . This determines a hyperplane

H ⊂ P(n+dd )−1 defined by

∑d cdZd. It is clear that H is the preimage of H under

the Veronese morphism and hence the latter identifies Pn −H with a closed subsetof the affine space P(n+d

d )−1 − H. So Pn −H is affine.For the rest of the argument we may, by passing to the Veronese embedding,

assume that H is a hyperplane. If dim(Z ∩H) ≤ dim(Z) − 2, then by Proposition7.3 there exists a linear subspace Q ⊂ H of dimension dim(H)− (dim(Z∩H)−1 ≤(n − 1) − (dim(Z) − 2) − 1 = n − dim(Z) which avoids Z ∩H. Then Q is a linearsubspace of Pn which avoids Z and we thus contradict Proposition 5.5. This provesthat dim(Z ∩H) ≥ dim(Z)− 1. If Z is irreducible and not contained in H, then wehave also dim(Z ∩H) ≤ dim(Z)− 1.

EXERCISE 67. Let d be a positive integer. The universal hypersurface of degree dis the hypersurface of Pn×P(n+d

d )−1 defined by F (X,Z) :=∑

d ZdXd00 Xd1

1 · · ·Xdnn .

We denote it by H and let π : H → P(n+dd )−1 be the projection.

(a) Prove that H is nonsingular.(b) Prove that projection π is singular at (X,Z) (in the sense that the deriv-

ative of π at (X,Z) is not a surjection) if and only the partial derivativesof FZ ∈ k[X0, . . . , Xn] have X as a common zero.

(c) Suppose from now on that d ≥ 2. Prove that the singular set of π is asmooth subvariety of Pn × P(n+d

d )−1 of codimension n+ 1.(d) Prove that the set of Z ∈ P(n+d

d )−1 over which π has a singular point is ahypersurface. This hypersurface is called the discriminant of π.

(e) For d = 2 we denote the coordinates of P(n+dd )−1 simply by Zij (where it

is understood that Zij = Zji). Prove that the discriminant of π is then thezero set of det(Zij).

8. GRASSMANNIANS 65

8. Grassmannians

Let P be a projective space of dimension n and let d ∈ 0, . . . , n. We wantto show that the collection Grd(P ) of linear d-dimensional subspaces of P is anonsingular projective variety. Let the projective structure on P be defined bythe pair (V, `) so that V is a (n + 1)-dimensional k-vector space and P has beenidentified with P(V ).

LEMMA 8.1. Let Q ⊂ P be a linear subspace of codimension d + 1. Then thecollection of linear d-dimensional subspaces of P contained in P −Q has in a naturalmanner the structure of an affine space AQ of dimension (n− d)(d− 1).

PROOF. Let Q correspond to the linear subspace W ⊂ V of dimension (n +1) − (d + 1) = n − d. Then the elements of AQ correspond to linear subspacesL ⊂ V of dimension d + 1 with L ∩W = 0. This means that V = L ⊕W . Theaffine structure is best seen by fixing some L0 with that property. Then every othersuch L is always the graph of a (unique) linear map L0 → W and vice versa. Sothis identifies AQ with Hom(L0,W ). It is easy to verify that this affine structureis independent of the choices (the translation vector space of AQ is canonicallyidentified with Hom(V/W,W )).

It can now be shown without much difficulty that Grd(P ) admits a uniquestructure of a variety for which a subset as in this lemma is affine open and itsidentification with affine space an isomorphism. We will however proceed in a moredirect manner to show that Grd(P ) admits the structure of a projective variety.

We recall that the exterior algebra∧•

V = ⊕∞p=0

∧pV is the quotient of the

tensor algebra on V , ⊕∞p=0V⊗p (here V ⊗0 = k by convention), by the two-sided

ideal generated by the ‘squares’ v ⊗ v, v ∈ V . It is customary to denote the productby the symbol ∧. So we can characterize

∧•V as (noncommutative) associative

k-algebra with unit element by saying that is generated by the k-vector space Vand is subject to the relations v ∧ v = 0 for all v ∈ V . It is a graded algebra(∧p

V is the image of V ⊗p) and ‘graded-commutative’ in the following sense: ifα ∈

∧pV and β ∈

∧qV , then β ∧ α = (−1)pqα ∧ β. If e0, . . . , en is a basis

for V , then a basis of∧p

V is indexed by the p-element subsets I ⊂ 0, . . . , n:I = 0 ≤ i1 < i2 < · · · < ip ≤ n is associated to the basis element eI = ei1∧· · ·∧eip(where the convention is that e∅ = 1 ∈ k =

∧0V ). So dim

∧pV =

(n+1p

). Notice

that∧n+1

V is one-dimensional and spanned by e0 ∧ · · · ∧ en, whereas∧p

V = 0for p > n + 1. Let us say that α ∈

∧pV is fully decomposable if the exist linearly

independent v1, . . . , vp in V such that α = v1 ∧ · · · ∧ vp. This is equivalent to theexistence of a p-dimensional subspace K ⊂ V such that α is a generator of ∧pK.

LEMMA 8.2. Let α ∈∧p

V be nonzero. Denote by K(α) the set of e ∈ V withe∧α = 0. Then dimK(α) ≤ p and equality holds if and only if α is fully decomposableand spans

∧pK(α).

PROOF. Let e1, . . . , er be a basis of K(α) and let V ′ ⊂ V be a subspace supple-mentary to K(α). Then we have a decomposition

•∧V =

⊕I⊂1,...,r

eI ∧•−|I|∧

V ′.


Wedging with ei kills all the summands in which ei appears and is injective on thesum of the others. So α ⊂ e1 ∧ · · · ∧ er ∧

∧p−rV ′. In particular, r ≤ p with equality

holding if and only if α is a multiple of e1 ∧ · · · ∧ ep.

If L is a linear subspace of V of dimension d + 1, then∧d+1

L is of dimension1 and will be thought of as a one dimensional subspace of

∧d+1V . We thus have

defined map δ : Grd(P ) → P(∧d+1

V ), [L] 7→ [∧d+1

L]. It is called the Pluckerembedding because of:

PROPOSITION 8.3. The map δ : Grd(P ) → P(∧d+1

V ) maps Grd(P ) bijectivelyonto a closed subset of P(

∧d+1V ).

PROOF. Let α ∈∧d+1

V . According to Lemma 8.2 , [α] is in the image of δif and only if K(α) is of dimension d + 1 and if that is the case, then δ−1[α] hasP(K(α)) as its unique element. In particular, δ is injective. Consider the linear map

d+1∧V → Hom(V,

d+2∧V ), α 7→ eα := α∧

The set of linear maps V →∧d+2

V with kernel of dimension ≥ d+ 1 are those ofrank < s := dim(

∧d+2V )−d. If we choose a basis for V , then this locus is given by

set a homogeneous equations in Hom(V,∧d+2

V ), namely the s × s-minors of thecorresponding matrices. Hence the set of α ∈

∧d+1V for which dimK(α) ≥ d+ 1

is also given by a set of homogeneous equations and thus defines a closed subset ofP(∧d+1

V ). This subset is the image of δ.

Proposition 8.3 gives Grd(P ) the structure of projective variety. In order tocomplete the construction, let Q ⊂ P be a linear subspace of codimension d. LetW ⊂ V correspond to Q and choose a generator β ∈

∧n−dW . Then we have a

nonzero linear form

eβ = β∧ :

d+1∧V →

n+1∧V ∼= k.

Its kernel defines a hyperplane in P(∧d+1

V ) and hence an affine subspace Ueβ ⊂P(∧d+1

V ).

LEMMA 8.4. The preimage of Ueβ under the Plucker embedding is the affine spaceAQ and δ maps AQ isomorphically onto its image.

PROOF. If α is the generator of∧d+1

L for some (d+ 1)-dimensional subspace,then L ∩W = 0 if and only if α ∧ β 6= 0: if L ∩W contains a nonzero vector vthen both α and β are divisible by v and so α ∧ β = 0; if L ∩W = 0, then wehave decomposition V = L⊕ V and it is then easily seen (by picking a compatiblebasis of V for example) that α ∧ β 6= 0. This implies that δ−1Ueβ = AQ.

Let us now express the restriction δ : AQ → Ueβ in terms of coordinates.Choose a basis e0, . . . , en for V such that ed+1, . . . , en is a basis for W and β =

ed+1 ∧ · · · ∧ en. Then eβ simply assigns to an element α of∧d+1

V the coefficient ofe0 ∧ · · · ∧ ed in α. If L0 ⊂ V denotes the span of e0, . . . , ed, then AQ ∼= Hom(L0,W )

is identified with the affine space A(d+1)×(n−d) of (d+ 1)× (n− d)-matrices via

(aji )0≤i≤d<j≤n 7→ k-span in V of the d+ 1 vectors ei +

n∑j=d+1

ajiejdi=0 ,

8. GRASSMANNIANS 67

so that δ is given by

(aji )0≤i≤d<j≤n 7→ (e0 +

n∑j=d+1

aj0ej) ∧ · · · ∧ (ed +

n∑j=d+1

ajdej).

The coefficient of ei0∧· · ·∧eid is a determinant of which entry is 0, 1 or some aji andhence is a polynomial in the matrix coefficients aji . It follows that this restrictionof δ is a morphism. Among the components of δ we find the matrix coefficientsthemselves: aji appears up to sign as the matrix coefficient of e0∧· · ·∧ei∧· · ·∧ed∧ej .So the image is really a graph of a morphism defined on A(d+1)×(n−d). Hence δmaps AQ isomorphically onto its image in Ueβ .

COROLLARY 8.5. The Plucker embedding realizes Grd(P ) as a nonsingular sub-variety of P(

∧d+1V ) of dimension (n − d)(d + 1). This structure is compatible with

the affine structure that we have on each AQ.

PROOF. Every two open subsets of the form AQ a have nonempty intersectionand so Grd(P ) is irreducible. The rest follows from the previous corollary.

REMARK 8.6. The image of Grd(P ) is a closed orbit of the natural SL(V )-actionon P(

∧d+1V ). It lies in the closure of any other SL(V )-orbit.

EXERCISE 68. Let P be a projective space. Given integers 0 ≤ d < e ≤ dimP ,prove that the set of pairs of linear subspaces R ⊂ Q ⊂ P with dimR = d anddimQ = e is a closed subvariety of Grd(P )×Gre(P ).

The Grassmannian of hyperplanes in a projective space is itself a projectivespace (see Exercise 65). So the simplest example not of this type is the Grassman-nian of lines in a 3-dimensional projective space.

Let V be vector space dimension 4. On the 6-dimensional space∧2

V we havea homogeneous polynomial F :

∧2V → k of degree two defined by

F (α) := α ∧ α ∈4∧V ∼= k

(the last identification is only given up to scalar and so the same is true for F ). Incoordinates F is quite simple: if e1, . . . , e4 is a basis for V , then (ei ∧ ej)1≤i<j≤4 isbasis for

∧2V . So if we label the homogeneous coordinates of P(

∧2V ) accordingly:

[X1,2 : · · · : X3,4], then F is given by

F (X1,2, . . . , X3,4) = X1,2X3,4 −X1,3X2,4 +X1,4X2,3.

Notice that F is irreducible. Its partial derivatives are the coordinates themselves(up to sign and order) and so F defines a smooth quadric hypersurface of dimension4 in a 5-dimensional projective space.

PROPOSITION 8.7. The image of the Plucker embedding of G1(P(V )) in P(∧2

V )is the zero set of F .

PROOF. The image of the Plucker embedding is of dimension 4 and so must bea hypersurface. Since the zero set of F is an irreducible hypersurface, it suffices toshow that the Plucker embedding maps to the zero set of F . For this, let α be agenerator of

∧2L for some linear subspace L ⊂ V of dimension 2. If e1, . . . , e4 is a

basis of V such that α = e1 ∧ e2, then it is clear that α∧α = 0. This proves that thePlucker embedding maps to the zero set of F .


REMARK 8.8. The image of the Plucker embedding Grd(P ) → P(∧d+1V ) is in factalways the common zero set of a collection of quadratic equations, called the Plucker re-lations. To exhibit these, we first recall that every φ ∈ V ∗ defines a linear ‘inner contrac-tion’ map ιφ : ∧•V → ∧•−1V of degree −1 characterized by the fact that for v ∈ V ,ιφ(v) = φ(v) ∈ k = ∧0V and for α ∈ ∧pV, β ∈ ∧•V , ιφ(α∧β) = ιφ(α)∧β+(−1)pα∧ ιφ(β).We then define a linear map B : ∧d+1V ×∧d+1V → ∧d+2V ⊗∧dV as follows: if (e0, . . . , en)is a basis of V and (e∗0, . . . , e

∗n) is the basis of V ∗ dual to (e0, . . . , en), then

B(α, β) :=

n∑r=0

(α ∧ er)⊗ (ιe∗rβ).

It is easy to check that this is independent of the choice of basis. In particular, if α is fullydecomposable, then we can choose (e0, . . . , en) in such a manner that α = e0 ∧ · · · ∧ ed. Wethen find:

B(α, α) =∑r

(e0 ∧ · · · ∧ ed ∧ er)⊗ (ιe∗r e0 ∧ · · · ∧ ed ∧ er) = 0,

because e0 ∧ · · · ∧ ed ∧ er = 0 for r ≤ d and ιe∗r e0 ∧ · · · ∧ ed ∧ er = 0 for r > d. This identitymight be called the universal Plucker relation. One can show that conversely, any α ∈ ∧d+1Vfor which B(α, α) = 0 is either zero or fully decomposable.

Let us rephrase this in terms of algebraic geometry: every nonzero linear form ` on∧d+2V ⊗∧dV , determines a quadratic form Q` on ∧d+1V defined by α 7→ `(B(α, α)) whosezero set is a quadratic hypersurface in P(∧d+1V ). This quadric contains the Plucker locusand the latter is in fact the common zero set of the Q`, with ` running over the linear formson ∧d+2V ⊗ ∧dV . It can be shown that the Q` generate the full graded ideal defined by thePlucker locus.

PROPOSITION-DEFINITION 8.9. LetX be a closed subvariety of the projective spaceP . If d is an integer between 0 and dimP , then the set of d-linear subspaces of Pwhich are contained in X defines a closed subvariety Fd(X) of Grd(P ), called theFano variety (of d-planes) of X.

PROOF. An open affine chart of Grd(P ) is given by a decomposition V = L⊕Wwith dimL = d + 1 and dimW = n − d and is then parametrized by Hom(L,W )by assigning to A ∈ Hom(L,W ) the graph of A. It suffices to prove that via thisidentification Fd(X) defines a closed subset of Hom(L,W ).

Choose homogeneous coordinates [Z0 : · · · : Zn] such that L resp. W is givenby Zd+1 = · · · = Zn = 0 resp. Z0 = · · ·Zd = 0. A linear map A ∈ Hom(L,W )is then given by A∗Zd+i =

∑j=0d aijZj , i = 1, . . . , n − d. It defines an element

of Fd(X) if and only for all G ∈ ∪m≥0Im(X), G(Z0, . . . , Zd, A∗Zd+1, . . . A

∗Zn) isidentically zero. This means that the coefficient of every monomial Zm0

0 · · ·Zmdd

in such an expression much vanish. Since this coefficient is a polynomial in thematrix coefficients aij of A, we find that the preimage of Fd(X) in Hom(L,W ) isthe common zero set of a set of polynomials and hence is closed therein.

EXAMPLE 8.10. Consider the case of a quadratic hypersurfaceX ⊂ P(V ) and as-sume for simplicity that char(k) 6= 2. So in terms of homogeneous coordinates [X0 :· · · : Xn],X can be given by an equation F (X0, . . . , Xn) = 1

2

∑0≤i,j≤n bijXiXj with

bij = bji (so 12bii is the coefficient of X2

i ). In more intrinsic terms: X is given by asymmetric bilinear form B : V × V → k (namely by B(v, v) = 0). Let us assumethat X is nonsingular. This means that the partial derivatives of F have no commonzero in P(V ). This translates into: B : V × V → k is nonsingular, that is, the linearmap b : v ∈ V 7→ B( , v) ∈ V ∗ is an isomorphism of vector spaces. A subspace

8. GRASSMANNIANS 69

L ⊂ V determines an element of the Fano variety of X precisely when B(v, v) = 0for all v ∈ L. This implies that B is identically zero on L × L. So b maps L to(V/L)∗ ⊂ V ∗. Since b is injective, this implies that dimL ≤ dim(V/L), in otherwords that dimL ≤ 1

2 dimV .This condition is optimal. If for instance dimV = 2m + 2 is even, then it

not difficult to show that we can find coordinates (X0, · · · , X2m+1) such that F =∑mi=0XiXm+1+i. Let L resp. L′ be the linear subspace defined by Xm+1 = · · · =

X2m+1 = 0 resp. X0 = · · · = Xm = 0, so that V = L⊕ L′. Notice that both [L] and[L′] are in Fm(X). The vector space Hom(L,L′) describes an affine open subset ofthe Grassmannian of m-planes in P(V ). An element A ∈ Hom(L,L′) is given byA∗Xm+i =

∑mj=0 aijXj , j = 0, . . . ,m. The corresponding m-plane is contained in

X precisely when∑mi,j=0 aijXiXj is identically zero, i.e., if (aij) is antisymmetric.

It follows that [L] ∈ Fm(X) has a neighborhood isomorphic to an affine space ofdimension 1

2m(m+ 1).

EXERCISE 69. Let X be a quadratic hypersurface P(V ) of odd dimension 2m+1and assume for simplicity that char(k) 6= 2. Prove that Fm+1(X) = ∅ and thatFm(X) 6= ∅ (Hint: use the fact that we can choose coordinates (X0, . . . , X2m+2) forV such that F is given by X2

0 +∑m+1i=1 XiXm+1+i.) Prove that Fm(X) is a smooth

variety and determine its dimension.

EXERCISE 70. Let X ⊂ Pn be a hypersurface of degree d and let 0 ≤ m ≤ n.Prove that the intersection of Fm(X) with a standard affine subset of Grm(Pn) isgiven by

(m+dd

)equations.

EXERCISE 71. Consider the universal hypersurface of degree d in Pn, H ⊂Pn × P(n+d

d )−1.

(a) For every m-plane Q ⊂ Pn, let Yz denote the set of z ∈ P(n+dd )−1 for which

the corresponding hypersurface Hz contains Q. Prove that Yz is a linearsubspace of P(n+d

d )−1 of codimension(m+dd

).

(b) Let Y ⊂ P(n+dd )−1 be the set of z ∈ P(n+d

d )−1 for which Hz contains anm-plane. Prove that Y is a closed subset of P(n+d

d )−1 of codimension atmost

(m+dd

)− (m+ 1)(n−m).

(c) Conclude that every hypersurface of degree d in Pn contains an m-planeif(m+dd

)≤ (m+ 1)(n−m). 1

Let P be as before and let X be a nonsingular (not necessarily closed) subvari-ety of P of dimension d. For every p ∈ X, we have defined the tangent space TpXas a d-dimensional subspace of TpP . There is precisely one d-dimensional linearsubspace T (X, p) of P which contains p and has the same tangent space at p as X.In terms of a standard affine chart: if U ⊂ P is the complement of a hyperplanewhich does not pass through p, then U has the structure of an affine space andT (X, p) ∩ U is then the affine subspace of U spanned by TpX.

PROPOSITION-DEFINITION 8.11. The map G : X → Grd(P ), p ∈ X 7→ T (X, p)is a morphism. This morphism is called the Gauss map.

1For instance, every cubic surface in P3 (so here n = 3, d = 3 and m = 1) contains a line. If itis nonsingular, then it contains in fact exactly 27 lines. This famous result due to Cayley and Salmonpublished in 1849 is still subject of research.


PROOF. Let p0 ∈ X. The tangent space T (X, p0) defines a linear subspaceL ⊂ V of dimension d + 1. This linear subspace contains the line `p0 definedby po. Choose homogeneous coordinates Z0, . . . , Zn on P such that p0 lies in thestandard open subset U0 ⊂ P defined by Z0 6= 0. We use the standard coordinateszi := Zi/Z0 to parametrize U0 ⊂ P , but this time we find it convenient to includez0, which is of course constant 1 on U0. In this way U0 is identified with the affinehyperplane H1 ⊂ V defined by z0 = 1. The tangent space TpU0 (p ∈ U0) is then justp+H, where H ⊂ V is defined by z0 = 0. In particular, if p ∈ X∩U0, then TpX canbe regarded as a linear subspace K(p) ⊂ H0 of dimension d displaced over p. Thespan of p and K(p) is then a linear subspace of V which corresponds to T (X, p).

According to Theorem 9.12 there exists a neighborhood U of p0 in U0, reg-ular functions f1, . . . , fn−d on U such that X ∩ U is their common zero set anddf1, . . . , dfn−d are linearly independent on all of U . Then for every p ∈ X ∩ U ,K(p) is the common zero set of df1(p), . . . , dfn−d(p). We regard (df1, . . . , dfn−d)as a matrix-valued function on U , denoted D : U → Hom(H0, k

n−d), so that forp ∈ U ∩X, K(p) is the kernel of D(p). We also assume our coordinates chosen insuch a manner that K(p0) is spanned by e1, . . . , ed and then write H = H ′ ⊕ H ′′with H ′ = K(p0) and H ′′ spanned by ed+1, . . . , en. We write

D = (D′, D′′), with D′ : U → Hom(H ′, kn−d), D′′ : U → Hom(H ′′, kn−d).

SinceD′′(p0) is an isomorphism, the set of p ∈ U for whichD′′(p) is an isomorphismdefines an open neighborhood of p0 and so we may as well assume that that this isthe case on all of U . Cramer’s rule shows that p ∈ U 7→ D′′(p)−1 ∈ Hom(kn−d, H ′′)has regular entries: it is a morphism. For every p ∈ U , the kernel of D(p) is thegraph of the linear map B(p) := −D′′(p)−1D′(p) ∈ Hom(H ′, H ′′). In particular,for p ∈ X ∩ U , TpX is parallel to the subspace spanned by e1 + B(p)(e1), . . . , ed +B(p)(ed).

It follows that a linear subspace of V which corresponds to T (X, p) is spannedby p = e0 +

∑ni=1 piei and the d vectors e1 + B(p)(e1), . . . , ed + B(p)(ed). This is

the graph of the linear map A(p) : ke0 ⊕H ′ → kn−d defined by

A(p)(e0) := −d∑i=1

zi(p)B(p)(ei) +

n∑j=d+1

pjej , A(p)(ei) := B(p)(ei) (i = 1, . . . d).

Since A is a morphism, this defines a morphism X ∩ U → Grd(P ).

For a closed irreducible subset X ⊂ P , the Gauss map is defined on its nonsin-gular part: G : Xreg → Grd(P ). The closure of the graph of G in X × Grd(P ) iscalled the Nash blowup of X and denoted X. The projection morphism X → X isan isomorphism over the open-dense subset Xreg and hence birational. A remark-able property of X is that the Zariski tangent space of each point contains a distin-guished d-dimensional subspace (prescribed by the second projection to Grd(P )) insuch a manner that these subspaces extend the tangent bundle in a regular manner.

9. Multiplicities of modules

Bezout’s theorem asserts that two distinct irreducible curves C,C ′ in P2 of de-grees d and d′ intersect in dd′ points. Strictly speaking this is only true if C and C ′

intersect as nicely as possible, but the theorem is true as stated if we count each

9. MULTIPLICITIES OF MODULES 71

point of intersection with an appropriate multiplicity. There is in fact a generaliza-tion: the common intersection of n hypersurfaces in Pn has cardinality the productof the degrees of these hypersurfaces, provided that this intersection is as nice aspossible and again, this is even true more generally when the locus of intersectionis finite and each point of intersection is counted with an appropriate multiplicity.One of our aims is to define these multiplicities. The commutative algebra toolsthat we use for this have an interest in their own right.

DEFINITION 9.1. We say that an R-module has length ≥ d if there exist a d-stepfiltration by submodules M = M0 )M1 ) · · · )Md = 0. The length of M is thesupremum of such d (and so may be∞).

EXERCISE 72. Suppose R is a noetherian local ring with maximal ideal m andresidue field K. Prove that the length of a finitely generated R-module M is finiteprecisely when mdM = 0 for some d and is then equal to

∑d−1i=0 dimK(miM/mi+1M).

Prove that if R is a K-algebra, then this is also equal to dimK(M).

We fix a noetherian ring R and a finitely generated R-module M .Recall that if p is a prime ideal of R, then Rp is a local ring with maximal ideal

pRp whose residue field can be identified with the field of fractions of R/p. Wedefine Mp := Rp ⊗RM . So this is a Rp-module.

REMARK 9.2. We can describe Mp and more generally, any localization S−1R ⊗R M ,as follows. Consider the set S−1M of expressions m/s with m ∈ M and s ∈ S with theunderstanding that m/s = m′/s′ if the identity s′′s′m = s′′sm holds in M for some s′′ ∈ S(so we are considering the quotient of S×M by an equivalence relation). Then the followingrules put on S−1M the structure of a R-module:

m/s−m′/s′ := (s′m− sm′)/(ss′), r ·m/s := rm/s.

The map S−1R×M → S−1M , (r/s,m)→ (rm)/s is R-bilinear and hence factors throughan R-homomorphism S−1R ⊗R M → S−1M . On the other hand, the map S−1M →S−1R ⊗R M , m/s 7→ 1/s ⊗R m is also defined: if m/s = m′/s′, then s′′(s′m = sm)for some s′′ ∈ S and so

1/s⊗R m = 1/(ss′s′′)⊗R s′s′′m = 1/(ss′s′′)⊗R ss′′m = 1/s′ ⊗R m.

It is an R-homomorphism and it immediately verified that it is a two-sided inverse of themap above. So S−1R⊗RM → S−1M is an isomorphism.

This description shows in particular that if N ⊂M is a submodule, then S−1N may beregarded as submodule of S−1M (this amounts to: S-localization is an exact functor).

DEFINITION 9.3. The multiplicity of M at a prime ideal p of R, denoted µp(M),is the length of Mp as an Rp-module. (For x ∈ Spec(R), we may also write µx(M)instead of µpx(M).)

REMARK 9.4. Let X be a variety, x ∈ X a closed point of X and I ⊂ OX,xan ideal with

√I = mX,x. So I ⊃ mrX,x for some positive integer r. Then

dimk(OX,x/I) is finite (since dimk(OX,x/mrX,x) is) and according to Exercise 72equal to the length of OX,x/I of as an OX,x-module. Hence it is the multiplicityof OX,x/I at the prime ideal mX,x: µx(OX,x/I) = dimk(OX,x/I). If X is affineand we are given an ideal I ⊂ A(X) whose image in OX,x is I, then OX,x/Iis the localization of A(X)/I at x and x is an isolated point of Z(I). Then stillµx(A(U)/I) = dimk(OX,x/I).


IfX is nonsingular at x of dimension n and I has exactly n generators f1, . . . , fn,then we will see that µp(OX,x/(f1, . . . , fn)) = dimk(OAn,p/(f1, . . . , fn)) can be in-terpreted as the multiplicity of p as a common zero of f1, . . . , fn.

We wish to discuss the graded case parallel to the ungraded case. This meansthat if R is graded: R = ⊕∞i=0Ri, then we assume M to be graded as well, that is,M is endowed with a decomposition as an abelian group M = ⊕i∈ZMi such thatRj sends Mi to Mi+j (we do not assume that Mi = 0 for i < 0). For example, ahomogeneous ideal in R is a graded R-module. In that case we have the notion ofgraded length of M , which is the same as the definition above, except that we onlyallow chains of graded submodules.

The annihilator of an R-module M , Ann(M), is the set of r ∈ R with rM =0. It is clearly an ideal of R. We denote by P(M) the set of prime ideals of Rwhich contain Ann(M) and are minimal for that property. According to Exercise15 these are finite in number and their common intersection equals

√Ann(M). In

the graded setting, Ann(M) is a graded ideal and then according to Lemma 2.3 themembers of P(M) are all graded.

We further note that the annihilator of any m ∈M , Ann(m) := r ∈ R : rm 6=0 is an ideal of R, which is graded when m is homogeneous.

LEMMA 9.5. Let M be a finitely generated and nonzero (graded) R-module. Thenthe collection of annihilators of nonzero (homogeneous) elements of M contains amaximal element and any such maximal element is a (graded) prime ideal of R.

PROOF. We only do the graded case. The first assertion follows from the noe-therian property of R. Let now Ann(m) be a maximal element of the collection(so with m ∈ M homogeneous and nonzero). It suffices to show that this is aprime ideal in the graded sense (see Exercise 58), i.e., to show that if a, b ∈ R arehomogeneous and ab ∈ Ann(m), but b /∈ Ann(m), then a ∈ Ann(m). So bm 6= 0and a ∈ Ann(bm). Since Ann(bm) ⊃ Ann(m), the maximality property of the latterimplies that this must be an equality: Ann(bm) = Ann(m), and so a ∈ Ann(m).

If l is an integer and M is graded, then M [l] denotes the same module M , butwith its grading shifted over l, meaning that M [l]i := Ml+i.

Let us call a (graded) R-module elementary if it is isomorphic to R/p ((R/p)[l]),where p is a (homogeneous) prime ideal (and l ∈ Z). So the above lemma says thatevery nonzero (graded) R-module contains an elementary submodule.

PROPOSITION 9.6. Every finitely generated (graded)R-moduleM can be obtainedas a successive extension of elementary modules in the sense that there exists a finitefiltration by (graded) R-submodules M = M0 ) M1 ) · · · ) Md = 0 such thateach quotient M i/M i+1 is elementary.

PROOF. We do the graded case only. Since M is noetherian, the collection ofgraded submodules of M that can be written as a successive extension of elemen-tary modules has a maximal member, M ′, say. We claim that M ′ = M . If M/M ′

were nonzero, then according to Lemma 9.5, it contains an elementary submodule.But then the preimage N of this submodule in M is a successive extension of anextension of elementary modules which strictly contains M ′. This contradicts themaximality of M .

9. MULTIPLICITIES OF MODULES 73

PROPOSITION 9.7. In the situation of the preceding proposition, let pi be the primeideal of R such that M i−1/M i ∼= R/pi. Then P(M) is the set of minimal membersof the collection pidi=1 and every p ∈ P(M) occurs precisely µp(M) times in thesequence (p1, . . . , pd).

PROOF. We first show that√

Ann(M) = p1∩· · ·∩pd. If r ∈ p1∩· · ·∩pd, then rmaps M i−1 to M i and so rd ∈ Ann(M). This proves that p1∩· · ·∩pd ⊂

√Ann(M).

Conversely, if r ∈ R and l ≥ 1 are such that rl ∈ Ann(M), then rl ∈ pi for all i.This implies that r ∈ pi for all i.

So the collection of minimal members of pidi=1 is just the set P(M) of minimalprime ideals containing Ann(M). In particular, no pi can be strictly contained in amember of P(M).

Fix p ∈ P(M). We then have a filtration Mp = M0p ⊃ · · · ⊃ Md

p = 0(for an inclusion of R-modules induces an inclusion of Rp-modules). We haveM i−1

p /M ip∼= Rp/p

iRp and the latter is the residue field Rp/pRp or trivial accordingto whether or not pi = p (if pi 6= p, then pi contains an element not in p; this givesin Rp an invertible element, and hence piRp = Rp). Following our definition thefirst case occurs precisely µp(M) times.

We can of course pass from the graded case to the nongraded case by justforgetting the grading. But more interesting and useful is the following construc-tion, where we pass from a projective setting to an affine one and vice versa. Theexample to keep in mind is when we associate to a homogeneous polynomial ofdegree d, F ∈ k[X0, . . . , Xn] the inhomogeneous polynomial f ∈ k[x1, . . . , xn] ofdegree ≤ d by f(x1, . . . , xn) := F (1, x1, . . . , xn) and go back via F (X0, . . . , Xn) :=Xd

0f(X1/X0, . . . , Xn/X0).Let p ⊂ R be a graded prime ideal. Then denote by Rp,0 the set of fractions r/s

with r ∈ Ri and s ∈ Ri − pi for some i. This is clearly a subring of Rp; it is in facta local ring whose maximal ideal mp is obtained by taking in the previous sentencer ∈ pi. For instance, if R = k[X0, . . . , Xn] and p = (X1, . . . , Xn), then p defines thepoint p = [1 : 0 : · · · : 0] ∈ Pn and Rp can be identified with OPn,p. This definitionmakes also sense for any graded R-module M : Mp,0 is the set of fractions m/s withm ∈Mi and s ∈ Ri − pi for some i. Note that this is a Rp,0-module.

Suppose now that p1 6= R1 and choose s1 ∈ R1 − p1 so that s−11 ∈ (Rp)−1.

Then R[l]p,0 is the the set of fractions r/s with r ∈ Ri+l and s ∈ Ri − pi for somei. For such a fraction we have r/s = sl1r/(s

l1s) ∈ sl1Rp,0, so that R[l]p,0 = sl1Rp,0,

which as an Rp,0-module is of course isomorphic to Rp,0. So the graded iteratedextension M = M0 )M1 ) · · · )Md = 0 of M by elementary R-modules yieldsan iterated extension of Mp,0 by trivial or by elementary Rp,0-modules: Mp,0 =M0

p,0 ⊃ M1p,0 ⊃ · · · ⊃ Md

p,0 = 0 and the number of times a nonzero successivequotient appears is µp(M). In other words, we have µp(M) = µmp

(Mp,0). We usethis observation mainly via the following example.

EXAMPLE 9.8. Let J ⊂ k[X0, . . . , Xn] be a homogeneous ideal and let p ∈ Pnbe a closed isolated point of ZJ . We take here M := k[X0, . . . , Xn]/J and take for pthe graded ideal Ip ⊂ k[X0, . . . , Xn] defining p. Then k[X0, . . . , Xn]Ip,0 can be iden-tified with the local k-algebraOPn,p. Moreover, J defines an ideal J ⊂ OPn,p and wecan identify MIp,0 with OPn,p/J. Notice that

√I = mPn,p. According to the above

discussion µI(M) = µp(OPn,p/I) and by Exercise 72 this is just dimk(OPn,p/J).


10. Hilbert functions and Hilbert polynomials

We shall be dealing with polynomials in Q[z] which take integral values on in-tegers. Such polynomials are called numerical. An example is the binomial functionof degree n ≥ 0: (

z

n

):=

z(z − 1)(z − 2) · · · (z − n+ 1)

n!.

It has the property that its value in any integer k is an integer, namely(kn

)if k ≥ n,

0 if 0 ≤ k ≤ n− 1 and (−1)n(−k+n−1

n

)if k ≤ −1.

Let ∆ : Q[z]→ Q[z] denote the difference operator: ∆(f)(z) := f(z+1)−f(z).It clearly sends numerical polynomials to numerical polynomials. It is also clear thatthe kernel of ∆ consists of the constants Q. Notice that ∆

(z

n+1

)=(zn

).

LEMMA 10.1. Every P ∈ Q[z] which takes integral values on sufficiently largeintegers is numerical and a Z-basis of the abelian group of numerical polynomials isprovided by the binomial functions.

If f : Z→ Z is a function such that for sufficiently large integers ∆(f) is given bya polynomial, then so is f .

PROOF. The first assertion is proved with induction on the degree d of P . Ifd = 0, then P is constant and the assertion is obvious. Suppose d > 0 and theassertion known for lower values of d. So ∆(P )(z) =

∑d−1i=0 ci

(zi

)for certain ci ∈ Z.

Then P (z) −∑d−1i=0 ci

(zi+1

)is in the kernel of ∆ and hence is constant. As this

expression takes integral values on large integers, this constant is an integer. Thisproves that P is an integral linear combination of binomial functions.

For the second assertion, let P ∈ Q[z] be such that P (i) = ∆(f)(i) ∈ Z forlarge i. By the preceding, P (z) =

∑i ai(zi

)for certain ai ∈ Z. So if we put

Q(z) :=∑i ai(zi+1

), then Q is a numerical polynomial with ∆(f − Q)(i) = 0 for

large i. This implies that f − Q is constant for large i, say equal to a ∈ Z. Sof(i) = Q(i) + a for large i and hence Q+ a is as required.

We shall see that examples of such functions are provided by the Hilbert func-tions of graded noetherian modules.

REMARK 10.2. A function f : Z→ Z which is zero for sufficiently negative integers de-termines a Laurent series Lf :=

∑k∈Z f(k)uk ∈ Z((u)). For the function k 7→ max0,

(kn

)

this gives∑k≥n

k(k − 1) · · · (k − n+ 1)

n!uk =

un

n!

dn

dun

∑k≥0

uk =un

n!

dn

dun1

1− u =un

n!(1− u)n.

So if we also know that for sufficiently large integers f is the restriction of a polynomialfunction, then Lemma 10.1 implies that Lf is a Z-linear combination of powers of u and therational functions un

n!(1−u)n . In particular, Pf ∈ Q[u][u−1, (1− u)−1].

In the rest of this section R stands for the graded ring k[X0, . . . , Xn] where eachXi has degree one. An R-module is always assumed to be graded and finitely gener-ated.

Let M be a graded noetherian R-module. We define the Hilbert function of M ,φM : Z→ Z, by φM (i) := dimkMi.

EXAMPLE 10.3. The Hilbert function of R is given by φR(i) =(i+nn

).

10. HILBERT FUNCTIONS AND HILBERT POLYNOMIALS 75

LEMMA 10.4. Let F ∈ Rd and denote by Ann(F,M) ⊂M the set of m ∈M withFm = 0 (this is a graded submodule of M). Then we have φM/FM (i) = φM (i) −φM (i− d) + φAnn(F,M)(i− d).

PROOF. In degree i, multiplication by F defines a k-linear map between finitedimensional k-vector spaces Mi−d →Mi whose kernel is Ann(F,M)i−d and whosecokernel is (M/FM)i. The lemma easily follows from this.

The zero set of the graded ideal Ann(M) in Pn will be called the support of Mand denoted supp(M).

THEOREM 10.5 (Hilbert-Serre). LetM be a finitely generated graded module overthe graded ring R = k[X0, . . . , Xn] (where each Xi has degree 1). Then there existsa unique numerical polynomial PM ∈ Q[z], the Hilbert polynomial of M , such thatφM (i) = PM (i) for i sufficiently large. The degree of PM is equal to dim(supp(M)) ifwe agree that the zero polynomial has the same degree as the dimension of the emptyset (namely −1).

PROOF. We proceed with induction on n ≥ −1. For n = −1, M is a finitedimensional graded k-vector space and hence Mi = 0 for i sufficiently large. Soassume n ≥ 0 and the theorem verified for lower values of n. We first reduceto the case when M is of the form R/p, with p a graded prime ideal. If N is agraded submodule of M , then Ann(M) = Ann(N)∩Ann(M/N) and so supp(M) =supp(N)∪ supp(M/N). Since dimk(Mi) = dimkNi+ dimk(Mi/Ni), we have φM =φN + φM/N . It follows that if the theorem holds for N and M/N , then it holds forM . As M is a successive extension of elementary modules, it suffices to do the caseM = (R/p)[l] with p a graded prime ideal. Since φM (i) = φR/p(l + i), it is enoughto do the case M = R/p.

So let M = R/p. Then supp(M) = supp(R/p) is defined by the graded idealp. In case p = (X0, . . . , Xn), the theorem holds trivially: we have dimkMi = 0for i > 0 (so that we may take PM to be identically zero) and supp(M) = ∅.Suppose therefore that p 6= (X0, . . . , Xn), say that Xn /∈ p. Now multiplication byXn sends M isomorphically onto XnM with quotient M := M/XnM and so fori ≥ 0, φM (i) = φM (i) − φM (i − 1) = ∆φM (i − 1). Since Ann(M) = Ann(M) +XnR, we have supp(M) = supp(M) ∩ Pn−1. According to Proposition 7.2 we thenhave dim supp(M) = dim supp(M) − 1. By regarding M as a finitely generatedmodule over R := k[X0, · · · , Xn−1], our induction hypothesis tells us that thereexists a polynomial PM of degree dim supp(M) such that φM and PM coincide onlarge integers. Since ∆φM (i − 1) = PM (i) for large i, Lemma 10.1 implies thatthere exists a polynomial PM of degree one higher than that of PM (so of degreedim(supp(M))) which coincides with φM for sufficiently large integers.

It follows from Lemma 10.1 that when PM is nonzero, then its leading term hasthe form cdz

d/d!, where d is the dimension of supp(M) and cd is a positive integer.

REMARK 10.6. For M as in this theorem we may also form the Laurent series LM (u) :=∑i dim(Mi)u

i (this is usually called the the Poincare series of M). It follows from Remark10.2 and Theorem 10.5 that LM (u) ∈ Q[u][u−1, (1− u)−1].

DEFINITION 10.7. If d = dim supp(M), then the degree deg(M) is d! times theleading coefficient of its Hilbert polynomial (an integer, which we stipulate to bezero in case supp(M) = ∅). For a closed subset Y ⊂ Pn, the Hilbert polynomial PYresp. the degree deg(Y ) of Y are those of R/IY as a R-module.


Let M be a finitely generated graded R-module. Recall that P(M) denotes theset of minimal prime ideals containing Ann(M). The support of M is defined bythe ideal Ann(M) and is graded. So it defines a closed subset in Pn which in ourprevious notation is just ZAnn(M). For every p ∈ P(M) not equal to (X0, . . . , Xn),Zp ⊂ Pn is an irreducible component of ZAnn(M) whose degree is deg(R/p) and allirreducible components of ZAnn(M) are so obtained (for p = (X0, . . . , Xn) we getZp = ∅, but deg(R/(X0, . . . , Xn) = 0).

PROPOSITION 10.8. Let M be a finitely generated graded R-module. Then

deg(M) =∑

p∈P(M)

µp(M) deg(R/p),

where we note that the right hand side can also be written as a sum over the irreduciblecomponents of ZAnn(M) of maximal dimension, where we then may replace deg(R/p)by the degree of that component.

PROOF. Write M as an iterated extension by elementary modules: M = M0 )M1 ) · · · )Md = 0 with M i−1/M i ∼= R/pi[li]. Then φM (i) =

∑di=1 φR/pi(i−li).

Now φR/pi is a polynomial of degree equal to the dimension of supp(R/pi) = Zpi ⊂Pn. This degree does not change if we replace the variable i by i − li. In view ofProposition 9.7 we only get a contribution to the leading coefficient of φM whenpi ∈ P(M) and for any given p ∈ P(M) this happens exactly µp(Mp) times. Weonly need to consider the cases p for which the degree of φR/p is maximal (i.e.,equals the degree of φM ). The proposition follows.

EXERCISE 73. Compute the Hilbert polynomial and the degree of

(a) the image of the d-fold Veronese embedding of Pn in P(n+dn )−1,

(b) the image of the Segre embedding of Pm × Pn in Pmn+m+n.

EXERCISE 74. Let Y ⊂ Pm and Z ⊂ Pn be closed and consider Y ×Z as a closedsubset of Pmn+m+n via the Segre embedding. Prove that the Hilbert function resp.polynomial of Y ×Z is the product of the Hilbert functions resp. polynomials of thefactors.

One can show that there exists a nonempty open subset of (n− d)-planes Q ⊂Pn which meet Y in exactly deg(Y ) points (see Exercise 75). This characterizationis in fact the classical way of defining the degree of Y .

Observe that if Y ⊂ Pn is nonempty, then deg(Y ) > 0. For then IY,d 6= Rdfor every d ≥ 0 and so the Hilbert function of S(Y ) is positive on all nonnegativeintegers. This implies that PY is nonzero with positive leading coefficient.

We also note that since deg(Y ) only depends on the dimensions of the gradedpieces S(Y )i of the homogenenous coordinate ring the degree of Y will not changeif we regard Y as sitting in a higher dimensional projective space Y ⊂ Pn ⊂ Pm(with m ≥ n).

We verify some other properties we expect from a notion of degree.

PROPOSITION 10.9. A hypersurface H ⊂ Pn defined by an irreducible polynomialof degree d has degree d.

PROOF. Let F ∈ Rd be the polynomial in question and apply Lemma 10.4 toM = R. Since the Hilbert function of R takes at the positive integer i the value

10. HILBERT FUNCTIONS AND HILBERT POLYNOMIALS 77(n+in

), the one of H takes at i the value

(n+in

)−(n+i−dn

). Its Hilbert polynomial is

therefore (z + n

n

)−(z − d+ n

n

)=

d

(n− 1)!zn−1 + lower order terms,

and so deg(H) = d.

PROPOSITION 10.10. Let Y1, . . . , Yr be the distinct irreducible components of Yof maximal dimension (= dimY ). Then deg(Y ) =

∑ri=1 deg(Yi).

PROOF. Put Y ′ := Y2∪ · · · ∪Yr. By proceeding with induction on r, we see thatit is enough to show that deg(Y ) equals deg(Y1)+deg(Y ′) or deg(Y1) depending onwhether r ≥ 2 or r = 1. Since IY = IY1 ∩ IY ′ , we have an exact sequence of gradedR-modules

0→ S(Y )→ S(Y1)⊕ S(Y ′)→M → 0,

where M := R/(IY1 + IY ′). This implies that PY = PY1 + PY ′ − PM . We havesupp(M) = Y1 ∩ Y ′ and so dim supp(M) < dimY . Hence deg(PM ) < dim(Y ). Theassertion then follows from comparing the coefficients of zdimY .

We can now state and prove a theorem of Bezout type.

COROLLARY 10.11. Let M be a graded R-module. Let F ∈ Rd \ Ann(M). ThenM/FM has degree ddeg(M) as a R-module and if we put m := dim(ZAnn(M)), then∑

Z irred. comp of ZAnn(M/FM) of dim. m− 1

µZ(M/FM) deg(Z) = ddeg(M).

PROOF. The exact sequence

0→M [−d]·F−−−−→ M →M/FM → 0

shows that PM/FM (z) = PM (z) − PM (z − d). Since PM is of degree m with lead-ing coefficient deg(M)/m!, it follows that PM/FM is of degree m − 1 with leadingcoefficient ddeg(M)/(m− 1)!. An irreducible component C of Ann(M) is not con-tained in ZF and so dim(C ∩ ZF ) = dim(C) − 1. This produces all the irreduciblecomponents of Ann(M/FM) = Ann(M) + FR. It remains to apply Proposition10.8 to M/FM .

COROLLARY 10.12. Let Y ⊂ Pn be closed with all irreducible components of thesame dimension m. If Q ⊂ Pn is a linear subspace of codimension m with the propertythat Q ∩ Y is finite, then the degree of Y is equal to

∑p∈Y ∩Q µp(S•(Y )/IQS•(Y )).

PROOF. Let H1, . . . ,Hm be hyperplanes in Pn such that Q = H1∩ · · · ∩Hm andput Y k := Y ∩H1 ∩ · · · ∩Hk (k = 0, . . .m). We are given that Y m = Q∩Y is finite.Since intersecting a subvariety of Pn with with a hyperplane makes its dimensiondrop by at most one, we see that each irreducible component of Y k has dimensionm− k. If we apply the first part of Corollary 10.11 to a defining linear form for Hk,we find that Y k has the same degree as Y k−1.

For p ∈ Q ∩ Y , let IY,p resp. IQ,p be the ideal in OPn,p defining Y resp. Q atthat point. Then we observed that

µp(S•(Y )/IQ) = dimkOPn,p/(IY,p + IQ,p).


EXERCISE 75. Prove that deg(Y ) is the cardinality of Q∩Y if and only if IY,p+IQ,p = mPn,p for all p ∈ Q ∩ Y . (We then say that Q and Y are in general positionwith respect to each other.)

THEOREM 10.13 (Theorem of Bezout). Let or i = 1, . . . n, Hi ⊂ Pn be a hyper-surface of degree di > 0 and assume that H1 ∩ · · · ∩Hn is finite. Each Hi determinesat p ∈ H1∩ · · · ∩Hn a principal ideal in OPn,p; denote by Ip ⊂ OPn,p the sum of theseideals. Then

d1d2 · · · dn =∑

p∈H1∩···∩Hn

dimk(OPn,p/Ip).

PROOF. Put Zk := H1∩· · ·∩Hk. Since Zn is finite, all irreducible components ofZk are of dimension n−k (by the same argument as in Corollary 10.12). Repeatedapplication of Corollary10.11 then yields that d1d2 · · · dn is the degree of Zn. Withthe help of Exercise 75 we see that this degree is computed as stated.

EXAMPLE 10.14. Assume char(k) 6= 2. We compute the intersection multiplic-ities of the conics C and C ′ in P2 whose affine equations are x2 + y2 − 2y = 0and x2 − y = 0. There are three points of intersection: (0, 0), (−1, 1) and (1, 1)(so none at infinity). The intersection multiplicity at (0, 0) is the dimension ofOA2,(0,0)/(x

2 + y2−2y, x2− y) as a k-vector space. But OA2,(0,0)/(x2 + y2−2y, x2−

y) = OA1,0/(x4 − x2) = k[x]/(x2) (for (x2 − 1) is invertible in OA1,0). Clearly

dimk(k[x]/(x2)) = 2 and so this is also the intersection multiplicity at (0, 0). Theintersection multiplicities at (−1, 1) and (1, 1) are easily calculated to be 1 and thusthe identity 2 + 1 + 1 = 2 · 2 illustrates the Bezout theorem.

REMARK 10.15. If Y ⊂ Pn is closed, then PY (0) can be shown to be an invariantof Y in the sense that it is independent of the projective embedding. In many ways,it behaves like an Euler characteristic. (It is in fact the Euler characteristic of OYin a sense that will become clear once we know about sheaf cohomology.) Forexample, PY×Z(0) = PY (0)PZ(0).

We have seen that for a hypersurface Y ⊂ Pn of degree d > 0, PY (z) =(z+nn

)−(

z−d+nn

)and so PY (0) = 1 −

(−d+nn

)= 1 − (−1)n

(d−1n

). For n = 2 (so that Y is a

curve), we get PY (0) = 1− 12 (d−1)(d−2). The number 1−PY (0) = 1

2 (d−1)(d−2)is then called the arithmetic genus of the curve. If the curve is smooth and k = C,then we may regard it as a topological surface (a Riemann surface) and g is thenjust the genus of this surface (and so PY (0) is half its Euler characteristic).

CHAPTER 3

Schemes

1. Presheaves and sheaves

The notion of a prevariety as well as its generalizations are best understood inthe general context of ringed spaces. This involves the even more basic notion of asheaf.

DEFINITION 1.1. Let X be a topological space. An abelian presheaf F on Xconsists of giving for every open subset U ⊂ X an abelian group F(U) (whoseelements are called sections of F over U) and for every inclusion of open subsetsU ⊂ U ′ a homomorphism of groups (called the restriction map) F(U ′) → F(U)such that:

(i) for the identity U = U we get the identity in F(U),(ii) if U ⊂ U ′ ⊂ U ′′ are open sets, then the homomorphism F(U ′′) → F(U)

is equal to the composite F(U ′′)→ F(U ′)→ F(U) and

If you are familiar with the language of categories, then you might observethat a presheaf is nothing but a contravariant functor F : O(X) → Ab from thecategory O(X) of open subsets of X (whose morphisms are inclusions of opensubsets of X) to the category of abelian groups Ab. The terminology ‘section overU’ and ‘restriction’ is suggestive, although it sometimes can be a bit misleading.This terminology is mirrored by the notation: if we are given an inclusion of opensubsets U ′ ⊂ U , then the image of s ∈ F(U) under the restriction map F(U) →F(U ′) is often denoted s|U ′ (a natural notation for the restriction map would beF(U ′ ⊂ U), but this is rarely used; more common is resU,U ′).

In case the groups F(U) come with the structure of a ring and the restrictionmaps are ring homomorphisms, then we say that F is a presheaf of rings (in otherwords, the contravariant functor takes values in the category Ring of rings). Like-wise, we have the notion of a presheaf of modules over a fixed ring R. In somesituations no structure on F(U) is imposed at all (the target category is then theone of sets), but we will only deal here with presheaves that are at least abelian.

EXAMPLE 1.2 (The constant presheaf). Given an abelian group G and a topo-logical space X, then we have an abelian presheaf defined on X if we take for everynonempty open U ⊂ X the group G and for each inclusion between two such setsthe identity map of G.

EXAMPLE 1.3. Given a topological space X and an abelian topological groupG, then assigning to every nonempty open U ⊂ X the group of continuous mapsU → G (with the obvious restriction maps) is an abelian presheaf CX,G. Of specialinterest areG = R andG = C, in which we get a presheaf of R-algebras (resp. of C-algebras). Another case of interest is when G has been given the discrete topology:then this assigns to U the space of locally constant maps U → G.

79

80 3. SCHEMES

EXAMPLE 1.4. For a smooth manifold M , we have defined the presheaf EMof R-algebras which assigns to every nonempty open U ⊂ M the R-algebra ofdifferentiable functions U → R.

EXAMPLE 1.5. On a complex manifold M , we have defined the presheaf OanM of

C-algebras which assigns to every nonempty open U ⊂M the C-algebra Oan(U) ofholomorphic functions U → C.

EXAMPLE 1.6. On a k-prevarietyX, we have defined the presheafOX of regularfunctions.

DEFINITION 1.7. Given a presheaf F on X and a point p ∈ X, then a germ ofa section of F at p, is a section of F on an unspecified neighborhood of p, withthe understanding that two such sections represent the same germ if they coincideon a neighborhood of p contained in their common domain of definition. The setof germs of sections of F at p is called the stalk of F at p and denoted by Fp.(In categorical language, this is expressed as an inductive limit Fp = lim−→U3p F(U),where one observes that the collection of neighborhoods of p in X form a projectivesystem: the intersection of two neighborhoods of p is another one.)

For instance in the case of Example 1.5, when M is an open subset of C, thenthe stalk Oan

M,p can be identified with the ring of convergent power series in z − p.Let us observe that an s ∈ F(U) determines an element sp ∈ Fp for every

p ∈ U . We will also write s+(p) for sp, where we view s+ as a map U →∏p∈U Fp.

The last three examples differ from the first in that they satisfy two additionalproperties:

DEFINITION 1.8. An abelian presheaf F on X is called an abelian sheaf if forevery open U ⊂ X and open covering Uii∈I of U ,

(iv) every system of sections si ∈ F(Ui)i∈I that is compatible in the sensethat for each pair (i, j) in I, si|Ui ∩ Uj = sj |Ui ∩ Uj comes from somesection s ∈ F(U) in the sense s|Ui = si for all i ∈ I, and

(v) two sections s, s′ ∈ F(U) coincide if s|Ui = s′|Ui for all i ∈ I.

In other words, the sequence

0→ F(U)→∏i∈IF(Ui)→

∏(i,j)∈I2

F(Ui ∩ Uj)

in which the first map is given by s 7→ (s|Ui)i∈I and the second by (si)i∈I 7→(si|Ui ∩ Uj − sj |Ui ∩ Uj)i,j is exact. This implies that F(∅) is the trivial group 0.The argument relies on a categorical notion of product1, but for our purposes wemay just as well stipulate that this is so.

A space X endowed with with a sheaf of rings is also called a ringed space.Examples are 1.3, 1.4, 1.5 and 1.6. A constant presheaf is usually not a sheaf: ifU1, U2 are disjoint open subsets and we take gi ∈ F(Ui) = G distinct for i = 1, 2,

1If we are given a product of abelian groups with index set I,∏i∈I Gi, then for every subset

J ⊂ I, this product maps isomorphically to the product (∏i∈J Gi) × (

∏i∈I−J Gi). By taking J = I,

we see that a product with empty index set must be the trivial group. Similarly, every union of sets∪i∈IXi, can be written as (∪i∈JXi) ∪ (∪i∈I−JXi), from which deduce (by taking J = I), that aunion with empty index set must be empty. If we then apply the above exact sequence to U = ∅ bytaking an open covering with empty index set, we find that F(∅) = 0.

1. PRESHEAVES AND SHEAVES 81

then clearly these elements cannot be the restriction of a single g ∈ F(U1∪U2) = G.There is however a simple modification which is a sheaf, namely the constant sheafGX which assigns to U ⊂ X the space of continuous maps from to U to G, wherewe endow G with the discrete topology (so these are the maps which are locallyconstant). This turns it into a special case of Example 1.3.

1.9. SHEAFIFICATION. This modification is a special case of a general construc-tion which produces a sheaf F+ out of a presheaf F with the same stalks: a sections of F+ over an open subset U is by definition a map from U to the disjoint unionof the stalks Fp, p ∈ U , with the property that locally this map is of the form s+: wemay cover U by open subsets Ui for which there exist si ∈ F(Ui) such that sp = si,pfor all p ∈ Ui. It is straightforward to verify that this is well-defined, defines a sheafand that F+ = F in case F happens to be a sheaf.

The sheaf F+ has a universal property. In order to explain this, we need thenotion of a homomorphism of presheaves. Let C be one of our categories. If Fand G are C-valued presheaves on X, then a homomorphism of presheaves φ : F →G is what is called in category theory a natural transformation of functors: forevery open U ⊂ X, we are given C-homomorphism φ(U) : F(U) → G(U) suchthat this collection is compatible with restriction: if U ′ ⊂ U and s ∈ F(U), thenφ(U)(s)|U ′ = φ(U ′)(s|U ′).

For example, the formation of F+ defines a homomorphism of presheaves F →F+. It is universal in the sense that if φ : F → G is a homomorphism of presheavesand G is a sheaf, then there is unique sheaf homomorphism φ+ : F+ → G such thatφ is the composite of F → F+ and φ+.

In general a homomorphism of presheaves φ : F → G induces a sheaf homo-morphism φ+ : F+ → G+.

EXERCISE 76 (Local nature of a sheaf). Let X be a topological space and Ua basis of open subsets of X. Prove that an abelian sheaf F on the space X isdetermined by its restriction to that basis, that is, by the collection F(U), and therestriction maps F(U ′)→ F(U), for the pairs (U,U ′) ∈ U × U with U ⊂ U ′.

Prove also a converse: suppose that the basis U is closed under finite intersec-tions and assume that for every U ∈ U is given an abelian group F(U) and for everyinclusion U ⊂ U ′ of members of U a homomorphism F(U ′) → F(U) such that theproperties (i) through (iii) are satisfied. Then F generates a sheaf F+ on X.

EXERCISE 77 (Direct image of a sheaf). Suppose we are given a continuousmap f : X → Y between topological spaces and an abelian presheaf F on X. Thena presheaf f∗F on Y is defined by assigning to an open V ⊂ Y the group F(f−1V ).Prove that if F is a sheaf, then so is f∗F . Is it true that (f∗F)+ = f∗(F+)?

EXAMPLE 1.10. Let f : S1 → S1 be defined by f(z) = z2 and let ZS1 be theconstant sheaf on S1 with stalks Z. Then f∗ZS1 is a sheaf on S1 which is locallylike the constant sheaf Z2. But (f∗ZS1)(S1) = ZS1(S1) = Z and so is of rank lessthan 2. Other examples in the same spirit are gotten as the sheaf on C − 0 thatis the direct image of the constant sheaf on C − 0 with stalk Z under the mapz ∈ C − 0 7→ zn ∈ C − 0 (n ∈ Z − 0) or of the direct image of the constantsheaf on Z with stalk C under the map z ∈ C 7→ ez ∈ C− 0.

1.11. KERNELS AND COKERNELS. We begin with some obvious definitions:given a presheaf F on X, then a subpresheaf of F is simply a presheaf F ′ on X

82 3. SCHEMES

with the property that F ′(U) ⊂ F(U) for every open subset U of X. If F ′ happensto be a sheaf, then we call it a subsheaf of F .

Suppose and F and G are abelian presheaves over X and let φ : F → Gbe a homomorphism of presheaves. So for every open U ⊂ X, we then have ahomomorphism φ(U) : F(U) → G(U). By assigning to U , Ker(φ)(U), Imφ(U),Coker(φ(U)) we have defined three presheaves, of first two being subpresheaves ofF and G respectively. If Ker(φ) is the zero sheaf, then clearly F can be thought of asubpresheaf of G. Suppose now that F and G are in fact sheaves. Then Ker(φ)(U)is easily seen to be a sheaf (and indeed denoted by Ker(φ)). But the other two neednot be (see Example 1.12 below) and so we define the image sheaf Im(φ) resp. thecokernel sheaf Coker(φ) as the sheaf associated to the corresponding presheaves.Since this process does not affect the stalks, we have that for every x ∈ X, Im(φ)xresp. Coker(φ)x is the image resp. the cokernel of φx : Fx → Gx. Moreover, Im(φ)xis a subsheaf of G. The possible failure of the image presheaf being a sheaf is ex-pressed by the fact that the image of φ(U) : F(U)→ G(U) is contained in Im(φ)(U)but need not equal it and a similar statement can be made for the cokernel sheaf.

We finally note that the homomorphisms from F to G form an abelian group;we denote that group by Hom(F ,G).

If F is a sheaf on X, then F(X) is called its group of global sections. Oneoften denotes this group by Γ(X,F) or H0(X,F) instead. The reason for the latternotation will become clear later.

An interesting twist to the notion of a constant sheaf is that of a local system(and called by some algebraic topologists a system of twisted coefficients): this is alocally constant sheaf. For instance, if G is an abelian group, then a local system oftype G on X is an abelian sheaf F on X with the property that X can be coveredby open subsets U such that F|U is isomorphic to the sheaf associated to the con-stant presheaf G on U . Examples were given in 1.10. Here is another one, whichillustrates some other phenomena as well.

EXAMPLE 1.12. Let X be C − 0 with its usual Hausdorff topology. On X wedefine a local system L of 2-dimensional complex vector spaces by letting L(U)be the space of C-valued functions that are of the form f + c with f : U → Ccontinuous and satisfying ef(z) = z. In other words, f is a branch of the logarithmfunction. But recall that branch need not defined and that

The constant functions define a constant subsheaf CX ⊂ L. The quotient sheafL/CX is also isomorphic to CX , for it admits as a global generating section thefunction “log” (note that the multivaluedness of this function disappears if we workmodulo constants). But the section “log” cannot be lifted to X: the only globalsections of L(X) are the constants. So the exact sequence of abelian sheaves

0→ CX → LX → LX/CX → 0

evaluated on X yields a sequence of abelian groups

0→ Γ(X,CX)→ Γ(X,LX)→ Γ(X,LX/CX)→ 0.

We observed that the last sequence can be identified with 0 → C = C 0−→ C → 0.So it fails to be exact at Γ(X,CX). This also shows that the image presheaf ofL → LX/CX and the cokernel presheaf of CX → L are not sheaves.

1.13. THE PREIMAGE OF A SHEAF. Here is a relative version of the sheafificationprocess. Suppose we are given a continuous map f : X → Y between topological

1. PRESHEAVES AND SHEAVES 83

spaces and an abelian sheaf G on Y . We want to define a sheaf f−1G on X with theproperty that the stalk of f−1G at x ∈ X is Gf(x). If U ⊂ X is open and s assignsto x ∈ X, sx ∈ Gf(x), then we stipulate that s defines a section of f−1F if for eachx ∈ U there exists a neighborhood Vx of y in Y and a section t ∈ G(Vy) such thatfor x′ in a neighborhood of x in f−1Vx ∩ U , we have sx′ = tf(x). This is indeed asheaf, called the sheaf pull-back of G. If Y is a single point y, so that G is givenby a single group G = Gy, then this reproduces the sheaf of locally constant mapsfrom open subsets of X to G.

If Y ⊂ X is a subspace, then we often write Γ(Y,F) for the group of globalsections of i−1F , where i : Y ⊂ X denotes the inclusion. Notice that when Y isopen, this is just F(Y ).

EXERCISE 78 (Adjunction). Suppose that f : X → Y is a continuous map andlet F resp. G be an abelian sheaf on X resp. Y .

(a) Prove that we may identify Hom(G, f∗F) with Hom(f−1G,F). (In cate-gorical language: G 7→ f−1G is the left adjoint of F 7→ f∗F and the latteris the right adjoint of the former, the convention being that ‘left’ acts onthe item before the comma in Hom and ‘right’ on what comes after it.)

(b) Deduce the existence of natural sheaf homomorphisms G → f∗f−1G and

f−1f∗F → F .(c) Give examples to show that neither of these needs to be an isomorphism.

(Hint for the second case: look at Example 1.10.)(d) Suppose that f is a homeomorphism. Show that under the identifica-

tion (a) a sheaf isomorphism from G → f∗F yields a sheaf isomorphismf−1G → F and vice versa. (If any such an isomorphism exists, we saythat (X,F) and (Y,G) are isomorphic.)

A space X endowed with a sheaf of rings OX is called a ringed space; the sheafin question is then often referred to as its structure sheaf. This notion becomes muchmore useful if we also have a notion of morphism of ringed spaces: if (X,OX) and(Y,OY ) are ringed spaces, then by a morphism from (X,OX) to (Y,OY ) we mean apair consisting of a continuous map f : X → Y and a sheaf homomorphism of ringsOY → f∗OX . According to Exercise 78 the latter is equivalent to giving a sheafhomomorphism of rings f−1OY → OX . (We often abuse notation by denoting thispair by the single symbol f .)

REMARK 1.14. Via Exercise 76 we can (at least formally) characterize certainextra structures on spaces in a uniform manner. For example, given a topologicalm-manifoldM , then a Ck-differentiable structure onM is simply given by a subsheaf ofthe sheaf R-valued continuous functions on M which has the property that locallyit is isomorphic to the sheaf of Ck-differentiable functions on Rm. It is a sheafof R-algebras. A Ck-map between two Ck-manifolds is then simply a morphismbetween the corresponding ringed spaces (or rather, spaces endowed with a sheafof R-algebras). This is a more conceptual (and perhaps also more concise) thanthe definition based on an atlas. A holomorphic structure on M (which turns Minto a complex manifold) and a holomorphic map between complex manifolds canbe defined in a similar manner. An affine variety is naturally a ringed space andindeed, a k-prevariety, is a a ringed space locally isomorphic as a ringed space to anaffine k-variety. This is why such sheaves are often referred to as structure sheaves.We shall define in the same spirit a structure sheaf on the spectrum of a ring.

84 3. SCHEMES

2. The spectrum of a ring as a locally ringed space

LetR be a ring. We recall that its spectrum Spec(R) is the set of its prime ideals.We agreed that for a prime ideal p of R, the corresponding element of Spec(R) isdenoted xp, and that the prime ideal associated to x ∈ Spec(R) is denoted px ⊂ R.

For s ∈ R, Z(s) ⊂ Spec(R) denotes the set of xp ∈ Spec(R) with s ∈ p andU(s) := Spec(R) − Z(s). The collection U(s)s∈R, is basis for a topology onSpec(R), the Zariski topology. We therefore refer to any U(s) as a basic open subsetof Spec(R). So the closed subsets of this topology are of the form Z(S) = ∩s∈SZ(s),where S ⊂ R is a subset (but note that Z(S) does not change if we replace S bythe radical of the ideal generated by S). We found in Remark 10.3 that Spec(R) isquasicompact.

The spectrum of a ring R is a singleton precisely when R has just one primeideal. Since

√(0) is the intersection of all the prime ideals of R, it follows that this

prime ideal equals√

(0) and must be a maximal ideal. (So if R is noetherian it willhave finite length.) We have

√(0) = (0) precisely when R is a field.

The spectrum of a local ring has just one closed point (namely the one definedby its maximal ideal), but can have plenty of nonclosed points, of course.

We observed that a ring homomorphism φ : R′ → R determines a map

Spec(φ) : Spec(R)→ Spec(R′), xp 7→ xp′ ,

where p′ := φ−1p. It has the property that Spec(φ)−1(U(r′)) = U(φ(r′)) so thatSpec(φ) is continuous. The ring homomorphism R′ → R → Rp factors throughR′p′ (by Exercise 12) and so we have an associated homomorphism of local ringsφp : R′p′ → Rp. We have φ−1

p (pRp) = p′Rp′ and so this homomorphism is local inthe sense of:

DEFINITION 2.1. A homomorphism ψ : A′ → A of local rings is said to be alocal homomorphism if ψ−1mA = mA′ (so that it defines an embedding of residuefields A′/mA′ → A/m).

Hence a homomorphism of local rings is local precisely if the induced map onspectra sends the closed point to the closed point. For instance, the inclusion of thelocal ring k[[t]] in its quotient field k((t)) is not local, for the maximal ideal of k((t))is the zero ideal, but the zero ideal is not equal to the maximal ideal tk[[t]] of k[[t]].Thus the singleton Spec(k((t))) maps to the nonclosed point of Spec(k[[t]]) definedby the zero ideal.

For every subset Y ⊂ Spec(R), we may consider the set I(Y ) of r ∈ R which‘vanish’ on Y , that is for which Z(r) ⊃ Y . This is clearly an ideal in R. In thiscontext, a Nullstellensatz comes easily as it is basically the content of Lemma 2.15:

PROPOSITION 2.2. Given a ring R, then for every ideal J ⊂ R, I(Z(J)) =√J .

PROOF. The property r ∈ I(Z(J)) means Z(r) ⊃ Z(J). This in turn meansthat every prime ideal which contains J must contain r. In other words, I(Z(J)) isthe intersection of all the prime ideals that contain J . According to Exercise 15 thisintersection is precisely

√J and so r ∈

√J .

PROPOSITION 2.3. Let R be a ring.

(i) The closure of the singleton xp consists of the xq ∈ Spec(R) with q ⊃ p.

2. THE SPECTRUM OF A RING AS A LOCALLY RINGED SPACE 85

(ii) A singleton xp is closed if and only if p is a maximal ideal (so that Spec(R)contains Specm(R) as the set of its closed points).

(iii) A closed subset of Spec(R) is irreducible if and only if it is the closure of asingleton.

(iv) A subset of Spec(R) is an irreducible component of Spec(R) if and only if itis of the form xp with p a minimal prime ideal of R.

PROOF. The proofs of (i) and (ii) are left to you as an exercise. As for (iii)and (iv), let C ⊂ Spec(R) be a closed irreducible subset. We first prove that I(C)is prime. For suppose that r1r2 ∈ I(C) for certain r1, r2 ∈ R. Then Z(r1r2) =Z(r1) ∪ Z(r2) contains C and since C is irreducible it follows that Z(r1) ⊃ C orZ(r2) ⊃ C or equivalently, that r1 ∈ I(C) or r2 ∈ I(C). By definition, x ∈ C ifand only if px ⊃ I(C) and so C is the closure of the singleton xI(C). On theother hand, a singleton is irreducible and hence so is its closure. Now (iii) and (iv)follow.

EXERCISE 79. Let φ : R′ → R be a ring homomorphism. Prove that if I ′ ⊂ R′

is a proper ideal, then the natural map Spec(R′/I ′) → Spec(R′) is injective withimage the closed subset of Spec(R′) defined by I ′. Show that the preimage of thisclosed subset under Spec(φ) : Spec(R) → Spec(R′) is the closed subset of Spec(R)defined by the ideal of R generated by φ(I).

When s is nilpotent, U(s) = ∅. Let us agree that then R[1/s] denotes thezero ring. Since a prime ideal must always be proper ideal, we then also haveSpec(R[1/s]) = ∅. The following proposition identifies basic open subsets withspectra of simple localizations and shows that inclusions between them come fromobvious ring homomorphisms.

PROPOSITION 2.4. Given s ∈ R, denote by js : R → R[1/s] the natural map.Then Spec(js) : SpecR[1/s] → Spec(R) is a homeomorphism onto U(s). We haveU(s) ⊂ U(s′) if and only if s ∈

√(s′) and the inclusion is then induced by the

associated homomorphism R[1/s′]→ R[1/s].

PROOF. When R[1/s] is the zero ring there is nothing to prove and so let usassume that this is not the case. Then s is not nilpotent. If p is a prime ideal R[1/r],then p = j−1

s p is a prime ideal with s /∈ p and we have p = p[1/s]. On the otherhand, for a prime ideal p of R with s /∈ p, p[1/s] is a prime ideal of R[1/r] andthe preimage of j−1

s p[1/s] = p. In other words, Spec(js) is injective with imageU(s). This map is continuous. It is also open: a nonempty basic open subset ofSpecR[1/s] is defined by a nonnilpotent s′ ∈ R[1/s]. If s′′ ∈ R denotes the productof s and the numerator of s′, then R[1/s][1/s′] = R[1/s′′] and so this basic open setis also a basic open subset of Spec(R).

We have U(s) ⊂ U(s′) if and only if Z(s) ⊃ Z(s′) and by Proposition 2.2 thismeans that sn = s′r for some r ∈ R and some integer n > 0. Hence R[1/s′] maps toR[1/(s′r)] = R[1/s] and this map of course defines the inclusion U(s) ⊂ U(s′).

Henceforth we will use Spec(js) to identify U(s) with SpecR[1/s].The following theorem is of fundamental conceptual importance. It interprets

in a geometric manner the localizations of R and the natural maps between them.

THEOREM-DEFINITION 2.1. Let R be a ring and put X := Spec(R). There is asheaf of rings OX on X characterized by the property that for a basic open set U(s),

86 3. SCHEMES

OX(U(s)) = R[1/s] and the restriction map defined by an inclusion U(s) ⊂ U(s′) isthe associated homomorphism R[1/s′]→ R[1/s]. Its stalk OX,x at x ∈ X is the localring Rpx . We call OX the structure sheaf of X.

PROOF. For the existence and uniqueness of OX , it suffices, in view of Exercise76, to prove the following:

Let be given a collection of nonempty basic open subsets U(sα)α∈A of X whoseunion is a basic open subset U(s) and let uα ∈ R[1/sα]α∈A be a collection with theproperty that for every pair α, β ∈ A, uα and uβ become equal in R[1/(sαsβ)]. Thenthe members of this collection come from a unique element u ∈ R[1/s].

The verification of this property is modeled after the proof of Proposition 10.2.Since U(s) = SpecR[1/s] is quasicompact, there is a finite subset K ⊂ A such thatU(s) = ∪κ∈KU(sκ). For κ ∈ K, we write uκ = rκ/s

Nκκ . We can take Nκ larger if

we want to and so we may assume that they are all equal to some N ′.For κ, λ ∈ K, rκ/sN

′

κ and rλ/sN′

λ become equal in R[1/(sκsλ)], and so rκsN′

λ −rλs

N ′

κ is annihilated by (sκsλ)Nκ,λ for some Nκ,λ ≥ 0. We can take Nκ,λ largerif we want to, and so we may assume that they are all equal to some N ′′. Nowput rκ := rκs

N ′′

κ and N := N ′ + N ′′. Then uκ = rκ/sN and rκs

Nλ = rλs

Nκ for all

κ, λ ∈ K.Since U(s) = ∪κ∈KU(sκ), we have

Z(s) = ∩κ∈KZ(sκ) = ∩κ∈KZ(sNκ ) = Z(∑κ∈K

RsNκ ).

Then sM of s will lie in the ideal∑κ∈K Rs

Nκ for some M > 0 by Proposition 2.2:

sM =∑κ∈K

tκsNκ

for certain tκ ∈ R. We put r :=∑κ∈K tκrκ ∈ R. Then for every λ ∈ K,

rsNλ =∑κ∈K

tκrκsNλ =

∑κ∈K

tκrλsNκ = rλs

M

This proves that u := r/sM ∈ R[1/s] maps to uκ in R[1/sκ] for κ ∈ K. This doesnot prove yet that u has this property for an arbitrary α ∈ A. Still we could inthe above argument replace K by K ∪ α and thus come up with a u′ ∈ R[1/s]which is such that it not only maps to uκ in R[1/sκ] for κ ∈ K, but also to uαin R[1/sα]. It therefore remains to show u = u′ (which then also would establishuniqueness). The fact that u − u′ ∈ R[1/s] maps to the zero element of R[1/sκ]means that snκκ (u − u′) = 0 for some nκ > 0. Since some power sn of s lies in theideal

∑κ∈K Rs

nκ , it follows that sn(u− u′) = 0. This means that u = u′ in R[1/s].It remains to prove the assertion about the stalks. An element of this stalk at xp

is represented by a fraction r/s ∈ R[1/s] with s /∈ p, with the understanding thattwo such fractions r/s ∈ R[1/s] and r′/s′ ∈ R[1/s′] are considered equal if theycoincide in some R[1/s′′] with s′′ /∈ p, where s and s′ divide s′′. This means thatrs′− r′s is annihilated by some power of s′′. This implies that they define the sameelement of Rp. In other words, the stalk at xp embeds in Rp. On the other hand,any element of Rp thus appears as it is represented by a fraction r/s with s /∈ p(and so nonnilpotent).

2. THE SPECTRUM OF A RING AS A LOCALLY RINGED SPACE 87

REMARK 2.5. If in the previous proposition we take s = 1, we find that R =Γ(X,OX) and so no information is lost when we pass from R to the associatedringed space.

EXERCISE 80. Determine the points and the stalks of the spectrum of the localrings K[[t]], K[t](t) (with K a field) and Z(p) (the fractions whose denominator isnot divisible by p).

EXERCISE 81 (Follows up on Exercise 32). Prove that any prime ideal p of thering Z[x] is one the following

(i) the zero ideal,(ii) the ideal generated by a positive degree polynomial whose coefficients

have no common divisor > 1 and which is irreducible in Q[x],(iii) the ideal generated by a prime number p and the cyclotomic polynomial

Φpr ∈ Z[x] for some r ≥ 1).

Conclude that every algebraic integer resp. every element of a finite field deter-mines an element of A1

Z with two such defining the same element if and only if theyhave the same minimum polynomial over their prime field (‘are Galois conjugate’)and that every element under (ii) resp. (iii) is obtained this way. (The spectrum ofZ[x] is called the affine line over the ring of integers and is usually denoted by A1

Z.)

Proposition 2.1 associated to a ring R a ringed space. We expect a ring homo-morphism φ : R′ → R to induce a morphism of ringed spaces. This is indeed thecase.

PROPOSITION 2.6. LetR andR′ be rings and putX := Spec(R), X ′ := Spec(R′).Every ring homomorphism φ : R′ → R defines a continuous map f = Spec(φ) :

X → X ′ and gives rise to a natural sheaf homomorphism of rings φ : OX′ → f∗OXwhose value on X ′, φ(X ′) : R′ = Γ(X ′,OX′)′ → Γ(X ′, f∗OX) = Γ(X,OX) = R, isφ. If we think of φ as a sheaf homomorphism f−1OX′ → OX , then its stalk at x = xpis given by the local sheaf homomorphism R′φ−1p → Rp induced by φ.

Conversely, any pair (f, φ) with f : X → Y a continuous map and φ : OX′ →f∗OX a sheaf homomorphism φ : OX′ → f∗OX such that the induced map on stalksis local, comes from the ring homomorphism R′ → R defined by φ(X ′).

PROOF. For the first assertion, we explain how φ is defined on a basic openU(r′) ⊂ X ′, i.e., we give the homomorphism Γ(U(r′),OX′) → Γ(U(r′), f∗OX).This will determine φ as a sheaf homomorpism. We have Γ(U(r′),OX′) = R′[1/r′].Put r := φ(r′). We have already seen that f−1U(r′) = U(r). The ring homomor-phism R′[1/r′]→ R[1/r] induced by φ can be understood as a ring homomorphism

Γ(U(r′),OX′) = R′[1/r′]→ R[1/r] = Γ(U(r),OX) = Γ(U(r′), f∗OX)

(when r is nilpotent, U(r) = ∅ and the right hand side is the zero ring). We thushave defined a homomorphism φ : OX′ → f∗OX of sheaves of rings. It is clear thatthis homomorphism is on the stalks as asserted.

Suppose now given (f, φ) is as in the proposition and let φ : R′ → R be thering homomorphism that is the value of φ on X ′. The sheaf property implies that

88 3. SCHEMES

for every x = xp ∈ X we have a commutative diagram

R′ = Γ(X ′,OX′)φ−−−−→ Γ(X ′, f∗OX) = Ry y

R′p′ = OX′,x′φp−−−−→ OX,x = Rp

where p′ is the prime ideal of R′ associated to x′ := f(x). The bottom map is localand so the preimage of pRp is p′R′p′ . The commutativity of the diagram implies thatthen φ−1p = p′. In other words, Spec(φ) and f take the same value on x and definethe same map on the stalk at this point. Hence φ induces the pair (f, φ).

Let us say that a locally ringed space is a ringed space whose stalks are localrings and let us agree that a morphism of locally ringed spaces is a morphism ofringed spaces that are locally ringed and which has the additional property that itgives local homomorphisms on the stalks. This defines the category of locally ringedspaces. To any ring R we have associated a locally ringed space (which we denoteSpec(R) by abuse of notation) and we found that ring homomorphims R′ → Rare in bijective correspondence with the morphisms Spec(R)→ Spec(R′) as locallyringed spaces. Thus the spec construction identifies the dual of the category ofrings as a full subcategory of the category of locally ringed spaces. We enlarge thiscategory by taking the full subcategory whose objects are locally like the spectrumof a ring and thus arrive at the central notions of algebraic geometry:

DEFINITION 2.7. An affine scheme is a ringed space that is isomorphic to Spec(R)for some ring R. A quasi-affine scheme is a ringed space that is isomorphic to anopen subset to Spec(R) for some ring R. A scheme is a ringed space that is locally anaffine scheme (hence this is always a locally ringed space). A morphism of schemesis a morphism between such objects in the category of locally ringed spaces.

It is often useful the fix a ‘base scheme’ S and to consider S-schemes, that is,schemes X endowed with a morphism X → S. We often refrain from giving thismorphism a name and write X/S instead. An S-morphism between S-schemes,X/S → Y/S, is then a morphism X → Y whose composite with the given mor-phism Y → S yields the given morphism X → S. But when S = Spec(R), weusually say that X is an R-scheme (rather than a Spec(R)-scheme) and we writeX/R accordingly. This simply means that OX is a sheaf of R-algebras and that theR-morphisms f : X/R→ Y/R are those with the property that the induced homo-morphisms on stalks are R-algebra homomorphisms. In particular, every scheme isa Z-scheme. We also observe that the category of k-prevarieties is a full subcategoryof the category of k-schemes.

3. The proj construction

We describe a way of producing a scheme out of a graded ring, such that therelation between the algebra and the geometry generalizes the way a projectivevariety defines a homogeneous coordinate ring. We start out with a graded ringR = ⊕∞k=0Rk. This has a distinguished ideal, R+ := ⊕∞k=1Rk. We define a schemeProj(R) whose underlying set is the set of homogeneous prime ideals strictly con-tained in R+ (briefly: strict homogeneous prime ideals). Given a homogeneouselement of positive degree d, s ∈ Rd, then the strict homogeneous prime ideals

3. THE PROJ CONSTRUCTION 89

of R which contain s define a subset Zs ⊂ Proj(R). We denote its complementby Us. We have Us ∩ Us′ = Uss′ as usual and so the subsets Us are a basis for atopology on Proj(R). We give Proj(R) with this topology. When s is not nilpo-tent, the monotone union R[1/s]0 := ∪k≥0Rdk/s

k is the subring of R[1/s] of degreezero. This monotone union makes R[1/s]0 a filtered ring: Rdk/sk.Rdl/sl ⊂ Rdkl/skl(think of the case of k[t] filtered by the subspaces of polynomials of degree ≤ k,k = 0, 1, 2 · · · ).

We claim that as a topological space, Us can be identified with Spec(R[1/s]0).For if p ⊂ R is a strict homogeneous prime ideal which does not contain s, thenyou easily check that ∪k≥0pdk/s

k is a prime ideal of R[1/s]0. Conversely, if p′ isa prime ideal of R[1/s]0, then denote by pk ⊂ Rk the set of a ∈ Rk for whichaRdl−k/s

dl ⊂ p′ for l large enough (if this inclusion holds for some l, then it holdsalso for any larger l). This is a homogeneous ideal which does contain s. It is alsoa prime ideal: if a ∈ Rk and a′ ∈ Rk′ are such that aa′Rdm−k−k′/sdm ⊂ p′ forsome m, then by enlarging m we may assume that we can write m = l + l′ suchthat k ≤ dl and k′ ≤ dl′. If neither aRdl−k/sdl ⊂ p′ nor a′Rdl−k′/sdl

′ ⊂ p′, thenchoose r ∈ Rdl−k and r′ ∈ Rdl−k′ such that neither ar/sdl nor a′r′/sdl

′is in p′.

But since their product is in p′ we get a contradiction. You can check that the twoconstructions are each others inverse.

Given s′ ∈ Rd′ , then we have an inclusion of basic open sets Us ⊂ Us′ if andonly s′ divides some positive power of s, and then such an inclusion is covered byan obvious ring homomorphism R[1/s′]0 → R[1/s]0. Hence Proj(R) is in a naturalway a scheme:

PROPOSITION-DEFINITION 3.1. Denote by X the set Proj(R) with the topologyas just defined. The map Us 7→ R[1/s]0 extends to a sheaf OX of rings on X withthe property that its restriction to Us equals Spec(R[1/s]0) as a ringed space. Thismakes X a scheme. If p is a strict homogeneous prime ideal of R, then the stalk ofOX at the point associated to p is the degree zero subring Rp,0 = ∪k≥1Rk/(Rk − pk)(a monotone union) of Rp. A scheme that arises from this construction is called aprojective scheme.

In case A is a ring and we take for R the graded ring A[X0, . . . , Xn] (withdeg(Xi) = 1 for all i), then the associated projective scheme Proj(A[X0, . . . , Xn])is called projective n-space over A and is denoted PnA. In case A = k, this indeedreproduces our Pnk .

algebraic geometry fall 2013looij101/ag2013.pdf · fall 2013 eduard looijenga. rings are always...

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