INTRODUCTION TO ALGEBRAIC GEOMETRY

CHRISTOPHER HACON AND STEFFEN MARCUS

Abstract. These notes are intended for the participants to the Undergraduate SummerCourse: Introduction to Algebraic Geometry held at the University of Utah May 16-27,2016.

Contents

1. Affine varieties and the Zariski topology 32. Morphisms of algebraic varieties 83. Rings and Ideals 94. Modules 135. C-algebras 166. Nakayama’s Lemma 187. Fields 208. Nullstellnsatz 229. The coordinate ring of an affine variety 2310. Morphisms of quasi-affine varieties 2611. Dimension 2612. Projective varieties 2813. Regular functions on quasi-projective varieties 3314. Rational maps 3415. Projective Morphisms are proper 3716. Normality 3817. Conics 4218. Smooth varieties 4419. Intrinsic definition of the tangent space 4620. Resolution of singularities 4721. Smooth projective varieties and generic smoothness 4922. Cubic surfaces 5023. Rational and unirational varieties 5324. Sheaves 5625. Sheaves of modules 5926. Bezout’s Theorem 6427. Riemann-Roch Theorem for Curves 6728. Tensor products 7029. Fiber products 7130. Monomial orders 7331. Monomial ideals 7632. Solutions and hints for the exercises 77

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References 77

Algebraic geometry is the study of solutions of polynomial equations. For example, if

p(x) = a0 + a1 + . . .+ anxn

with ai ∈ R and an 6= 0 is a polynomial of degree n, then it is well known that p(x) hasat most n real solutions. A real solution is any number a ∈ R such that p(a) = 0. To seethat there are at most n solutions, note that if a is a solution then (x− a) divides p(x) andhence p(x) = (x − a)p1(x) where deg p1(x) = n − 1. Repeating this process, one sees thatif a1, . . . , ar are distinct solutions, then (x− a1)(x− a2) · · · (x− ar) divides p(x). But thenn = deg p(x) ≥ r.

Of course it can happen that p(x) has less than deg p(x) solutions. For example if p(x) =x2 + bx + c, then p(x) may have 0, 1, 2 solutions depending on wether the value of b2 − 4cis < 0, 0 or > 0. On the other hand, if we look for complex solutions a ∈ C, then p(x) hasexactly deg p(x) solutions when counted with multiplicity, and so we may write

p(x) = a0(x− a1) · · · (x− an)

where ai ∈ C.It is also natural to look for other kinds of solutions to polynomial equations. Consider

for example the famous equation

xn + yn = zn, n ≥ 2

then it is natural to look for solutions in Z, Q, R or C. Let’s start with the case n = 2:x2 + y2 = z2, then any solution x, y, z ∈ N is a Pythagorean triple. For example (3, 4, 5).Note that looking for solutions in N, Z or Q is essentially the same thing as looking forrational solutions to

x2 + y2 = 1

i.e. rational points on the unit circle. Given a solution a, b, c ∈ N (or ∈ Q), then a/c, b/cis a rational solution of x2 + y2 = 1. Conversely given a rational solution a/c, b/c ∈ Q ofx2 + y2 = 1, then |a|, |b|, |c| ∈ N is a solution to x2 + y2 = z2. It is not hard to see thatx2 + y2 = 1 has infinitely many solutions ∈ Q (cf. Exercise 0.4).

The bahaviour of the solutions to the equation xn+yn = zn when n ≥ 3 is drammaticallydifferent: there are no interesting solutions. More precisely, if a, b, c ∈ N satisfy an+ bn = cn

for some n ≥ 3, then (a, b, c) ∈ (0, 0, 0), (1, 0, 1), (0, 1, 1). This result is the famousFermat’s last Theorem conjectured by the french mathematician Fermat in 1637 (Fermatfamously wrote that he could prove this deceptively simple theorem but it would not fit in themargin of the book). The Theorem was finally proved more than 3 centuries later by AndrewWiles in 1994 after intense effort by many mathematichians. This example illustrates thateven though natural numbers N are the simplest numbers, looking for solutions to polynomialequations that are natural numbers can be very complicated or even hopeless. However, bythe Fundamental Theorem of Algebra, every polynomial p(x ∈ C[x] admits a solution a ∈ Cand this suggests that the problem is more tractable when we look for solutions that arein the complex numbers C. We will adopt this point of view throughout this course. Wewill often be more interested in qualitative results (eg. a degree 2 polynomial can have 0,

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1, 2 real solutions and a polynomial of degree n has exactly n solutions when counted withmultiplicities) rather than finding the exact solution set.

Exercise 0.1. Show that if p(x) ∈ C[x] is a non-zero polynomial and a is a solution of p(x),then (x− a) divides p(x).

Exercise 0.2. Use the FTA (fundamental Theorem of Algebra) to show that if p(x) ∈ C[x]is a non-zero polynomial, then p(x) has exactly deg p(x) complex solutions when countedwith multiplicity.

Exercise 0.3. Show that if a, b ∈ Z is a solution of x2 + y2 = 1, then a2 − b2, 2ab is also asolution. Compute the first few solutions obtained from 3/5, 4/5.

Exercise 0.4. Show that if a, b ∈ N, then [(a2− b2), 2ab, (a2 + b2)] solves x2 + y2 = z2. Usethis to show that the point (0, 1) is an accumulation point of solutions to x2 +y2 = 1. Finallydeduce that the rational solutions to x2 + y2 = 1 are dense in S1 = (x, y)|x2 + y2 = 1.

Exercise 0.5. (Hard) State and prove the FTA (Fundamental Theorem of Algebra).

1. Affine varieties and the Zariski topology

Definition 1.1. An affine algebraic set X ⊂ CN is the set of common zeroes of a set ofpolynomial equations pii∈I where pi ∈ C[x1, . . . , xN ] so that

X = V(pii∈I) = a = (a1, . . . , aN) ∈ CN |pi(a) = 0 for all i ∈ I.

In particular V(pi) denotes the set of points at which pi vanishes. We have the followingexamples:

• ∅ = V(1) ⊂ CN , and CN = V(0).• (a1, . . . , aN) ∈ CN is given by V(x1 − a1, . . . , xN − aN).• N = 1, then the only affine algebraic sets in C are ∅, C and finite union of pointsa1, . . . , ar ⊂ C.• N = 2, then the only affine algebraic sets in C2 are ∅, C2 and finite unions of points

(a, b) ∈ C2 and curves given by zero sets of one polynomial equation (see Exercise11.14). For example lines V(ax+ by+ c), quadrics (given by zero sets of polynomialsof degree 2), cubics (given by zero sets of polynomials of degree 3) etc.• Zero sets of linear equations are called hyperplanes and zero sets of one polynomial

are called hypersurfaces.• The sets a ∈ C| 0 < |a| < 2, 1− 1

n|n ∈ N ⊂ C, R ⊂ C, (a, b) ∈ C2|aa+bb ≤ 1

and (a, exp(a))|a ∈ C ⊂ C2 are all subsets of C or C2 that are not affine algebraicsets.

Exercise 1.2. Prove the last statement above.

Exercise 1.3. Suppose that X = V(p1, . . . , pr) ⊂ Cn. Let

I = (p1, . . . , pr) := r∑i=1

piri|ri ∈ C[x1, . . . , xn].

Show that X = V(p|p ∈ I). We will denote V(p|p ∈ I) by V(I).

Exercise 1.4. Suppose that if ∅ 6= I ⊂ J ⊂ C[x1, . . . , xn], then V(I) ⊃ V(J).3

Definition 1.5. Let I ⊂ C[x1, . . . , xn] be a non empty subset, then I is an ideal if I isclosed under addition and multiplication by arbitrary polynomials so that if i, j ∈ I andr ∈ C[x1, . . . , xn], then

i+ j ∈ I, and ri ∈ I.Exercise 1.6. Show that if pii∈J is a collection of polynomials pi ∈ C[x1, . . . , xn], thenthe set I = (pii∈J) given by all finite linear combinations

∑ripi where ri ∈ C[x1, . . . , xn],

is closed under addition and multiplication by elements of C[x1, . . . , xn] (and hence I ⊂C[x1, . . . , xn] is an ideal; we say that I is the ideal generated by pii∈J).

Notice that the polynomial equations determining an affine set are not uniquely deter-mined. For example V(x) = V(x2) = V(x3) . . . ⊂ C and V(x, y) = V(x + y, x − y) ⊂ C2.However given an affine algebraic set (or more generally any subset) X ⊂ Cn, we mayconsider

I(X) := p ∈ C[x1, . . . , xn]|p(x) = 0 ∀x ∈ X.For example I(V(x2)) = (x) ⊂ C[x].

Exercise 1.7. Show that I(V(pp∈J)) ⊃ (pp∈J).

Exercise 1.8. Show that if X ⊂ Cn is a subset, then I(X) ⊂ C[x1, . . . , xn] is an ideal.

It is not hard to see that any affine set X ⊂ Cn can be defined by finitely many polyno-mials.

Theorem 1.9. Let X ⊂ Cn be an affine algebraic set, then there exists finitely manypolynomials p1, . . . , pr ∈ C[x1, . . . , xn] such that

X = V(p1, . . . , pr).

Proof. Let X = V(pp∈I) ⊂ Cn be an affine algebraic set. We may assume that I = I(X).We must show that X is defined by finitely many polynomial equations. By Exercise 1.8we may assume that I is and ideal (i.e. it is closed under addition and multiplication byarbitrary polynomials). We will proceed by induction on n.

Suppose that n = 1, i.e. pi ∈ C[x]. Pick q ∈ I an element of minimal degree. We claimthat (q) = I. To see this, pick any element p ∈ I and write p = qs + r where s, r ∈ C[x]and deg r < deg q (long division). Since p, q ∈ I, then r = p− qs ∈ I (I is an ideal). Sincedeg r < deg q, we have r = 0 proving the claim.

Suppose now that the claim holds in C[x1, . . . , xn−1] and consider pi ∈ C[x1, . . . , xn] =C[x1, . . . , xn−1][xn] as a polynomial in the variable xn with coefficients in C[x1, . . . , xn−1].Let qi ∈ C[x1, . . . , xn−1] be the leading coefficient so that

pi = qixdin + (terms of lower degree in xn).

In particular the degree of pi with respect to xn is di. Now pick p1 ∈ I of minimal degreewith respect to xn, then inductively pick pi ∈ I \ (p1, . . . , pi−1) of minimal degree withrespect to xn. Let qi ∈ C[x1, . . . , xn−1] be the corresponding leading terms. Consider thesequence of ideals

(q1) ⊂ (q1, q2) ⊂ (q1, q2, q3) ⊂ . . . ⊂ Q = (qi|i ∈ N) ⊂ C[x1, . . . , xn−1].

By our inductive assumption, Q is generated by finitely many elements and soQ = (q1, . . . , qk)for some k > 0. We claim that I = (p1, . . . , pk). To see this, consider pk+1. Then, the degreewith respect to xn, satisfies

dk+1 = deg pk+1 ≥ dk = deg pk ≥ dk−1 = deg pk−1 ≥ . . . ≥ d1 = deg p1.4

But then qk+1 =∑k

i=1 siqi for some si ∈ C[x1, . . . , xn−1] and so

p′ := pk+1 −k∑i=1

sipixdk+1−din ∈ I

has degree < dk+1 (with respect to xn). But then p′ ∈ (p1, . . . , pk). But this easily implies

that pk+1 = p′ +∑k

i=1 sipixdk+1−dii ∈ (p1, . . . , pk) which is impossible.

Corollary 1.10. Let Zi|i ∈ I be affine algebraic sets, then Z = ∩i∈IZi is an affine setand in fact there exists a finite subset I0 ⊂ I such that Z = ∩i∈I0Zi .

Proof. Assume for simplicity that Zi = V(pi), then Z = V(pi|i ∈ I). By Theorem 1.9,there is a finite subset I0 ⊂ I such that Z = V(pi|i ∈ I0). Thus Z = ∩i∈I0V(pi).

Exercise 1.11. Show by example that the arbitrary union of algebraic sets is not in generalan algebraic set.

Next we will show that affine algebraic sets are the closed subset of the Zariski topologyon CN . Recall that

Definition 1.12. A topology on a space (or set) X is a collection of subsets (called opensubsets) Ω = Uii∈I such that Ui ⊂ X and

(1) ∅, X ∈ Ω,(2) finite intersections of open subsets are open: Ui1 ∩ . . . ∩ Uir ∈ Ω, and(3) arbitrary unions of open subsets are open: for any J ⊂ I we have ∪j∈JUj ∈ Ω.

The closed subsets are the complements of open subsets.

Exercise 1.13. Show that Ω is a topology on X if and only if ∅, X are closed, the union oftwo closed subsets is closed and arbitrary intersections of closed subsets are closed.

Exercise 1.14. The open subset of the (usual) Euclidean topology on Rn are the arbitraryunions of open balls (an open ball of radius r > 0 centered at a1, . . . , an is just the set(b1, . . . , bn) ∈ Rn|

∑(bi − ai)2 < r2). Show that this defines a topology on Rn.

Exercise 1.15. Show that the subsets of C given by ∅,C and the complements of finitelymany points, define a topology on C.

Exercise 1.16. Given any (non-empty) set X, show that one can consider the two followingtopologies:

(1) Ω consists of all subsets of X,(2) Ω = ∅, X.

Exercise 1.17. Show that V(p) ∪ V(q) = V(pq) as subsets of Cn thus the union of twohypersurfaces is a hypersurface.

Exercise 1.18. Show that V(p) ∩V(q) = V(p, q) thus the intersection of two hypersurfacesin CN is an affine algebraic set (but may not be a hypersurface).

We have the following important fact.

Theorem 1.19. Let Ω be the set of all complements of all algebraic sets in CN , then Ωdefines a topology (the Zariski topology) on CN .

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Proof. Since ∅,Cn are algebraic sets (as observed above), then their complements Cn, ∅ areopen subsets.

Suppose that V(p1, . . . , pr) and V(q1, . . . , qs) are closed subsets of CN where pi, qj ∈C[x1, . . . , xN ], then

V(p1, . . . , pr) ∪ V(q1, . . . , qs) = V (piqj|1 ≤ i ≤ r, 1 ≤ j ≤ s).

To see the inclusion ⊂, note that if a = (a1, . . . , an) ∈ V(p1, . . . , pr), then pi(a) = 0 for1 ≤ i ≤ r and so (piqj)(a) = pi(a)qj(a) = 0 for 1 ≤ i ≤ r, 1 ≤ j ≤ s so that a ∈ V(piqj|1 ≤i ≤ r, 1 ≤ j ≤ s). Thus we have shown that V(p1, . . . , pr) ⊂ V(piqj|1 ≤ i ≤ r, 1 ≤ j ≤ s).The inclusion V(q1, . . . , qs) ⊂ V(piqj|1 ≤ i ≤ r, 1 ≤ j ≤ s) is similar. To see the inclusion⊃, suppose that a 6∈ V(p1, . . . , pr) ∪ V(q1, . . . , qs), then there exists i, j such that pi(a) 6= 0and qj(a) 6= 0. Thus (piqj)(a) = pi(a)qj(a) 6= 0 so that a 6∈ V(piqj|1 ≤ i ≤ r, 1 ≤ j ≤ s).

Suppose that Vjj∈J is an arbitrary collection of affine algebraic sets in Cn, then eachVj is defined by a collection of polynomials pj ⊂ C[x1, . . . , xn]. But then

∩j∈JVj = V(pjj∈J).

Thus arbitrary intersection of closed subsets is closed.

Remark 1.20. More generally, if X ⊂ Cn is an algebraic set, then we define the Zariskitopology by letting the closed subsets of X corresponding to the intersection of closed sub-sets of Cn with X (so these closed sets are the sets defined by the vanishing of polynomialequations). It is immediate from Theorem 1.19 that this is in fact a topology.

Exercise 1.21. Show that if X ⊂ CN is an affine algebraic set (these are the closed sets inthe Zariski topology), then it is closed in the Euclidean topology. Therefore the Euclideantopology has ”more” closed (and open) subsets than the Zariski topology. we say that theEuclidean topology is finer than the Zariski topology.

Definition 1.22. An algebraic set X ⊂ Cn is reducible if it can be written as the unionof two algebraic sets X = X1 ∪X2 such that X1 6⊂ X2 and X2 6⊂ X1. If X is not reducible,then we say that X is irreducible. Irreducible affine algebraic sets will be called affinevarieties.

We notice the following:

(1) Cn is irreducible.(2) V(xy) ⊂ C2 is reducible.(3) A hypersurface V(p) ⊂ Cn is irreducible if and only if p = qn for some irreducible

polynomial q ∈ C[x1, . . . , xn]. In this case I(V(p)) = q.(4) If f : X → Y is a surjective continuous map and Y is reducible, then X is reducible.

(Recall that a map is continuous if inverse images of open subsets are open.)

Definition 1.23. Let X ⊂ Cn be an algebraic variety, then the dimension of X is themaximum length of a chain of irreducible non-empty closed subsets X0 ⊂ X1 ⊂ . . . ⊂ Xn ⊂X. In this case we say that dimX = n.

We will see later that affine varieties are finite dimensional and dimCn = n.

Exercise 1.24. Show that Cn is irreducible.

Exercise 1.25. Show that dimCn ≥ n.

Exercise 1.26. Use the remarks after Definition 1.1 to show that dimC2 = 2.6

Definition 1.27. A topological space X is Noetherian if any decreasing sequence of closedsubsets

X ⊃ X1 ⊃ X2 ⊃ . . .

is eventually constant (so that Xj = ∩i∈NXi for all j 0).

We will see later that affine varieties are Noetherian topological spaces.

Proposition 1.28. let X be a Noetherian topological space, then any closed subset V ⊂ Xis a finite union of irreducible closed subsets

V = V1 ∪ . . . ∪ Vr.

If moreover, we require that there are no inclusions Vi ⊂ Vj for i 6= j, then the Vi areuniquely determined.

Proof. See [Hartshorne, 1.5].

Lemma 1.29. If X ⊂ Cn is an algebraic set and f ∈ C[x1, . . . , xn], then X \ V(f) is analgebraic set.

Proof. Let a = I(X) ⊂ C[x1, . . . , xn], and consider the ideal

a = (a, x0f − 1) ⊂ C[x0, x1, . . . , xn]

and the corresponding affine variety Xf = V(a) ⊂ Cn+1. The projection Cn+1 → Cn

defined by (x0, x1, . . . , xn) → (x1, . . . , xn) gives a bijection Xf → X \ V(f). (In fact, if(a0, . . . , an) ∈ V(a), then clearly (a1, . . . , an) ∈ V(a) and since 1 = a0f(a1, . . . , an), thenf(a1, . . . , an) 6= 0 and conversely if (a1, . . . , an) ∈ V(a) and f(a1, . . . , an) 6= 0, then leta0 = 1/f(a1, . . . , an) so that (a0, . . . , an) ∈ V(a)). In order to complete the proof, it sufficesto check that the induced bijections X \V(f)→ Xf and Xf → X \V(f) are continuous.

Remark 1.30. We can think of the process of realizing the complement of a hypersurfacein X as a hypersurface in an algebraic set of dimension one more. Another perspective isgiven by localization of rings (see Section 3).

Exercise 1.31. Show that the induced bijections X \ V(f) → Xf and Xf → X \ V(f) arecontinuous. (See Lemma 2.2.)

We say that the open subsets Xp where p ∈ C[x1, . . . , xn] are principal open subsets.Next we show that the principal open subsets generate the Zariski topology.

Lemma 1.32. let U ⊂ X ⊂ Cn be an open subset of an affine variety, then U is a finiteunion of principal open subsets U = ∪Xpi where p ∈ C[x1, . . . , xn].

Proof. Let Z = X \ U , then since Z is closed, for any x ∈ U , there is a polynomialpx ∈ I(Z) ⊂ C[x1, . . . , xn], such that px(x) 6= 0. It is then clear that U = ∪x∈UXpx . Notethat Z = ∩X \Xpx is a closed subset and hence there exists a finite subset Xi = Xpi suchthat Z = ∩X \Xpi (see Corolary 1.10) or equivalently U = ∪ri=1Xi.

Exercise 1.33. Let V1 = V(x + 2y + 3), V2 = V(3x− 2y + 9) ⊂ C2. Compute V1 ∪ V2 andV1 ∩ V2.

Exercise 1.34. Write C2 \ (0, 0) as a union of two principal open subsets of C2.7

Exercise 1.35. Show that the Euclidean topology on R2 is induced by the product of theEuclidean topologies on R1 (every open subset for the Euclidean topology on R2 is a unionof products U1 × U2 where U1, U2 are open subsets in R).

Exercise 1.36. Show that the Zariski topology on C2 is not the product of the Zariskitopologies on C.

Exercise 1.37. Let Cn = (1, t, t2, . . . , tn)|t ∈ C ⊂ Cn+1 be the rational normal curve ofdegree n. Show that Cn is defined by the 2× 2 minors of the matrix with rows (x0, . . . , xn−1)and (x1, . . . , xn).

Exercise 1.38. Use Corollary 1.10 to show that if X is an affine variety and Uii∈I is acollection of open subsets Ui ⊂ X such that ∪i∈IUi = X, then there is a finite subset I0 ⊂ Isuch that X = ∪i∈I0Ui. In other words X is compact.

Exercise 1.39. Show that if X is an irreducible subset of a topological space W , then theclosure X of X in W (defined as the intersection of all closed subsets containing X) isirreducible.

Exercise 1.40. (Hard) Show that a subset X ⊂ Cn is compact in the Euclidean topology(meaning that if Uii∈I is an open cover of X, then there is a finite subset I0 ⊂ I suchthat Uii∈I0 is an open cover of X) if and only if X is bounded and any sequence of pointsxi ∈ X admits a convergent subsequence xik whose limit belongs to X i.e. limxik ∈ X.

2. Morphisms of algebraic varieties

We will now discuss maps between affine algebraic sets. The most natural maps to considerare the ones induced by polynomials: given P = (p1, . . . , pm) a collection of polynomialspi ∈ C[x1, . . . , xn] we define the morphism of affine spaces

F : Cn → Cm, a = (a1, . . . , an)→ (p1(a), . . . , pm(a)).

Given affine varieties X ⊂ Cn and Y ⊂ Cm such that F (X) ⊂ Y , then we say thatf = F |X : X → Y is a morphism of affine varieties.

Exercise 2.1. Show that the composition f g of two morphisms of affine varieties g : X →Y and f : Y → Z is a morphism of affine varieties.

Lemma 2.2. Morphisms of algebraic varieties are continuous in the Zariski topology.

Proof. Recall that a map is continuous if the inverse image of every open subset is anopen subset or equivalently the inverse image of every closed subset is a closed subset. Letf : X → Y be a morphism of algebraic varieties given by a morphism of affine spacesF = (p1, . . . , pm) : Cn → Cm. If V ⊂ Y is a closed subset, then V = V(qi|i ∈ I) is the zeroof a collection of polynomials qi ∈ C[y1, . . . , ym]. The inverse image of V is

f−1(V ) = V(qi(p1(x1, . . . xn), . . . , pm(x1, . . . xn))|i ∈ I)

which is clearly a closed subset.

Definition 2.3. Let X ⊂ Cn and Y ⊂ Cm, then we say that X and Y are isomorphic ifthere are morphisms f : X → Y and g : Y → X such that f g = idY and g f = idX .

Exercise 2.4. Show that V(x2 − y, x3 − z) ⊂ C3 is isomorphic to C.8

Exercise 2.5. Show that the projection of V(y = x2) on to the x axis is an isomorphismbut the projection on to the y axis is not an isomorphism.

Exercise 2.6. (Hard) Show that V(x2 − y3) is not isomorphic to C, although they arehomeomorphic. (Recall that two sets X, Y are homeomorphic if there is a bijection f :X → Y such that f and f−1 are continuous.)

Exercise 2.7. Show that V(xy) is not isomorphic to C.

Exercise 2.8. Show that V(xy− 1) ⊂ C2 is homeomorphic to C∗ but not isomorphic to C∗(as affine varieties; in fact C∗ is not an affine subvariety of C. We will see later that theyare isomorphic as quasi-projective varieties).

Exercise 2.9. Show that C2 is not isomorphic to C.

Exercise 2.10. Show that dim is an isomorphism invariant.

Exercise 2.11. Give an example of a morphisms of algebraic varieties f : X → Y suchthat f(X) ⊂ Y is not a closed subset.

3. Rings and Ideals

In this section we will re-interpret the definitions and results of the previous sections inmore algebraic terms. We begin by recalling the following algebraic definitions and results.

Definition 3.1. A commutative ring with identity (R,+, ∗) is a nonempty set R withtwo operations

+ : R×R→ R and ∗ : R×R→ R

(addition and multiplication) such that (R,+) is an abelian group (i.e. for any a, b, c ∈ Rwe have (a+ b) + c = a+ (b+ c), there exists an element (the additive identity) 0 ∈ R suchthat a + 0 = 0 + a = a, there exists additive inverses −a ∈ R such that a + (−a) = 0, anda + b = b + a), multiplication is associative, commutative and has a multiplicative identity1 6= 0 (a ∗ (b ∗ c) = (a ∗ b) ∗ c, a ∗ b = b ∗ a and 1 ∗ a = a ∗ 1) and the distributive law holds(a ∗ (b+ c) = a ∗ b+ a ∗ c).

Since the rings that we consider will almost always be commutative with identity, we willjust refer to these as rings. A typical example of a ring is the polynomial ring in n variablesC[x1, . . . , xn]. Other examples are Z,R,Q,C and the corresponding polynomial rings.

Exercise 3.2. Show that the set of n× n matrices with complex coefficients Mn×n(C) is anon-commutative ring. Its identity is the identity matrix (i.e. the matrix I defined by theentries Ii,j = 0 if i 6= j and Ii,i = 1).

Definition 3.3. If I ⊂ R is a non-empty subset of a ring, then we say that I is an idealif it is closed under addition, additive inverses (so that I ⊂ R is an additive subgroup) andri ∈ I for all i ∈ I and r ∈ R.

Exercise 3.4. Given r ∈ R, (r) := sr|s ∈ R ⊂ R is an ideal. These ideals are known asprincipal ideals.

Exercise 3.5. Let I, J be ideals. Show that I ∩J and I+J = i+ j|i ∈ I, j ∈ J are ideals.

Exercise 3.6. Let I1 ⊂ I2 ⊂ . . . be an increasing sequence of ideals. Show that ∪i>0Ij is anideal.

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Exercise 3.7. Let I ⊂ R be a subset containing 0 and 1 such that if a, b ∈ I, then sa+tb ∈ Ifor any s, t ∈ R, then I is an ideal.

The importance of ideals is given by the fact that we can quotient rings by ideals toobtain new rings. This is acomplished as follows: Let I ⊂ R be an ideal, then we may definethe equivalence relation r ∼ r′ if and only if r − r′ ∈ I. It is easy to check that this is anequivalence relation with equivalence classes

R/I = r + I|r ∈ R.We can then define addition and multiplication on R/I via

(r + I) + (s+ I) = (r + s) + I, (r + I) ∗ (s+ I) = (r ∗ s) + I.

Exercise 3.8. Show that + and ∗ defined above, are well defined (i.e. if r+ I = r′ + I ands+I = s′+I, then (r+I)+(s+I) = (r′+I)+(s′+I) and (r+I)∗(s+I) = (r′+I)∗(s′+I)).

Theorem 3.9. Let I ⊂ R be an ideal, then R/I with addition and multiplication defined asabove is a ring.

Proof. Exercise.

Exercise 3.10. Let m ∈ N be a positive integer. Show that Z/(m) is a ring.

Exercise 3.11. Let R be a ring and rii∈I a collection of elements of R, then (rii∈I) =∑ri1si1 + . . .+ riksik |si ∈ R is the smallest ideal containing rii∈I .

Exercise 3.12. Show that if I is an ideal in C[x], then I = (p(x)) for some p(x) ∈ C[x].

Exercise 3.13. Show that if I is an ideal in Z, then I = (k) for some k ∈ Z.

Exercise 3.14. Show that there are ideals in C[x, y] such that I is not generated by oneelement.

Exercise 3.15. Show that there are ideals of Z[x] that are not generated by one element.

Most rings that we will consider are Noetherian. This means that for any ideal I ⊂ R,there exist finitely many elements r1, . . . , rk ∈ R such that I = (r1, . . . , rk). We have thefollowing:

Lemma 3.16. Let R be a ring then the following are equivalent.

(1) R is Noetherian.(2) If rii∈N is a sequence of elements of R, then there exists k > 0 such that (rii∈N) =

(r1, . . . , rk)(3) If I1 ⊂ I2 ⊂ . . . is an increasing sequence of ideals, then Ik = I = ∪j>0Ij for all

k 0 (or equivalently the sequence stabilizes so that there exists k0 such that Ik = Ik0

for all k ≥ k0).

Proof. (2) implies (1): If (1) does not hold, then we can find a sequence of elements ri ∈ Isuch that the inclusions

(r1, . . . , rk) ⊂ (r1, . . . , rk+1)

are strict for all k > 0. By (2), this is impossible.(1) implies (3): Suppose that R is Noetherian and let I = ∪j>0Ij. Let I = (r1, . . . , rk),

then each ri belongs to some Iji and hence I = Ij1 ∪ . . . ∪ Ijk .10

(3) implies (2): Let Ii = (r1, . . . , ri), then Ii ⊂ Ii+1 . . . and so by (3) Ik = ∪Ii so that(r1, . . . , rk) = (rii∈N).

Theorem 3.17. Let R be a ring (commutative with 1). If R is Noetherian, then so is R[x].

Proof. Exercise (follow the arguments in the proof of Theorem 1.9).

Definition 3.18. φ : R→ S is a homomorphism of rings if R, S are rings and φ(a+b) =φ(a)+φ(b) and φ(a∗b) = φ(a)∗φ(b). The kernel of φ is the subset ker(φ) = r ∈ R|φ(r) =0. φ is injective if ker(φ) = 0 and φ is surjective if φ(R) = S.

Exercise 3.19. Show that ker(φ) ⊂ R is an ideal and φ(R) ⊂ S is a subring such thatφ(R) ∼= R/ker(φ).

Exercise 3.20. Show that if I ⊂ φ(R) is an ideal, then φ−1(I) ⊂ R is an ideal.

Exercise 3.21. Show that if φ : R→ S is surjective and I ⊂ R is an ideal, then φ(I) ⊂ Sis an ideal. Show by example that φ being surjective is necessary.

Exercise 3.22. Show that the ideals of φ(R) are in one to one correspondence with theideals of R that contain K = ker(φ).

Exercise 3.23. Show that if R is Noetherian, then so is φ(R).

Exercise 3.24. Show that if R′ is a finitely generated ring over R (meaning that there is asurjective homomorphism R[x1, . . . , xn]→ R′) and R is Noetherian, then so is R′.

Definition 3.25. An ideal P ⊂ R is prime if given x, y ∈ R such that x ∗ y ∈ P , thenx ∈ P or y ∈ P .

Exercise 3.26. Show that if m ≥ 2, then (m) ⊂ Z is prime if and only if m is a primenumber.

Exercise 3.27. Show that if f ∈ C[x1, . . . , xn], then (f) ⊂ C[x1, . . . , xn] is prime if and onlyif f is irreducible (i.e. it can not be written as the product of two non constant polynomials).

Exercise 3.28. If f : A → B is a ring homomorphism and P ⊂ B is a prime ideal, thenshow that f−1(P ) ⊂ A is also a prime ideal.

Definition 3.29. An ideal m ⊂ R is maximal if given any other ideal m ⊂ I ⊂ R theneither m = I or I = R.

Exercise 3.30. Show that the ideals (x1 − a1, . . . , xn − an) ⊂ C[x1, . . . , xn] are maximal(here ai ∈ C).

Exercise 3.31. Show by example that if f : A→ B is a ring homomorphism and m ⊂ B isa maximal ideal, then f−1(m) ⊂ A may not be maximal.

Definition 3.32. Let R be a ring (commutative with 1), then R is a domain if givenr, s ∈ R, then r ∗ s = 0 implies r = 0 or s = 0 (in other words R has no zero divisors).

Exercise 3.33. Let I ⊂ R be an ideal, then I is prime if and only if R/I is a domain.

Definition 3.34. Let R be a ring (commutative with 1), then R is a field if given 0 6= r ∈ R,there exists s ∈ R such that r ∗ s = 1 (we will say that r is invertible and denote s = r−1).

11

Exercise 3.35. Show that multiplicative inverses are unique (if they exist)

Exercise 3.36. Show that any field is a domain.

Exercise 3.37. Let I ⊂ R be an ideal, then I is maximal if and only if R/I is a field.(I.e. show that any element r+ I 6= 0 + I is invertible in R/I; see §5 for the definition andproperties of fields.)

Definition 3.38. Let I ⊂ R be an ideal, then the radical of I is√I = f ∈ R|fm ∈ I, for some m > 0.

Exercise 3.39. Show that√I ⊂ R is an ideal.

Exercise 3.40. Show that R/I has no nilpotent elements (i.e. no elements 0 6= f ∈ R/Isuch that fm = 0 for some m > 0) if and only if I is a radical ideal (i.e.

√I = I).

Lemma 3.41. If I ⊂ R is an ideal, then there is a 1-1 correspondence between the idealsof R/I and the ideals of R that contain I.

Proof. Exercise.

Definition 3.42. A subset of a ring 1 ∈ S ⊂ R is a multiplicative subset if for anys, t ∈ S we have st ∈ S. If R is a domain and 0 6∈ S, then we define

S−1R := (a, s)|a ∈ R, s ∈ S/ ∼where ∼ is the equivalence relation defined by (a, s) ∼ (a′, s′) iff as′ = a′s.

Exercise 3.43. Show that (a, s) ∼ (a′, s′) if and only if as′ = a′s is an equivalence relation.We denote the equivalence class of (a, s) by a/s.

Lemma 3.44. Define (a, s) + (a′, s′) = (as′ + a′s, ss′) and (a, s)(a′, s′) = (aa′, ss′), thenS−1R is a ring.

Proof. Exercise.

Note that there is a natural inclusion of rings R→ S−1R.

Exercise 3.45. Let I ⊂ S−1R be an ideal, then I ∩ R is also an ideal and if J ⊂ R is anideal, then S−1J ⊂ S−1R is an ideal.

Exercise 3.46. Show that if I ⊂ S−1R is an ideal, then I = (I ∩ R)S−1R and so the mapI → I ∩R is an injection from the set of all ideals in S−1R to the set of ideals in R. Showthat this injection takes prime ideals to prime ideals.

Exercise 3.47. Show that if I ⊂ R is an ideal, then I = J ∩ R for some ideal J ⊂ S−1Rif and only if I = (IS−1R) ∩ R. Note that if I ∩ S 6= ∅, then (IS−1R) ∩ R = R and so wehave a bijection between the ideals of S−1R and the ideals I ⊂ R such that I ∩ S = ∅.

Exercise 3.48. Let R be a domain and S a multiplicative subset. Show that if R is Noe-therian, then so is S−1R.

Exercise 3.49. Show that if P ⊂ R is a prime ideal, then S = R \P is a multiplicative set.We denote S−1R = RP .

Exercise 3.50. Show that if m ⊂ R is the unique maximal ideal, then every element ofR \m is invertible.

12

Exercise 3.51. Show that PRP is the unique maximal ideal of RP .

Exercise 3.52. Let S ⊂ T be multiplicative subsets of a domain R. Show that there is anatural inclusion S−1R ⊂ T−1R and an isomorphism T−1(S−1R) ∼= T−1R.

Definition 3.53. A ring R is local if it has a unique maximal ideal. In particular if P ∈ Ris a prime ideal in a domain, then RP is a local ring.

By 3.50, we know that a ring R with a maximal ideal m is local if and only if everyelement of R \m is invertible.

Exercise 3.54. Let f ∈ R be a non zero element in a domain, then S = (f) is a multi-plicative subset and we denote S−1R by Rf . Show that the ideals of Rf are in one to onecorrespondence with ideals I ⊂ R such that I ∩ (f) = ∅.

Proposition 3.55. Let R be a domain and Q(R) its field of fractions, then

R = ∩PRP = ∩mRm

where P ⊂ A are prime ideals (and m ⊂ A are prime ideals).

Proof. The inclusions R ⊂ ∩PRP ⊂ ∩mRm are immediate. Suppose that z ∈ ∩mRm \R. LetI = r ∈ R| rz ∈ R, then I is a proper ideal of R (since 1 6∈ I) and so I ⊂ m for somemaximal ideal m of R. But then z 6∈ Rm since otherwise z ∼ r/s where s 6∈ m and so sz = ri.e. s ∈ I ⊂ m which is the required contradiction.

Remark 3.56. If R is a ring (not necessarily a domain), one can define the localizationwith respect to a multiplicative subset 1 ∈ S ⊂ R by using the equivalence relation

(r, s) ∼ (r′, s′) iff t(rs′ − r′s) = 0

for some t ∈ S. Addition and multiplication are defined in the same way.

Exercise 3.57. Show that this is an equivalence relation and multiplication and additionare well defined.

Exercise 3.58. Show that if φ : R → R′ is a homomorphism of rings and S ⊂ R amultiplicative subset such that φ(s) ∈ R′ is a unit for all s ∈ S, then φ extends uniquely toa homomorphism of rings S−1R→ R′.

Exercise 3.59. Show that the natural map R → S−1R is not injective (this can happenonly when R is not a domain).

4. Modules

Modules over rings are a generalization of vector spaces over a field. The formal definitionis as follows:

Definition 4.1. Let (M,+) be an abelian group and R a ring and R×M →M an operationdenoted by (r,m) → rm ∈ M , then M is a (left) R-module if the following propertieshold.

(1) 1 ·m = m for all m ∈M ,(2) (r + s)m = rm+ sm for all r, s ∈ R, m ∈M ,(3) (rs)m = r(s(m)) for all r, s ∈ R, m ∈M , and(4) r(m+m′) = rm+ rm′ for all r ∈ R, m,m′ ∈M .

13

Exercise 4.2. Show that if R is a field and M is an R-module, then M is an R-vectorspace.

Exercise 4.3. Show that every abelian group is a Z-module.

Exercise 4.4. Show that if I ⊂ R is an ideal, then I is an R-module and R/I is anR-module.

Exercise 4.5. Show that if S ⊂ R is a multiplicative subset of a domain, then S−1R is anR-module. More generally if M is an R module, define S−1M similarly to S−1R and showthat S−1M is both an R and an S−1R module.

Exercise 4.6. Let R be a ring, show that R[x1, . . . , xn] is an R-module.

Definition 4.7. Let M be an R-module and N ⊂ M a subgroup, then N is a sub R-module if rn ∈ N for all n ∈ N and r ∈ R. Let M/N be the quotient of abelian groups,then M/N is naturally an R-module.

Exercise 4.8. If N ⊂M , L ⊂M are submodules, then L ∩N ⊂M is a submodule.

Exercise 4.9. If N ⊂ M is a submodule, then show that the submodules of M/N are inone to one correspondence with the submodules N ⊂ L ⊂M .

Definition 4.10. Let M,N be R modules. A homomorphism of R-modules f : M → Nis a map such that

f(rm+ sm′) = rf(m) + sf(m′) ∀ r, s ∈ R, m,m′ ∈M.

If f is bijective, we say that f is an isomorphism of modules. The kernel of f is theset ker(f) = m ∈ M |f(m) = 0 ∈ N. It is a submodule of M . The image of f is the setim(f) = n ∈ N |n = f(m) for some m ∈M. It is a submodule of M .

Exercise 4.11. Show that the kernel and image of a homomorphism of R-modules is anR-module.

If N ⊂ M is a submodule, then M → M/N is a homomorphism of modules. It is easyto see that if f : M → N is a homomorphism of R-modules, then we have a short exactsequence of R modules

0→ ker(f)→M → im(f)→ 0.

As usual, this means that the first map is injective, the last one is surjective and ker(M →im(f)) = im(ker(f)→M).

Exercise 4.12. Show that if 0 → M ′ → M → M ′′ → 0 is a short exact sequence ofR-modules, then M ′′ ∼= M/M ′.

Lemma 4.13. Let 0 → M1 → M2 → M3 → 0 and 0 → M ′1 → M ′

2 → M ′3 → 0 be short

exact sequences of R modules and fi : Mi → M ′i be compatible homomorphisms (so that

the induced maps Mi → Mi+1 → M ′i+1 and Mi → M ′

i → M ′i+1 coincide). If f1 and f3 are

isomorphisms, then so is f2.

Proof. We must show that f2 is injective and surjective. If f2(m) = 0 for some m ∈M2, thenthe image in M ′

3 is 0, and since f3 is an isomorphism, the image of m in M3 is also 0 and byexactness of the first sequence, m is the image of an element m1 in M1. Let m′1 = f1(m1),then the image of m′1 in M ′

2 is f2(m) = 0. But since M ′1 → M ′

2 is injective, m′2 = 0 andhence m2 = 0 but then m = 0 and so f2 is injective.

14

The proof that f2 is surjective is similar and we leave it as an exercise. This line ofreasoning is called diagram chasing (for obvious reasons).

Definition 4.14. A Noetherian module is a module M such that for any subsequence ofmodules N1 ⊂ N2 ⊂ N3 ⊂ . . . ⊂ M we have Ni = ∪j∈NNj for all i 0. We say that Msatisfies the ascending chain condition.

Exercise 4.15. Show that a ring R is Noetherian if and only if it is a Noetherian R-module.

Exercise 4.16. Show that C[x] is not a Noetherian C module (but of course it is a Noe-therian C[x] module).

Definition 4.17. Let M be an R-module satisfying the descending chain condition,then we say that M is an Artinian module. Equivalently if M ⊃ N1 ⊃ N2 ⊃ . . ., thenNi = ∩j∈NNj for all j 0.

Exercise 4.18. Show that if 0 → M ′ → M → M ′′ → 0 is a short exact sequence ofR-modules such that M ′,M ′′ are Noetherian, if and only if so is M .

Definition 4.19. An R-module M is finitely generated if there are elements m1, . . . ,mn ∈M such that any element m ∈M can be expressed as a linear combination m =

∑rimi with

ri ∈ R. In other words, the homomorphism φ : R⊕n →M given by φ(r1, . . . , rm)→∑rimi

is surjective.

Exercise 4.20. Show that a submodule N ⊂ M of a finitely generated Noetherian moduleis also finitely generated.

Exercise 4.21. Show that M is Noetherian if and only if every submodule N ⊂M is finitelygenerated.

Exercise 4.22. Give an example of a homomorphism φ : M → M of finitely generated Rmodules with φ injective but not surjective. (Note that if φ is surjective, then it is injective.See Exercise 6.10.)

Exercise 4.23. Give an example of a non Notherian finitely generated module (hint let Rbe a polynomial ring with infinitely many variables).

Theorem 4.24. Let M be a finitely generated R-module. If R is a Noetherian ring, thenM is a Noetherian R-module.

Proof. Let φ : R⊕n →M be the corresponding homomorphism. If N1 ⊂ N2 ⊂ N3 ⊂ . . . ⊂Mis an increasing sequence of submodules, then so is f−1(N1) ⊂ f−1(N2) ⊂ f−1(N3) ⊂ . . . ⊂R⊕n. It suffices to show that f−1(Ni) = ∪j∈Nf−1(Nj) for any i 0. Therefore we mayassume that M = R⊕n.

We now reason by induction on n. If n = 1, the statement is clear. In general we havea homomorphism g : R⊕n → R⊕n−1. Let N ′′j = g(Nj) and N ′j = ker(Nj → N ′′j ). Therefore,we have a short exact sequence 0 → N ′j → Nj → N ′′j → 0. By our inductive assumptionN ′i = N ′ := ∪j∈NN ′j and N ′′i = N ′′ := ∪j∈NN ′′j for all i 0. But then we have short exactsequences 0→ N ′ → Ni → N ′′ → 0 for all i 0 and applying Lemma 4.13 to the inclusionNi ⊂ Ni+1, one sees that Ni = Ni+1 for all i 0 and the result follows.

Corollary 4.25. If R is Noetherian, then M is Noetherian if and only if it is a finitelygenerated R module.

15

Proof. Exercise.

Theorem 4.26. Let M be a finitely generated Z module, then M ∼= Zr0 ⊕ Z/pr11 Z ⊕ . . . ⊕Z/prtt Z where pi are primes and ri ∈ N. (More generally the same result holds if R is a PIDand M is a finitely generated R module.)

Proof. Since M is finitely generated, there is a surjection f : Z⊕n → M with kernel K =ker(f). By Exercise 4.20, K is also finitely generated and so there is a surjection Z⊕m → K.Let g : Z⊕m → Z⊕n be the induced homomorphism, then M is the cokernel of g (i.e.M ∼= Z⊕n/im(g)). The map g is given by a n × m matrix A. After performing row andcolumn operations (which simply corresponds to choosing another basis for Z⊕n and Z⊕m)we may assume that the only non-zero entries are diagonal entries a1,1, . . . , ad,d (one canshow that ai,i divides ai+1,i+1). It then follows that M ∼= Zn−d ⊕ Z/a1,1Z ⊕ . . . ⊕ Z/ad,dZ.The result now follows by applying unique factorization; i.e. if a = pr11 · · · prtt , then Z/aZ ∼=Z/pr11 Z⊕ . . .⊕ Z/prtt Z.

Lemma 4.27. Let M be an R module, then m ∈ M is zero if and only if 0 = m ∈ Mm forany m ⊂ R maximal ideal.

Proof. If 0 = m ∈ M , then clearly 0 = m ∈ Mm. Suppose now that 0 = m ∈ Mm for anym ⊂ R maximal ideal. Thus vm = 0 for some v ∈ R \ m. Let I = r ∈ R|rm = 0 ∈ M,then I is an ideal and hence I ⊂ m for some maximal ideal m ⊂ R. this is the requiredcontradiction.

Lemma 4.28. Let M be an R module, then M = 0 if and only if Mm = 0 for all maximalideals m ⊂ R.

Proof. Exercise.

Lemma 4.29. Let M,N be an R modules and φ : M → N a homomorphism of R modules,then φ is injective (resp. surjective) if and only if the induced map φm : Mm → Nm isinjective (resp. surjective) for all maximal ideals m ⊂ R.

Proof. Exercise.

5. C-algebras

Definition 5.1. A ring R is a C-algebra if there is an inclusion of rings C → R. Ahomomorphism φ : R → S of C-algebras is a C-linear ring homomorphism i.e. a ringhomomorphism such that φ(λr) = λφ(r) for all λ ∈ C.

Exercise 5.2. Find a ring homomorphism (of C algebras) that is not a C-algebra homo-morphism

Note that a C-algebra is automatically a C-vector space. It is easy to see that ifI ⊂ C[x1, . . . , xn] is a proper ideal, then C[x1, . . . , xn]/I is a C-algebra. Given elementsr1, . . . , rn ∈ R where R is a C-algebra, we can always define a unique C algebra homomor-phism

φ : C[x1, . . . , xn]→ R

by letting φ(p(x1, . . . , xn)) = p(r1, . . . , rn).16

Definition 5.3. We say that a C-algebra R is finitely generated if there are elementsr1, . . . , rn ∈ R such that the homomorphism

C[x1, . . . , xn]→ R, φ(p(x1, . . . , xn)) = p(r1, . . . , rn)

is surjective (and hence R ∼= C[x1, . . . , xn]/ker(φ)). If ker(φ) = 0, then R ∼= C[x1, . . . , xn]and we say that r1, . . . , rn are algebraically independent (over C).

Remark 5.4. In this case R is ”polynomially” finitely generated over C (i.e. finitely gen-erated as a C-algebra) but it is not finitely generated as a C module.

Definition 5.5. Let B ⊂ A be an inclusion of C algebras. we say that A is a finite Balgebra if there are finitely many elements a1, . . . , an such that for any a ∈ A there existselements b1, . . . , bn ∈ B such that a =

∑ni=1 biai. (Thus A is finitely generated as a B

module.)

Note that finite dimensional vector spaces are finite C algebras, but finitely generated Calgebras are typically infinite dimensional C vector spaces (eg. C[x]).

Lemma 5.6. If A ⊂ B is a finite A algebra and B ⊂ C is a finite B-algebra, then A ⊂ Cis a finite A algebra.

Proof. By definition, we may fix elements b1, . . . , bn generating B over A and c1, . . . , cmgenerating C over B. We claim that bicj1≤i≤n,1≤j≤m generate C over A. To see this, pickany element c ∈ C, then we may write c =

∑cjβj where βj ∈ B. Similarly each βj can be

written as βj =∑biαi,j for appropriate elements αi,j ∈ A. But then

c =∑j

cjβj =∑j

cj(∑i

biαi,j) =∑i,j

αi,jbicj.

Recall that a polynomial p ∈ A[x] is monic if its leading term is 1. We have the following.

Lemma 5.7. Let A ⊂ B be an inclusion of C-algebras. An element b ∈ B satisfies a monicpolynomial p ∈ A[x] if and only if A[b] is a finite A algebra.

Proof. Suppose that b satisfies a monic polynomial p(x) = xd + a1xd−1 + . . . + ad−1x + ad

with ai ∈ A. Then A[b] is generated by 1, b, b2, . . . , bd−1. In fact any element of A[b] maybe written as f(b) where f ∈ A[x] is an arbitrary polynomial. Divide f(x) by p(x) to getf(x) = p(x)q(x) + r(x) where deg r(x) ≤ d − 1 (we can always do this when the leadingcoefficient of p(x) is invertible in B). But then f(b) = p(b)q(b) + r(b) = r(b).

Replacing B by A[b], we may now suppose that B is finite over A. Let b1, . . . , bn be agenerating set. For any x ∈ B, we write

xbi =n∑j=1

ai,jbj.

This equation can be rewritten as

(xIn − A) ·

b1

.

.bn

= 0,

17

where In is the n × n identity matrix and A is the n × n matrix with entries ai,j. Recallfrom linear algebra that

adj(xIn − A)(xIn − A) = det(xIn − A)In

and so

0 = adj(xIn − A)(xIn − A)

b1

.

.bn

= det(xIn − A)In

b1

.

.bn

= det(xIn − A)

b1

.

.bn

But this implies that det(xIn − A) = 0 i.e. that x satisfies a monic polynomial.

Exercise 5.8. Show that det(xIn − A) is a monic polynomial.

Exercise 5.9. Show that Q[√

2] is a finite Q algebra. Show that 3√

2 + 1 satisfies a monicpolynomial.

Definition 5.10. An element a ∈ A is algebraic if C[a] is a finite C algebra (or equivalentlyif a satisfies a monic polynomial p ∈ C[x]). If a ∈ A is not algebraic, then we say that it istranscendental.

Theorem 5.11 (Noether normalization lemma). Let A be a C algebra generated by finitelymany elements a1, . . . , an ∈ A, then there exist transcendental elements x1, . . . , xm ∈ A suchthat C[x1, . . . , xm] ⊂ A is a finite extension.

Proof. (Cf. [Reid, 3.13]) Consider the surjective homomorphism of C-algebras C[y1, . . . , yn]→A = C[a1, . . . , an] with kernel I. If I = 0 there is nothing to prove. Suppose therefore thatthere is an element 0 6= f ∈ I. If f is a monic polynomial of y1 with coefficients inC[y2, . . . , yn], then we conclude that A′ = C[a2, . . . , an] ⊂ A is finite. Proceeding by induc-tion on n we also have that there are transcendental elements x1, . . . , xm ∈ A′ ⊂ A withm ≤ n− 1 such that C[x1, . . . , xm] ⊂ A′ is finite and by the above lemma we conclude thatC[x1, . . . , xm] ⊂ A is finite as required.

Therefore, we must show that if 0 6= I, then I contains a monic polynomial of y1. To seethis start with any element 0 6= f ∈ I. we may assume that deg(f) = d > 0 and we writef = fd+g where deg g ≤ d−1. We now consider elements a2, . . . , an ∈ C and new variablesy′2 = y2 − a2y1, . . . , y

′n = yn − any1 so that

f(y1, y2, . . . , yn) = f(y1, y2 + a2y1, . . . , yn + any1) = fd(1, a2, . . . , an)yd1 + h,

where h has degree ≤ d− 1 with respect to y1. Since fd(1, y2, . . . , yn) 6= 0, we may assumethat fd(1, a2, . . . , an) 6= 0 as required.

Exercise 5.12. Apply Noether’s normalization lemma to C[x, y]/(x2 − y2 + 1).

Exercise 5.13. Apply Noether’s normalization lemma to C[x, y, z]/(xy + z2, x2y − xy3 +z4 − 1).

6. Nakayama’s Lemma

Let R be a Noetherian ring and M a finitely generated R-module.

Lemma 6.1. If I ⊂ R is an ideal and φ : M → M is a homomorphism of R modules suchthat φ(M) ⊂ IM , then there exists a monic polynomial p(x) = xn+a1x

n−1 + . . .+an−1x+anwhere ai ∈ I i and p(φ) = 0.

18

Proof. Exercise. (See the proof of Lemma 5.7.)

Corollary 6.2. If IM = M , then there exists an element r ∈ 1 + I ⊂ R such that rM = 0.

Proof. We apply Lemma 6.1 with φ = id. Let p(x) be the corresponding polynomial, thenlet r = p(1) = 1 +

∑i≥1 ai. Clearly r ∈ 1 + I. It follows easily that for any m ∈M , we have

0 = p(1)m = rm.

Corollary 6.3. If (R,m) is a local ring and M is a finitely generated R module such thatmM = M , then M = 0.

Proof. Let r ∈ 1 +m be defined by Corollary 6.2, then r is invertible (see Exercise 3.50) andso M = (r−1r)M = 0.

Exercise 6.4. Show that the finitely generated hypothesis is needed. (Hint: R = Z(p) andM = Q.)

Corollary 6.5. If (R,m) is a local ring and M is a finitely generated R module and N ⊂Ma sub module such that M = N + mM , then N = M .

Proof. By Corollary 6.3 applied to M/N , since M/N = (N + mM)/N = mM/N , it followsthat mM/N = 0 and hence M = N .

Corollary 6.6. If (R,m) is a local Noetherian ring and M is a finitely generated R moduleand m1, . . . ,mn ∈ M are elements such that the R module M/mM is generated by theimages of m1, . . . ,mn, then M is generated by m1, . . . ,mn.

Proof. Apply Corollary 6.5 to the submodule N =∑miR.

Corollary 6.7 (Nakayama’s Lemma). If (R,m) is a local Noetherian ring and M is a finitelygenerated R module, then R/m = k is a field and M/mM is a k-vector space. There is a oneto one correspondence between any basis v1, . . . , vn of M/mM and the minimal generatingsets m1, . . . ,mn of M (over R). In particular two basis of M/mM (resp. two minimalgenerating sets of M) are related by an invertible matrix with coefficients in k (resp. R).

Proof. Exercise.

Theorem 6.8. [Going Up Theorem] Let R ⊂ S be an integral extension of rings. If P ⊂ Ris a prime ideal, then there is a prime ideal Q ⊂ S such that Q∩R = P . If moreover Q1 isan ideal of S such that R ∩Q1 ⊂ P , then we may assume that Q1 ⊂ Q.

Proof. Replacing S by S/Q1 and R by R/(Q1∩R), we may assume that Q1 = 0. Localizingat P , we may assume that (R,P ) is a local ring. Let Q ⊂ S be any maximal ideal containingPS, since clearly R ⊃ Q∩R ⊃ P , it suffices to show that Q 6= S or equivalently that PS 6= S.If PS = S, then 1 =

∑pisi with pi ∈ P and si ∈ S. Since this sum is finite, S ′ =

∑Rsi ⊂ S

is a finitely generated R-module such that PS ′ = S ′. By Nakayama’s Lemma, S ′ = 0 whichis a contradiction.

Corollary 6.9. With the above notation, if P1 ⊂ P2 ⊂ . . . ⊂ Pn is a chain of prime idealsin R and Q1 ⊂ Q2 ⊂ . . . ⊂ Qm is a chain of prime ideals in S with 1 ≤ m < n andPi = Qi ∩R, then there exists a Qm ⊂ Qm+1 ⊂ . . . ⊂ Qn a chain of prime ideals in S wherePi = Qi ∩R.

Proof. Exercise. 19

Exercise 6.10. Let M be a finitely generated R module and φ : M → M a surjectivehomomorphism. Show that φ is injective. (Hint: View M as an R[x] module, with x ·m =φ(m).)

7. Fields

Definition 7.1. A field F is a ring (commutative with 1) such that every non-zero elementhas a multiplicative inverse: if a ∈ F ∗ = F \ 0, then there is an element b ∈ F such thatab = 1.

It is easy to see that multiplicative inverses are unique and hence we denote the multi-plicative inverse of a by a−1. Typical examples of fields are Q,R,C,Fp = Z/pZ.

Definition 7.2. If F is a field, then a subfield of F is any subset 0, 1 ⊂ E ⊂ F such thatE is closed under addition and multiplication, additive inverses and multiplicative inverses(of non-zero elements). It is easy to see that E with the operations induced by the additionand multiplication on F is also a field.

Note that if Eii∈I is a collection of subfields of a field F then ∩i∈IEi ⊂ F is a subfield.In particular if S ⊂ F is a subset, then (∩F⊃E⊃SE) ⊂ F is a field (where the intersection isover all subfields E ⊂ F containing S).

Definition 7.3. Let E ⊂ F be an inclusion of fields, and fii∈I be elements of F , then welet E(fii∈I) be the smallest field containing E and the fi.

Definition 7.4. Let E ⊂ F be an inclusion of fields and f ∈ F , then we say that fis algebraic over E if dimE E(f) is finite. If f is not algebraic, then we say that f istranscendental over E.

Lemma 7.5. Let E ⊂ F be an inclusion of fields and f ∈ F , then the following areequivalent:

(1) f is algebraic over E,(2) E(f) = E[f ] where E[f ] = p(f)|p ∈ E[x],(3) f is the solution of a polynomial p ∈ E[x].

Proof. (1) implies (3): Suppose that f is algebraic over E, then since dimE E(f) is finite,there exists d ∈ N such that 1, f, f 2, . . . , fd are not linearly independent and so there existse0, . . . , ed ∈ R such that

∑di=0 eif

i = 0. But then f is the solution of the polynomialp =

∑eix

i ∈ E[x].(3) implies (2): The inclusion ⊃ is clear and so it suffices to show that E[f ] is a field,

i.e. that every non-zero element of E[f ] is invertible. Suppose that f is the solution of apolynomial p ∈ E[x]. We may assume that p is irreducible. Let 0 6= g ∈ E[f ], then g = q(f)for some polynomial q ∈ E[x]. Since 0 6= q(f), one sees that p does not divide q, but sincep ∈ E[x] is irreducible, then gcd(p, q) = 1 and so there exist polynomials a, b ∈ E[x] suchthat ap+ bq = 1. but then

(gb)(f) = q(f)g(f) = 1− a(f)p(f) = 1

and so g is invertible in E[f ] and hence E[f ] is a field.(2) implies (1): Since E[f ] is a field, there exists a polynomial q ∈ E[x] such that f ·q(f) =

1. Let p(x) = xq(x)− 1 ∈ E[x], then p(f) = 0 and so f is algebraic. 20

Corollary 7.6. If E ⊂ F is an inclusion of fields and f ∈ F is transcendental over E, thenE(f) ∼= E(x), E[f ] ∼= E[x] and dimE E(f) is infinite.

Proof. Exercise.

We next show how to associate a field to certain rings in a fashion similar to the con-struction of the rational numbers Q.

Definition 7.7. A ring R (commutative with 1) is a domain if it has no zero divisorsi.e. there are no pairs of elements x, y ∈ R such that xy = 0.

Lemma 7.8. If I ⊂ C[x1, . . . , xn] is an ideal, then C[x1, . . . , xn]/I is a domain if and onlyif I is prime. Thus, if X is an affine variety, then C[x1, . . . , xn]/I(X) is a domain if andonly if X is irreducible.

Proof. If I is not prime, then there exist x, y 6∈ I such that xy ∈ I, but then x+ I, y+ I 6= Iand (x+I)(y+I) = (xy+I) = I so that x+I, y+I 6= 0 ∈ R/I but (x+I)(y+I) = 0 ∈ R/I,i.e. R/I is not a domain. We leave the converse as an excercise.

Definition 7.9. Let R be a domain, then the fraction field of R is the set

Q(R) = (f, g)|f ∈ R, g ∈ R∗/ ∼, (f, g) ∼ (h, k), iff fk = gh.

Addition is defined by (f, g) + (h, k) = (fk + gh, gk) and multiplication is defined by(f, g)(h, k) = (fh, gk). It will be convenient to denote the equivalence class of (f, g) simplyby f/g. In other words Q(R) = S−1R where S = R∗.

Definition 7.10. We say that an extension of fields F ⊂ E is a finite field extension ifdimF E < +∞ (i.e. if the dimension of E as an F vector space is finite).

Note that if e1, . . . , er are algebraic over F , then F ⊂ F (e1, . . . , er) is finite.

Theorem 7.11 (Primitive element theorem). Let F ⊂ E be a finite extension of a fieldof characteristic 0. Then there exists an element e ∈ E such that E = F (e). In fact, ifE = F (e1, . . . , er), then we may assume that e =

∑fiei where fi ∈ F .

Proof. Unluckily the standard proof relies on Galois Theory. It is necessary to know thatthere are only finitely many intermediate field extensions F ⊂ Fi ⊂ E. If this is the case,then it suffices to pick an element α ∈ E \ ∪Fi.

Remark 7.12. It is not necessary to assume that F have characteristic 0. It suffices toassume that F is infinite and E/F is separable. The primitive element theorem also holdsfor all finite fields (by a different proof).

Exercise 7.13. Show that Q(Z) = Q.

Exercise 7.14. Show that addition and multiplication are well defined in Q(R), 1 = (1, 1)and 0 = (0, 1) are the multiplicative and additive identities. The inverses are −(a, b) =(−a, b) and (a, b)−1 = (b, a). Conclude that Q(R) is a field.

Exercise 7.15. Let F be a field, then either m · 1 = 1 + · · · + 1 6= 0 for all m ∈ N orthere exists a prime number p such that p · 1 = 0. In the first case we say that F hascharacteristic 0 and in the second case we say that F has characteristic p > 0.

Exercise 7.16. If F is a finite field (meaning that its cardinality |F | is finite), then |F | = pe

for some e ∈ N and some prome number p > 0. It turns out that all fields of order pe areisomorphic. We call such a field Fpe.

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Exercise 7.17. Show that Q[√

2] is a field and dimQ Q[√

2] = 2.

Exercise 7.18. Let E ⊂ F be an inclusion of fields, p(x) ∈ E[x] be an irreducible polynomialof degree d, and f ∈ F a solution of p(x), then show that E(f) ∼= E[x]/(p) and dimE E(f) =d.

Exercise 7.19. Let Q = f ∈ C| dimQ Q(f) < ∞. We say that Q is the algebraicclosure of Q. Show that Q is a field and dimQ Q =∞.

Exercise 7.20. Show that Q(x, i) is not generated over Q by a single element and that thereare infinitely many intermediate field extensions Q ⊂ F ⊂ Q(x, i).

Exercise 7.21. Find a primitive element for Q(√

2,√

3)/Q.

Exercise 7.22. Show that E = F3[x]/(x2 + 2x + 1) is a field and find a primitive elementfor E/F3.

Exercise 7.23. Show that E = F3[x]/(x2 + x+ 1) is not a field.

Exercise 7.24. Let p be a prime and q ∈ Fp[x] a polynomial. Show that E = Fp[x]/(q(x))is a field if and only if q(x) is irreducible.

Exercise 7.25. Let p be a prime number. Show that there is always a degree 2 irreduciblepolynomial in Fp[x] and hence Fp[x]/(q(x)) is a field with q2 elements.

8. Nullstellnsatz

Theorem 8.1. [Weak Nullstelensatz] If m ⊂ C[x1, . . . , xn] is a maximal ideal, then m =(x− a1, . . . , x− an) for a1, . . . , an ∈ C.

Proof. Note that (x − a1, . . . , x − an) is a maximal ideal and C ⊂ C[x1, . . . , xn]/m is aninclusion of fields (cf. Exercise 3.37).

Suppose that the field C[x1, . . . , xn]/m is an algebraic extension of C. Since C is al-gebraically closed, then C[x1, . . . , xn]/m ∼= C. Let φ : C[x1, . . . , xn] → C be the in-duced homomorphism with kernel m and ai = φ(xi), then clearly φ(xi − ai) = 0 so thatm ⊃ (x1 − a1, . . . , xn − an). Since (x1 − a1, . . . , xn − an) ⊂ C[x1, . . . , xn] is maximal (cf.3.30), we have m = (x1 − a1, . . . , xn − an) as required.

Therefore, we may assume that the field C[x1, . . . , xn]/m is not an algebraic extension ofC. Thus there exists a transcendental element t ∈ C[x1, . . . , xn]/m and hence

C[x1, . . . , xn]/m ⊃ C(t) ⊃ C.

Note that the dimension of C[x1, . . . , xn]/m (as a C vector space) is clearly countable i.e. itdoes not contain an uncountable set of linearly independent elements. However it is easy tosee that the set

1

t− a|a ∈ C

is uncountable and linearly independent.

Exercise 8.2. Show that 1t−a |a ∈ C is linearly independent over C.

Exercise 8.3. Let V be a vector space over C with a countable set of generators. Show thatany uncountable set of elements of V is linearly dependent.

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Corollary 8.4. If V(p1, . . . , pr) = ∅ for some polynomials (p1, . . . , pr) ∈ C[x1, . . . , xn], then(p1, . . . , pr) = C[x1, . . . , xn] or equivalently there exist q1, . . . , qr ∈ C[x1, . . . , xn] such that1 =

∑ri=1 piqi.

Proof. Let I = (p1, . . . , pr) be the corresponding ideal. If I 6= C[x1, . . . , xn] then I iscontained in a maximal ideal I ⊂ m. By Theorem 8.1, m = (x − a1, . . . , x − an) fora1, . . . , an ∈ C. But then each polynomial q ∈ I ⊂ m must vanish at (a1, . . . , an) ∈ Cn.This is a contradiction and so I = R. In particular 1 ∈ I and so 1 =

∑ri=1 piqi where

q1, . . . , qr ∈ C[x1, . . . , xn].

Theorem 8.5. [Nullstelensatz] Let I ⊂ C[x1, . . . , xn] be an ideal and V = V(I) ⊂ Cn thecorresponding zero set. If f ∈ C[x1, . . . , xn] vanishes along V , then fm ∈ I for some m > 0.Equivalently, we have

I(V(I)) =√I.

Proof. (Rabinowitsch’s trick) Let I = (f1, . . . , fr). The polynomials f1, . . . , fr, 1− xn+1f ∈C[x1, . . . , xn+1] have no common zeroes and so by Corollary 8.4,

1 = f1q1 + . . .+ frqr + (1− xn+1f)q

where q1, . . . , qr, q ∈ C[x1, . . . , xn+1]. Now substitute xn+1 = 1/f to get

1 = f1(x1, . . . , xn)q1(x1, . . . , xn, 1/f) + . . .+ fr(x1, . . . , xn)qr(x1, . . . , xn, 1/f).

Let t be the biggest power of (1/f) appearing in the above expression, then q′i := qi · f t ∈C[x1, . . . , xn] and so

f t = f1q′1 + . . .+ frq

′r ∈ I.

Corollary 8.6. Let I, J ⊂ C[x1, . . . , xn] be two ideals. Then V(I) = V(J) if and only if√I =√J .

Proof. Exercise.

Exercise 8.7. Show that p1, . . . , pk ∈ Q[x1, . . . , xn] have a common root in Cn if and onlyif (p1, . . . , pk) 6= Q[x1, . . . , xn].

Exercise 8.8. Use the Nullstellensatz to show that two distinct irreducible monic polyno-mials p, q ∈ C[x1, . . . , xn] define different hypersurfaces in Cn. Show that this fails in Rn.

Exercise 8.9. Show that there are irreducible polynomials f ∈ R[x, y] such that V(f) ⊂ R2

is reducible.

9. The coordinate ring of an affine variety

Definition 9.1. The coordinate ring of an affine algebraic set X ⊂ Cn is the ring A(X) =C[x1, . . . , xn]/I(X). Note that this may be identified with the ring of functions on X inducedby restricting polynomials from Cn to X. In particular A(Cn) = C[x1, . . . , xn].

Exercise 9.2. Show that the coordinate ring of an affine variety (resp. algebraic set)is a C algebra that has no zero divisors (resp. no nilpotents). Equivalently A(X) =C[x1, . . . , xn]/I(X) where I(X) is prime (resp. I(X) is radical).

Exercise 9.3. Compute the coordinate rings of V(y2− x3) and V(y2− x2(x− 1)) and showthat they are not integrally closed.

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It is not hard to see that the coordinate rings actually determine the affine variety.

Lemma 9.4. There is a bijection between finitely generated C algebras without nilpotents(resp. zero divisors) and affine algebraic sets (resp. varieties).

Proof. Let A be a finitely generated C algebra without nilpotents, then we must show thatthere exists an affine algebraic set X ⊂ Cn such that A = A(X). Fix g1, . . . , gn generatorsof A (as a C-algebra) and consider the surjective homomorphism of C-algebras

φ : C[x1, . . . , xn]→ A, p(x1, . . . , xn)→ p(g1, . . . , gn).

If I = ker(φ), then A ∼= C[x1, . . . , xn]/I. Since A has no nilpotents, then I is radical and soI = I(V(I)). let X = V(I). We have shown that A = A(X).

Consider now F : X → Y a morphism of affine varieties X ⊂ Cn and Y ⊂ Cm (see §2),then we have an induced homomorphism of C-algebras

F ∗ : A(Y )→ A(X), F ∗(g) = g F.For example if F : C → C3 is defined by F (t) = (t, t2, t3), then F ∗(x) = t, F ∗(y) = t2 andF ∗(z) = t3 and so we have an induced homomorphism of C algebras

F ∗ : C[x, y, z]→ C[t], g(x, y, z)→ g(t, t2, t3).

Notice also that F ∗ actually determines F ; in fact F ∗(x) = t, F ∗(y) = t2 and F ∗(z) = t3

implies that F (t) = (t, t2, t3). One easily sees that:

Lemma 9.5. There is a bijection between homomorphisms of finitely generated C-algebraswithout nilpotents and morphisms or affine algebraic sets such that X ⊂ Cn, Y ⊂ Cm areaffine varieties, then

α : Hom(X, Y )→ Hom(A(Y ), A(X))

is defined as follows: for any morphism f : X → Y let f ∗ ∈ Hom(A(Y ), A(X)) be thehomomorphism of C-algebras defined by f ∗(g) = g f .

Proof. Let y1, . . . , ym be coordinates on Cm. Given Φ ∈ Hom(A(Y ), A(X)), we definef : X → Cm by

f(p) = (Φ(y1)(p), . . . ,Φ(ym)(p)).

Note that Φ(yj) ∈ A(X) and in order to check that we have a map f : X → Y , we mustcheck that if g ∈ I(Y ), then g(Φ(y1)(p), . . . ,Φ(ym)(p)) = 0 for all p ∈ X. This is immediatesince if yi = yi|Y , then for any g ∈ I(Y ) we have

0 = Φ(g(y1, . . . , ym)) = g(Φ(y1), . . . ,Φ(ym)) ∈ A(X).

We now check that f ∗ = Φ. It suffices to check that this holds for the generators y1, . . . , ymof A(Y ). Again this is clear since

f ∗(yi) = fi = Φ(yi).

The claim now follows easily.

Let x ∈ X be a point on an affine variety, then A(x) = C since functions on points areclearly constant. From the lemma above, it follows that any point x ∈ X is determined bya non-zero homomorphism φx : A(X) → C. We may think of this as being determined byevaluating functions f ∈ A(X) at x, i.e. φx(f) = f(x). Note that φx is determined by mx :=ker(φx) where mx is the ideal of functions vanishing at x. Since A(X)/mx = C, it followsthat mx is a maximal ideal. For example, the points of a ∈ Cn are determined by n-tuples

24

a = (a1, . . . , an), φa is given by p(x1, . . . , xn) → p(a1, . . . , an), ma = (x1 − a1, . . . , xn − an)and so

φa : C[x1, . . . , xn]→ C[x1, . . . , xn]/ma = (x1 − a1, . . . , xn − an) ∼= C.

Exercise 9.6. Show that C2 \ (0, 0) is not affine.

Exercise 9.7. Suppose that X is an affine variety and f ∈ A(X) is not zero at every pointx ∈ X i.e. f(x) 6= 0. then 1/f ∈ A(X).

Definition 9.8. Let X ⊂ Cn be an irreducible affine variety, then the ring of rationalfunctions on X is C(X) = Q(A(X)) the quotient field of A(X).

Definition 9.9. Let X be an affine variety, then we let C(X) = Q(A(X)) be the field ofrational functions on X. If X is a quasi-affiine variety (i.e. an open subset of an affinevariety), then we let C(X) = C(U) for some non-empty open affine subset of X.

Exercise 9.10. Check that the definition of C(X) does not depend on the choice of thenon-empty open subset U ⊂ X.

Definition 9.11. Let p ∈ X be a point on an affine variety, then we let

Op = OX,p = A(X)mp

be the local ring of p ∈ X. Here mp denotes the maximal ideal of p ∈ X and A(X)mp isthe localization of the coordinate ring of X at the maximal ideal corresponding point p ∈ X.

Proposition 9.12. Let p ∈ X be a point on an affine variety, then there are naturalinclusions

A(X) ⊂ OX,p ⊂ Q(X).

Moreover we have the equality

A(X) = ∩P∈Spec(A(X))A(X)P = ∩m∈Specmax(A(X))A(X)m

where the spectrum of A(X) is Spec(A(X)) is the set of all prime ideals in A(X) and themaximal spectrum of A(X) is Specmax(A(X) the set of all maximal ideals in A(X).

Proof. The inclusions are clear and the equalities follow from Proposition 3.55.

Definition 9.13. We say that a morphism of affine varieties f : X → Y is finite ifA(Y )→ A(X) is an integral extension of rings.

Exercise 9.14. Use the Going up Theorem 6.8 to show that if f : X → Y is a finitemorphism of affine varieties, then dimX = dimY .

Exercise 9.15. Show that the inclusion C \ 0 → C is not a finite morphism.

Exercise 9.16. Show that the projection C2 → C on to the first factor is not a finitemorphism.

Exercise 9.17. Show that the inclusion X ⊂ W of an irreducible closed subset in an affinevariety is a finite morphism.

Exercise 9.18. Show that the composition of two finite morphisms is a finite morphism.

Exercise 9.19. Apply Noether’s Normalization Lemma to show that if X is an affine vari-ety, then there exists a finite morphism X → Cn.

25

10. Morphisms of quasi-affine varieties

Definition 10.1. A quasi-affine variety is an open subset of an affine variety X ⊂ Cn.

Definition 10.2. A regular function on a quasi-affine variety X ⊂ Cn is a functionf : X → C such that for any point p ∈ X there is an open subset p ∈ U ⊂ X such thatf |U = g/h where g, h ∈ C[x1, . . . , xn] are polynomials and h(x) 6= 0 for all x ∈ U .

The set of all regular functions on a quasi-affine variety is denoted by O(X). It is easyto see that O(X) is a ring (commutative with 1).

Exercise 10.3. Show that a regular function is continuous in the Zariski topology.

Exercise 10.4. Show that if X ⊂ Cn is an affine variety and I = I(X), then O(X) =C[x1, . . . , xn]/I.

Definition 10.5. A morphism f : X → Y between two quasi-affine varieties X ⊂ Cn andY ⊂ Cm, is a continuous map such that f = (f1, . . . , fm) where fi ∈ O(X).

Exercise 10.6. Show that if f : X → Y and g : Y → Z are two morphisms of quasi-affinevarieties, then g f is a morphism of quasi-affine varieties.

Proposition 10.7. Let X be a quasi-affine variety and Y be an affine variety, then thereis a natural equivalence

α : Hom(X, Y )→ Hom(A(Y ),O(X)).

Here, an element f ∈ Hom(X, Y ) is a morphism of quasi-affine varieties f : X → Y andg ∈ Hom(A(Y ),O(Y )) is a homomorphism of C algebras and α(f) = f ∗ where f ∗(p) =p f ∈ O(X) for any p ∈ A(X).

Proof. Similar to Lemma 9.5.

11. Dimension

Definition 11.1. Let R be a ring, then the Krull dimension of R is the maximum lengthn of a chain of distinct prime ideals

P0 ⊂ P1 ⊂ . . . ⊂ Pn ⊂ R.

For example the dimension of C[x1, . . . , xn] is n. The inequality ≥ n follows easily fromthe inclusion of prime ideals

0 ⊂ (x1) ⊂ (x1, x2) ⊂ . . . ⊂ C[x1, . . . , xn].

Note that since prime ideals correspond to irreducible closed subsets, the above definitionmatches Definition 1.23, so that if R = A(X)mx is the local ring of X at x ∈ X, then dimxXis equal to the Krull dimension of (A(X)mx ,mx).

Exercise 11.2. Show that if R is Noetherian, then the dimension of R always exists and isfinite.

Exercise 11.3. Show that if R is a field, then the Krull dimension of R is 0.

We will denote the Krull dimension by Krdim(R). We have the following important fact.

Theorem 11.4. Let (R,m) be a local Notherian domain, then

Krdim(R) = maxx∈m\0Krdim(R/xR) + 1.26

Proof. We begin by proving the inequality ≥. Suppose that x ∈ m \ 0 and pick P0 ⊂ P1 ⊂. . . ⊂ Pk ⊂ R/xR a chain of distinct prime ideals where k = Krdim(R/xR). Let Pi be theinverse image of Pi in R. Then Pi are prime and the inclusions Pi ⊂ Pi are strict. Thus wehave a sequence

(x) ⊂ P0 ⊂ P1 ⊂ . . . ⊂ Pk ⊂ R.

Since x 6= 0, the inclusion 0 = P−1 ⊂ P0 is also strict and hence Krdim(R) ≥ Krdim(R/xR)+1.

We now prove the reverse inequality. Let P−1 ⊂ P0 ⊂ . . . ⊂ Pk ⊂ R be a chain of distinctprime ideals where k+1 = Krdim(R). Then there is an element x ∈ P0∩(m\0) and so lettingPi = Pi/(x) for i ≥ 0, we obtain a chain of distinct prime ideals P0 ⊂ P1 ⊂ . . . ⊂ Pk ⊂ R/xRand so Krdim(R) ≤ Krdim(R/xR) + 1.

Definition 11.5. If R is a Noetherian ring, then we let

dimR = supPKrdimRP

where P ⊂ R is a prime ideal and RP is the corresponding localized ring.

Definition 11.6. Let P ⊂ R be a prime ideal, then the height of P is the maximal lengthof distinct prime ideals P0 ⊂ P1 ⊂ . . . ⊂ P .

We next recall the following result.

Theorem 11.7. Let R be an integral domain which is a finitely generated C-algebra. Then

(1) Krdim(R) = tr.deg.CQ(R), and(2) If P ⊂ R is prime, then height(P ) + Krdim(R/P ) = Krdim(R).

In particular the length of of all maximal chains of prime ideals is given by dimR.

Proof. [Hartshorne, Theorem 1.8A]. Here is however an intuitive argument for (1). We willsee later that if X is an affine variety, R = A(X) and the transcendence degree of Q(X) is d,then there an open subset U ⊂ X which is isomorphic to an open subset of a hypersurfaceV(f) ⊂ Pd+1. But then by Exercise 11.14

dimX = dimU = dimV(f) = dimPd+1 − 1 = d.

Exercise 11.8. Show that dimCn = n.

Exercise 11.9. What is the dimension of C[x, y, z]/x(y, z) localized at (x, y, z)?

Exercise 11.10. Show that dimR = dimR/√R.

Exercise 11.11. Let R be a principal ideal ring (so that any ideal is generated by oneelement). Show that dimR ≤ 1.

Exercise 11.12. Show that if X is a quasi-affine variety, then dimX = dim X. (See[Hartshorne, I.1.10].

Theorem 11.13 (Krull’s Hauptidealsatz). Let f ∈ R be an element in a Noetherian ringwhich is neither a unit nor a zero divisor, then every minimal prime ideal P containing fhas height 1.

27

Proof. We will assume that R = A(X) where X ⊂ Cn is an affine variety. Let z1 = f . Wemay assume that z1, . . . , zr ∈ C[x1, . . . , xn] are algebraically independent in A(X) wherer = dimX. Let W = V(P ) ⊂ V(f) = V so that W is an irreducible component of V .Pick g 6∈ P such g vanishes along V \ W . We will show that z2, . . . , zr are algebraicallyindependent in A(W ). If this where not the case, then there is a nonzero polynomialF ∈ C[t2, . . . , tr] such that F (z2, . . . , zr) = 0 ∈ A(W ) and hence F (z2, . . . , zr)g = 0 ∈ A(X).By the Nullstellensatz, z1 = f divides (F (z2, . . . , zr)g)n in A(X), but then f divides gn inA(X) and so g vanishes along V which is a contradiction.

Exercise 11.14. Let Y ⊂ Cn be an irreducible subset . Show that Y is the zero of anirreducible polynomial if and only if it has codimension 1 in Cn.

Exercise 11.15. Show that if X ⊂ Cn is defined by r equations, then dimX ≥ n− r (butstrict inequality is possible).

12. Projective varieties

Let 0 = (0, . . . , 0) ∈ Cn+1 and C∗ = C \ 0. Consider the equivalence relation on Cn+1 \ 0defined by

(a0, . . . , an) ∼ (b0, . . . , bn) iff ∃λ ∈ C∗ s.t. a0 = λb0, . . . , an = λbn.

In other words, if a = (a0, . . . , an) and b = (b0, . . . , bn), then a ∼ b if and only if a = λb forsome λ ∈ C∗. We denote the equivalence class of (a0, . . . , an) by [a0 : . . . : an].

Exercise 12.1. Check that ∼ defined above is an equivalence relation whose equivalenceclasses correspond to lines through the origin in Cn+1.

Definition 12.2. We define n-dimensional projective space via

Pn := (Cn+1 \ 0)/ ∼ .

Consider the natural inclusion Cn → Pn given by

(a1, . . . , an)→ [1 : a1, . . . , an],

which gives a one to one correspondence between the set Cn and the set x0 6= 0 ⊂ Pn.This correspondence can be inverted as follows: if [a0 : . . . : an] ⊂ Pn and a0 6= 0, then welet

[a0 : . . . : an]→ (a1/a0, . . . , an/a0).

Note that the subset x0 = 0 ⊂ Pn can be identified with Pn−1 by the map φ : Pn−1 → Pndefined by [x1 : . . . : xn]→ [0 : x1 : . . . : xn]. In this way we have

Pn = Cn ∪ Pn−1 = Cn ∪ Cn−1 ∪ . . . ∪ C ∪ pt..We may think of Pn−1 as the set of points at infinity. For example if n = 2 and ax+by+c = 0is a line l ⊂ C2, then the homogeneous equation aX + bY + cZ = 0 defines a subset L ⊂ P2

such that L∩C2 = l (where C2 corresponds to the subset Z 6= 0). Notice that the intersectionwith the line at infinity is given by

L ∩ P1 = aX + bY + cZ = 0 ∩ Z = 0 = [b : −a : 0]

is determined by the slope of l and two lines l, l′ ⊂ C2 will meet at infinity if and only ifthey have the same slope. In this way, we have that two distinct lines in P2 always meet atexactly one point.

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Lemma 12.3. Pn is an n-dimensional compact complex manifold.

Proof. We define local charts Ui ∼= Cn for 0 ≤ i ≤ n by letting Ui = xi 6= 0 ⊂ Pn. Thenatural inclusions φi : Ui → Pn are given by

φi(a0, . . . , ai−1, ai+1, . . . , an) = [a0 : . . . : ai−1 : 1 : ai+1 : . . . : an]

and therefore on Ui,j = Ui ∩ Uj we have (holomorphic) transition functions ψij = φ−1j φi

given by

ψij(a0, . . . , ai−1, ai+1, . . . , an) = (a0/aj, . . . , ai−1/aj, 1/aj, ai+1/aj, . . . , aj−1/aj, aj+1/aj, . . . , an).

To see that Pn is compact in the Euclidean topology, consider a sequence ai ∈ Pn. Writingai = [ai0 : . . . , : ain], after dividing by ||ai|| = (

∑i=0n a

ij aij)

1/2, we may assume that the ai

are represented by points in S2n+1 ⊂ Cn+1 which is compact. Therefore, after passing to asubsequence we may assume that the limit a = limj→∞ a

j ∈ S2n+1 exists. Since a ∈ S2n+1,one of the coordinates of a = (a0, . . . , an) is non-zero and hence [a0 : . . . : an] represents apoint in Pn. Finally we must check that lim ai = [a0 : . . . : an] in Pn. Since the topologyon Pn is induced by the topology on the charts Ui, it suffices to check this on such a chart.Assume that a0 6= 0, then ai0 6= 0 for all i 0 and so we may consider the correspondingpoints (1, ai1/a

i0, . . . , a

in/a

i0) ∈ U0. Since the induced map f : S2n+1 \ x0 = 0 → U0 defined

by (x0, . . . , xn) → (x1/x0, . . . , xn/x0) is continuous in a neighborhood of a, it follows thatlim f(ai) = f(a) as required.

We would like to define projective varieties in a similar fashion to what we did for affinevarieties. Notice however that it does not make sense to evaluate a polynomial at a pointof projective space. For example what should the value of x2 − y3 + z be at the point[1 : 1 : 0] ∈ P2? You may want to say 12−13+0 = 0 however, recall that [1 : 1 : 0] = [2 : 2 : 0]and 22 − 23 + 0 = −4 6= 0. Notice however that if p ∈ C[x0, . . . , xn] is a homogeneouspolynomial of degree d, then

p(λa0, . . . , λan) = λdp(a0, . . . , an)

and so p(λa0, . . . , λan) = 0 if and only if p(a0, . . . , an) = 0 (since λ 6= 0). Therefore, we canuse homogeneous polynomials to define projective varieties.

Given an affine variety we can always define a corresponding projective variety. Forexample if X = V(x2 − y3) ⊂ C2, then we can consider X = V(x2z − y3) ⊂ P2. Notice thatX ∩ z 6= 0 = X and X ∩ z = 0 = [1 : 0 : 0]. We have compactified the curve X byadding a point at infinity. In general if p ∈ C[x1, . . . , xn] is any polynomial of degree d, welet

p = xd0p(x1/x0, . . . , xn/x0) ∈ C[x0, . . . , xn]

be the corresponding homogeneous polynomial. So for example if p = x21 − x3

2, then

p = x30((x1

x0

)2 − (x2

x0

)3) = x0x21 − x3

2.

Notice thatp(1, x1, . . . , xn) = p(x1, . . . , xn)

and so we can recover the original polynomial p from its homogeneization p. Similarly, ifX = V(I) ⊂ Cn for some ideal I = (f1, . . . , fr) ⊂ C[x1, . . . , xn], then we consider the ideal

I = (f1, . . . , fr) ⊂ C[x0, . . . , xn].

As before, we can recover I from I and we have X = X ∩ x0 6= 0 where X = V(I) ⊂ Pn.29

Exercise 12.4. Let X, Y be lines in C2. Show that the corresponding lines X, Y ⊂ P2 alwaysmeet in exactly one point. If X, Y are not parallel, then X ∩ Y = X ∩ Y ⊂ C2 = x0 6= 0and if X, Y are parallel (but distinct), then X ∩ Y = [0, 1,−m] ∈ P1 = x0 = 0 is the pointat infinity corresponding to the slope m of the lines.

Exercise 12.5. Let X ⊂ Pn be a projective variety. Show that X is compact in the Euclideantopology.

Given an affine variety, there are several advantages to considering the correspondingprojective variety X ⊂ Pn.

(1) X is compact,(2) Intersection theory: As we have seen above, in P2 two distinct lines always intersect

at a point. We will see later that two distinct curves of degree d, d′ intersect at dd′

points in P2 (when the points are counted with the correct multiplicity).

Exercise 12.6. Show that if C ⊂ P2 is a curve of degree d > 0 and L ⊂ P2 is a line notcontained in C, then L ∩ C consists of d points when counted with multiplicity.

Exercise 12.7. Show that if C ⊂ P2 is a curve of degree d > 0 and Q ⊂ P2 is a quadricand Q ∩ C is finite, then Q ∩ C consists of 2d points when counted with multiplicity.

Definition 12.8. An ideal I ⊂ C[x0, . . . , xn] is a homogeneous ideal if for any f ∈ I,then writing f =

∑i≥0 fi where fi is homogeneous of degree i, then fi ∈ I. Notice that I is

generated by finitely many homogeneous polynomials I = fii∈I where fi is homogeneous.

Exercise 12.9. Show that any homogeneous ideal I ⊂ C[x0, . . . , xn] is generated by finitelymany homogeneous polynomials.

Definition 12.10. A subset X ⊂ Pn is a projective variety if it is defined by the vanishingof finitely many homogeneous polynomials or equivalently X = V(I) where I ⊂ C[x0, . . . , xn]is a homogeneous ideal.

Given any projective variety X ⊂ Cn we let I(X) ⊂ C[x0, . . . , xn] be the ideal correspondingto the closure of X in Pn. We note the following:

(1) I(X) ⊂ C[x0, . . . , xn] is a homogeneous ideal. If f ∈ I(X), then f vanishes at eachpoint a = (1, a1, . . . , an) ∈ X. Since X ⊂ Pn is a projective variety, for any λ ∈ C∗,f vanishes at each point λ · a = (λ, λa1, . . . , λan) ∈ X, but then one sees that iff =

∑fi where deg(fi) = i, then

0 = f(λ, λa1, . . . , λan) =∑

fi(λ, λa1, . . . , λan) =∑

λifi(1, a1, . . . , an)

holds for any λ ∈ C∗. But then fi(1, a1, . . . , an) = 0 for all i and so fi ∈ I(X) asrequired.

(2) I(X) ⊂ C[x0, . . . , xn] is a radical ideal, i.e.√

I(X) = I(X).(3) The set of homogeneous radical ideals are in one to one correspondence with the

set of projective varieties with the exception of (x0, . . . , xn) that corresponds to theemptyset.

Exercise 12.11. Intersections of projective varieties are projective varieties and the unionof two projective varieties is a projective variety.

30

Definition 12.12. The Zariski topology on a projective variety X ⊂ Pn is defined byletting the closed subsets correspond to the zero sets of homogeneous ideals.

Lemma 12.13. Let X ⊂ Pn be a projective variety and X = X ∩ x0 6= 0 ⊂ Cn be thecorresponding affine subvariety. Then the natural inclusion X → X is continuous in therespective Zariski topologies.

Proof. It suffices to show that if Z ⊂ X is a closed subset then so is Z ∩X. Suppose thatZ = V(I) for some homogeneous ideal I = (f1, . . . , fr) ⊂ C[x0, . . . , xn], then Z ∩X = V(I ′)where I ′ = (f1(1, x1, . . . , xn), . . . , fr(1, x1, . . . , xn)) ∈ C[x1, . . . , xn] and so Z ∩ X ⊂ Cn isclosed.

Exercise 12.14. Show that projective varieties are compact in the Euclidean topology.

We now wish to define morphisms of projective varieties. The main difficulty is that pointson projective varieties a ∈ X ⊂ Pn are only defined up to multiplication by a non-zero scalarλ ∈ C∗. However, consider the following example defined by

φ : P1 → P3, φ(t0, t1) = (t30, t20t1, t0t

21, t

31).

Notice that

φ(λt0, λt1) = (λ3t30, λ3t20t1, λ

3t0t21, λ

3t31) = λ3(t30, t20t1, t0t

21, t

31)

and therefore φ gives a well defined map φ : P1 → P3. One would be tempted to concludethat given any set of homogeneous polynomials of (the same) degree d say

p1, . . . , pr ∈ C[x0, . . . , xn]d

then we obtain a morphism

φ : Pn → Pr

however note that if a ∈ Pn is contained in V(p1, . . . , pr), then φ(a) = [0 : . . . : 0] is notdefined! Therefore we only have a map

φ : (Pn \ V(p1, . . . , pr))→ Pr.

Definition 12.15. A quasi-projective variety is an open subset of a projective varietyX ⊂ Pn.

Note that if X ⊂ Pn is a projective variety and Cn ∼= Ui = xi 6= 0 ⊂ Pn, thenXi := X ∩ Ui is a quasi-projective variety which is an affine variety.

Definition 12.16. Let F : X → Y be a map of quasi-projective varieties X ⊂ Pn andY ⊂ Pm, then F is a morphism if for any x ∈ X there exists an open subset x ∈ U ⊂ Xsuch that

F |U = [p0 : . . . : pm]

for some homogeneous polynomials of degree d in C[x0, . . . , xn]d such that V(p0, . . . , pn)∩U =∅. We say that F is an isomorphism if there is a morphism G : Y → X such thatF G = idY and G F = idX .

Definition 12.17. Let X be a quasi-projective variety. If X is isomorphic to an affinevariety Y ⊂ Cn ⊂ Pn, then we say that X is an affine variety.

Here are some examples:31

(1) d-Uple embedding of Pn. Let M0, . . . ,MN be the monomials of degree d in x0, . . . , xn,so that N =

(n+dn

)− 1 and define

ρd : Pn → PN , ρd(x) = [M0(x) : . . . : MN(x)].

So for example if n = 1, d = 3, then ρ3(x, y) = [x3 : x2y : xy2 : y3] and if n = 2,d = 2 then ρ2(x, y, z) = [x2 : y2 : z2 : xy : xz : yz].

(2) Segre embedding. Define ψ : Pr × Ps → PN where N = (r + 1)(s+ 1)− 1 by

ψ([x0 :, . . . : xr]× [y0 : . . . : ys]) = [x0y0 : . . . x0ys : x1y0 : . . . : x1ys : . . . : xry0 : . . . : xrys].

(3) ψ : P2 \ xyz = 0 → P2 defined by ψ([x : y : z]) = ([yz : xz : xy]).(4) Mobius transformations: Let

M =

(a bc d

)∈ GL(2,C)

be an invertible 2× 2 matrix, then we define the map

φM : P1 → P1, [x : y]→ [ax+ by : cx+ dy].

Notice that since M is invertible, if (x : y) 6= (0, 0), then (ax+ by : cx+ dy) 6= (0, 0)and so the map is well defined. It is also clear that φM = φtM for any t ∈ C∗ andone can check that φM 6= φN if M,N ∈ GL(2,C) are linearly independent. Noticealso that φ−1

M = φM−1 . Therefore the group PGL(2,C) = GL(2,C)/C∗ acts on P1.Finally we remark that in affine coordinates (on C = P1 \ V(y)) we obtain

[x : 1]→ [ax+ b

cx+ d: 1].

(5) P2 99K P1 × P1 defined by [x : y : z] 99K ([x : z], [y : z]).

Exercise 12.18. Show that the product of two affine varieties is an affine variety (in par-ticular irreducible).

Definition 12.19. If X ⊂ Pn is a projective variety, then S(X) = C[x0, . . . , xn]/I(X) isthe homogeneous coordinate ring of X.

Since A(X) is Noetherian, S(X) is also Noetherian (see Exercise 3.23). Note that S(Pn) =C[x0, . . . , xn] and if Ui = xi 6= 0, then A(Ui) ∼= C[x0/xi, . . . , xi−1/xi, xi+1/xi, . . . , xn/xi]which we think of as the set of degree 0 elements in the ring C[x0, . . . , xn][1/xi].

Consider now the following example of an affine curve X = V(x2 − y3) ⊂ C2 correspond-ing to the projective curve X = V(zx2 − y3) ⊂ P2. Then S(X) = C[x, y, z]/(x2z − y3)and A(X) = C[x, y]/(x2 − y3) which we identify with the set of degree 0 elements inC[x, y, z, z−1]/(x2z−y3) via the inclusion φ : C[x, y, z]/(x2z−y3)→ C[x, y, z, z−1]/(x2z−y3)defined by φ(x) = x/z, φ(y = y/z) and hence φ(p(x, y)) = p(x/z, y/z). In particularφ(x2 − y3) = (x/z)2 − (y/z)3 = z−3(zx2 − y3) = 0 ∈ C[x, y, z, z−1]/(x2z − y3).

Lemma 12.20. Let X ⊂ Pn be a projective variety, Xi = X ∩ Ui, then A(Xi) correspondsto the elements of degree 0 in S(X)[1/xi] and S(X)[1/xi] = A(Xi)[xi, x

−1i ].

Proof. Exercise.

Theorem 12.21. Let X, Y ⊂ Cn be affine varieties, then every irreducible component ofX ∩ Y has dimension ≥ dimX + dimY − n.

32

Proof. Suppose that Y = V(f) is a hypersurface. If X ⊂ Y , then the claim is obvious,so suppose that X 6⊂ Y and pick an irreducible component W of X ∩ Y . If A(X) is thecoordinate ring of X, then Y ∩ X corresponds to A(X)/(f) and W to a minimal primeideal P containing (f). By Krull’s Hauptidealsatz, P has height 1 and hence dimW =dimA(X)/P ≥ dimX − 1 as required.

For the general case, we let Cn ∼= ∆ ⊂ Cn × Cn be the diagonal and we observe that∆ ∩ (X × Y ) = X ∩ Y . The result now follows by applying the hypersurface case n times,since ∆ = V(x1 − y1, . . . , xn − yn).

Theorem 12.22. Let X, Y ⊂ Pn be projective varieties, then every irreducible componentof X∩Y has dimension ≥ dimX+dimY −n and if dimX+dimY −n ≥ 0, then X∩Y 6= ∅.Proof. To see that X ∩ Y has dimension ≥ dimX + dimY −n it suffices to observe that Pnis covered by affine varieties Ui ∼= Cn and apply Theorem 12.21.

Suppose now that dimX+dimY−n ≥ 0 and consider the corresponding cones C(X), C(Y ) ⊂Cn+1. Since the intersection is non-empty (it contains 0), it follows that there is an irre-ducible component W ⊂ C(X) ∩ C(Y ) of dimension

≥ dimC(X) + dimC(Y )− (n+ 1) = dimX + 1 + dimY + 1− (n+ 1) ≥ 1.

But then W = C(Z) for some non empty subset Z ⊂ Pn and hence X ∩ Y ⊃ Z is notempty.

Exercise 12.23. Show that the homogeneous coordinate ring is not an invariant of X (eg.consider X = P1 embedded in P3 via [s : t]→ [s4 : s3t : st3 : s4] and X = P1 embedded in P4

via [s : t]→ [s4 : s3t : s2t2 : st3 : s4]).

Exercise 12.24. Show that the rational normal curve in P3 is not a complete intersection.

Exercise 12.25. Show that if H ∈ C[x0, . . . , xn] is homogeneous of degree d > 0, thenPn \ V(H) is affine.

13. Regular functions on quasi-projective varieties

Definition 13.1. Let X ⊂ Pn be a quasi-projective variety and f : X → C a continuousfunction, then f is regular at x ∈ X if there is an open affine subset x ∈ U ⊂ X such thatf |U is regular.

For the above definition to be useful, we will need the following observation:

Lemma 13.2. Let x ∈ X ⊂ Pn be a point on a quasi-projective variety, then there existsan open affine subset x ∈ U ⊂ X.

Proof. Let x = [a0 : . . . : an] and assume that ai 6= 0, then x ∈ Xi = X \ V(xi) ⊂ X is aquasi-affine variety. By Lemma 1.32, there is an affine open subset x ∈ U ⊂ Xi.

We make the following observations: Regular functions on X define a sheaf, so that

(1) It is easy to see that if U ⊂ X is an open subset, then the set of regular functionson U given by

OX(U) = f : U → C, s.t. f is regular at each point of Uis a C algebra.

(2) If U ⊂ V ⊂ X are two open subsets and f ∈ OX(V ), then f |U ∈ OX(U). The readershould check that this defines a homomorphism of C-algebras OX(V )→ OX(U).

33

(3) If U, V ⊂ X are two open subsets and f ∈ OX(V ), g ∈ OX(U) agree on U ∩ V(i.e. f |U∩V = g|U∩V ), then there exists a regular function h ∈ OX(U ∪ V ) such thath|U = g and h|V = f .

Lemma 13.3. Let f : X → Y be a map of quasi-projective varieties then f is a morphismif and only if for any p ∈ X there are affine neighborhoods p ∈ U ⊂ X and f(p) ∈ V ⊂ Ywith f(U) ⊂ V such that f |U is a regular map of affine varieties.

Proof. Exercise.

Definition 13.4. A morphism of quasi-projective varieties f : X → Y is a projectivemorphism if X is isomorphic to a closed subset of Y ×Pn (which we identify with X so thatwe write X ⊂ Y × Pn) and f is induced by the projection p : Y × Pn → Y (i.e. f = p|X).

Note that for any point p ∈ Y , we have that f−1(p) ⊂ Pn is a projective variety andhence is compact in the Euclidean topology. In fact, it follows that for any compact (in theEuclidean topology) subset V ⊂ Y , the inverse image f−1(V ) is also compact.

14. Rational maps

Definition 14.1. Let X ⊂ Pn be a projective variety and p, q ∈ C[x0, . . . , xn] be non-zerohomogeneous polynomials of the same degree d. If p 6∈ I(X), then p/q defines a rationalfunction on X \ V(q). Note that if [a0 : . . . : an] ∈ Pn and λ ∈ C∗, then

p(λa0 : . . . : λan)

q(λa0 : . . . : λan)=λdp(a0 : . . . : an)

λdq(a0 : . . . : an)=p(a0 : . . . : an)

q(a0 : . . . : an)

so that p/q is a well defined function on Pn \ V(g) and hence on X \ V(g).

Exercise 14.2. Check that p/q and p′/q′ define the same function on (a non-empty opensubset of) X if and only if pq′ − qp′ ∈ I(X).

Definition 14.3. Let X ⊂ Pn be a projective variety. We let

C(X) = p/q|p, q ∈ C[x0, . . . , xn]d, q 6∈ I(X)/ ∼be the field of rational functions of X. Here ∼ is the equivalence relation given byp/q ∼ p′/q′ if pq′ − qp′ ∈ I(X).

Exercise 14.4. Check that ∼ is an equivalence relation.

Exercise 14.5. Check that C(X) is a field.

Exercise 14.6. Let X ⊂ Cn be an affine variety and X ⊂ Pn be its projective closure. Showthat C(X) is the field of fractions of A(X).

Definition 14.7. Let f be a function defined on an open subset U ⊂ X of a projectivevariety. For any point p ∈ U , we say that f is regular at p if there exists an open subsetp ∈ V ⊂ U and a rational function p/q ∈ C(X) such that V ∩ V(q) = ∅ and f |V = (p/q)|V .We let Reg(f) ⊂ X be the set of all regular points of f . Note that this is an open subset.

Definition 14.8. Let X ⊂ Cn (resp. X ⊂ Pn) be an affine (resp. projective) variety. Arational map φ : X 99K Cm (resp. φ : X 99K Pm) is given by

φ : x→ (f1(x), . . . , fm(x)), x→ [f0(x) : . . . : fm(x)]34

where x ∈ X and fi ∈ C(X). Note that φ is not defined if fi(x) is not defined (i.e. ifx 6∈ ∩Reg(fi)) resp. if fi(x) is not defined or if all f0(x) = . . . = fm(x) = 0 (since[0 : . . . : 0] is not a point of Pm). This is why φ is denoted with a broken arrow as it isonly parrtially defined. We consider two rational maps φ, φ′ : X 99K Pm to be equivalent ifthey agree on a common open subset U ⊂ X. In this case we have fi(x) = gi(x)f ′i(x) for allx ∈ U and so gi(x) ∈ C(U). Given a rational map φ : X 99K Pm we may then consider thedomain of definition Dom(φ) = ∪Ui where φ|Ui

: Ui → Pm is defined by fi ∈ C(X) suchthat Ui ∩mi=0 V(fi) = ∅ and Ui ⊂ ∩Reg(fi).

Exercise 14.9. Show that Dom(φ) is a non empty open subset of X.

Remark 14.10. Note that given φ : X 99K Pm represented by x → [f0(x) : . . . : fm(x)],we may also consider the representative x → [1 : f1(x)/f0(x) : . . . : fm(x)/f0(x)]. It isthen easy to see that there is a one to one correspondence between the set of rational mapsφ : X 99K Cm and the set of rational maps φ : X 99K Pm.

Remark 14.11. Note that if f0, . . . , fm ∈ C[x1, . . . , xn] are homogeneous of the same degreesay d, then we may define a rational map φ : Pn → Pm by [f0/x

d0 : . . . : fm/x

m0 ] and this

map is clearly equivalent to [f0/xdi : . . . : fm/x

mi ] for any 0 ≤ i ≤ m.

We have the following examples:

(1) Rational normal curve of degree n: φ : P1 99K Pn defined by [x : y] → [xn :xn−1y : . . . : xyn−1 : yn].

(2) Projection from a linear subspace: Let L ⊂ pn be a linear subspace of dimensionr (i.e. L ∼= Pr), then we may define the projection φ : X 99K Pn−r−1. In appropriatehomogeneous coordinates, we may assume that L = V(xr+1, . . . , xn) and φ(x) =[xr+1 : . . . : xn]. It follows easily that Dom(φ) = X \ L.

(3) φ : P2 99K P2 defined by [x : y : z] 99K [1/x : 1/y : 1/z]. check that Dom(f) =P2 \ [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1].

(4) Projection of the cusp from the origin: Let X = V(zx2− y3) and φ : P2 99K P1

defined by [x : y : z]→ [x : y] be the projection from [0 : 0 : 1], then φ|X : X \ [0 :0 : 1] → P1 \ [0 : 1] is a bijection. The rational map ψ : P1 99K X defined by[x : y]→ [x : y : y3/x2] is the inverse of φ|X . Thus X and P1 are birational.

(5) Projection of a nodal cubic from the origin: Let X = V(y2z − x3 − x2z) andφ : P2 99K P1 defined by [x : y : z] → [x : y] be the projection from [0 : 0 : 1], thenφ|X : X \ [0 : 0 : 1] → P1 \ [1 : 1], [1 : −1] is a bijection. The rational mapψ : P1 99K X defined by [x : y]→ [x : y : x3/(y2− x2)] is the inverse of φ|X . Thus Xand P1 are birational.

We will see later that neither the cusp nor the nodal cubic are isomorphic to P1 (since theyare singular, but P1 is smooth).

Definition 14.12. Let φ : X 99K Y be a rational map of projective varieties. We sat thatφ is dominant if f(Dom(φ)) ⊂ Y is dense i.e. f(Dom(φ)) = Y .

Definition 14.13. Let φ : X 99K Y be a dominant rational map of projective varietiesthen there is a naturally induced map of function fields φ∗ : C(Y ) → C(X) defined byφ∗(g) = g φ. More explicitely, we may assume that X ⊂ Pn (with coordinates x0, . . . xn)and Y ⊂ Pm (with coordinates y0, . . . ym) so that if φ is represented by [f0 : . . . : fm] for somerational functions fi ∈ C[x0, . . . , xn] whose regular locus intersects X and who do not all

35

vanish along X, then for any rational function g ∈ C(Y ), we may consider g = g(y0, . . . , ym)and we let φ∗(g) = g(f0(x0, . . . , xn), . . . , (f0(x0, . . . , xn)). Of course one must check that thisis all well defined. For example if g′ ∈ C[y0, . . . , ym] is another representative of g so that(g−g′)|Y = 0, then g = g′+h where h ∈ I(Y ). But then since φ∗(h) ∈ I(X) it easily followsthat φ∗(g)− φ∗(g′) ∈ I(X) so that φ∗(g) and φ∗(g′) define the same element in C(X).

Definition 14.14. A rational map of projective varieties φ : X 99K Y is birational if thereexists a rational map ψ : Y 99K X such that φ ψ = idY and ψ φ = idX (note that thismeans that φ ψ and ψ φ are the identities on some open subsets).

Proposition 14.15. Let φ : X 99K Y be a dominant rational map of projective varietiesthen the following are equivalent.

(1) φ is birational,(2) φ∗ : C(Y )→ C(X) is an isomorphism,(3) there exists open subsets X0 ⊂ X and Y0 ⊂ Y such that f |X0 : X0 → Y0 is an

isomorphism.

Proof. We will first show that (1) and (2) are equivalent. Assume that φ is birational, thenby assumption there is a rational map ψ : Y 99K X such that ψ φ and φ ψ induce theidentity on some open subsets of X and Y respectively. But then it is clear that ψ is alsodominant and φ∗ ψ∗ and ψ∗ φ∗ are the identity on C(X) and C(Y ). Suppose now thatφ∗ : C(Y ) → C(X) is an isomorphism. Since X ⊂ Pn and Y ⊂ Pm are projective, we mayfix coordinates x0, . . . , xn and y0, . . . , ym. If fi = φ∗(yi), then φ : X → Y is representedby [f0 : . . . : fm]. Since φ∗ is an isomorphism we may pick g0, . . . , gm ∈ C(Y ) such thatφ∗(gi) = xi and we define ψ : Y 99K X via [g0 : . . . : gn]. Note that

(φ ψ)∗(xi) = φ∗(ψ∗(xi)) = φ∗(gi) = xi

and so φ ψ is the identity on an open subset.We next show that (1) and (3) are equivalent. Since (3) clearly implies (1), suppose that

(1) holds i.e. there is a rational map ψ : Y 99K X such that ψ φ and φ ψ induce theidentity on X and Y respectively. Let X0 = Dom(φ) and Y0 = Dom(ψ); φ0 = φ|X0 andψ0 = ψ|Y0 ; X1 = φ−1

0 (Y0) and Y1 = ψ−10 (X0). Thus we have a map

idX1 = (ψ φ)|X1 : X1 → Y0 → X.

We claim that φ0(X1) ⊂ Y1. In other words we must show that if x ∈ X1, then ψ0(φ0(x)) ∈X0, but this is obvious since ψ0(φ0(x)) = x ∈ X1 ⊂ X0. Clearly we also have ψ0(Y1) ⊂ X1.But then it is clear that the induced maps X1 → Y1 and Y1 → X1 are isomorphisms.

Exercise 14.16. Use the Primitive Element Theorem and Noether’s normalization lemma,to show that any r dimensional variety is birational to a hypersurface in Pr+1.

Exercise 14.17. Show that P2 and P1 × P1 are birational.

Theorem 14.18. If X is a projective variety, then OX(X) = C.

Proof. Let Xi = X ∩ xi 6= 0 ⊂ Cn be the corresponding affine varieties, then f |Xi∈

OX(Xi) ⊂ C[x0, . . . , xn, x−1i ]/I(Xi) so f = gi/x

Nii for some gi ∈ C[x0, . . . , xn] homogeneous

of degree Ni and so fxNii ∈ S(X)Ni

(where S(X)Nidenotes the subset corresponding to

homogeneous polynomials of degree Ni).36

Let N =∑Ni, then S(X)N · f ⊂ S(X)N . To see this, it suffices to observe that if xI is

homogeneous of degree N , then xNii divides xI for some i and so xI ·f = xI/xi ·(xi ·f) ∈ S(X)

is homogeneous of degree N . But then,

S(X)N · xN00 f t ⊂ S(X)N · xN0

0 f t−1 ⊂ . . . xN00 ⊂ S(X)N

and so S(X)[f ] ⊂ x−N00 S(X). Since S(X) is a Noetherian ring and x−N0

0 S(X) is a finitelygenerated S(X) module, it follows that S(X)[f ] is finitely generated S(X) module (seeTheorem 4.24). But then we may find a monic polynomial for f (see the proof of Lemma5.7)

fd + a1fd−1 + . . .+ ad−1f + ad = 0

where ai ∈ S(X). Since f has degree 0, the above equation also holds if we replace eachai by their homogeneous component of degree 0 and so we may assume that f satisfies amonic polynomial in C[x]. Since C is algebraically closed, then f ∈ C.

Exercise 14.19. Explain why there are(n+dn

)monomials of degree d in n variables.

Exercise 14.20. Let X = V(xy − zw) ⊂ P3. Show that X is isomorphic to P1 × P1.

Exercise 14.21. Show that P2 is not isomorphic to C2 nor to P1 × P1.

Exercise 14.22. If an affine variety is isomorphic to a projective variety, then it is a point.

15. Projective Morphisms are proper

Roughly speaking, a projective morphism is a morphism f : X → Y of quasi projec-tive varieties that factors through PnY := Pn × Y , more precisely f is induced by a closedimmersion X → PnY followed by the projection p : PnY → Y (a closed immersion is a mapf : X → P which is a homeomorphism of topological spaces and induces a surjectionf ]OP → OX . See [Hartshorne].)

Theorem 15.1. Let Y be a quasi-projective variety. Let X ⊂ Pn×Y be a closed subset andp : Pn × Y → Y the projection on to the second factor, then p(X) ⊂ Y is a closed subset.

Proof. Since the claim is local on Y , we may assume that Y is affine and hence Y ⊂ Cn.Notice that if p(X) ⊂ Y ⊂ Cm is a closed subset of Cm, then it is a closed subset of Y .Therefore we may assume that Y = Cm.

Suppose now that X = V(f1(x, y), . . . , fr(x, y)) where x = [x0 : . . . : xn] are homoge-neous coordinates on Pn and y = (y1, . . . , ym) are coordinates on Cm such that fi(x, y) ishomogeneous of degree di with respect to the coordinates [x0 : . . . : xn]. It suffices to showthat if y ∈ Cm \ p(X), then there is an open subset y ∈ U ⊂ Cm \ p(X). Note that ify ∈ Cm \ p(X), then V(f1(x, y), . . . , fr(x, y)) = ∅ or equivalently there is a k > 0 such that(f1(x, y), . . . , fr(x, y)) ⊃ (x0, . . . , xn)k. Consider now the ideal

I = (f1(x, y), . . . , fr(x, y)) ⊂ R = C[x0, . . . , xn, y1, . . . , ym],

and Let Rk ⊂ C[x0, . . . , xn, y1, . . . , ym] be the set of polynomials which are homogeneous ofdegree N with respect to x0, . . . , xn. By what we have seen above, we have

I ∩R/myR ⊃ (x0, . . . , xn)k ⊗R/my.

It is easy to see that this implies I ∩Rk + myR = Rk and hence that

my(RN/I ∩Rk) = Rk/(I ∩Rk).37

Since RN/(I∩RN) is a finitely generated k[y1, . . . , ym] module, it follows that (Rk/(I∩Rk))my

is a finitely generated C[y1, . . . , ym]my module, and so by Nakayama’s Lemma (Corollary 6.3)we have By Nakayama’s lemma it follows that (Rk/(I ∩ Rk))my = 0. But then there existsan f ∈ C[y1, . . . , ym] \ my such that f(Rk/(I ∩ Rk)) = 0 i.e. Rk = I ∩ Rk over Cm \ V(f).But then (f1(x, y), . . . , fr(x, y)) ⊃ (x0, . . . , xn)k for any y ∈ Cm \ V(f). This completes theproof.

Exercise 15.2. Show that Theorem 15.1 fails if the morphism is not projective.

We use the above result to show the following result on the semicontinuity of the fiberdimension of a projective morphism.

Theorem 15.3. Let Y be a quasi-projective variety. Let X ⊂ Pn × Y be a closed subsetand p : Pn × Y → Y the projection on to the second factor, then

Yk = y ∈ Y | dim(p−1(y)) ≤ k

is open.

Proof. Since the question s local on Y , we may assume that Y is afffine. If dim(p−1(y)) ≤ kand H1, . . . , Hk+1 are general hyperplanes in |OPn

Y(1)|, then (p−1(y))∩H1 ∩ . . .∩Hk+1 = ∅.

Let Z = H1 ∩ . . . ∩ Hk+1 and W = p(Z) ⊂ Y . By Theorem 15.1, W is closed. Sincep−1(y′) ∩ H1 ∩ . . . ∩ Hk+1 = ∅ for any y′ ∈ Y \W , it follows that dim p−1(y′) ≤ k for anyy′ ∈ Y \W and the theorem is proven.

Exercise 15.4. Show that if f : X → Y is a surjective projective morphism, k = dimX −dimY , then Yk−1 = ∅, Yk = Y and dimYk+t ≤ dimY − t− 1.

16. Normality

Definition 16.1. Let A ⊂ B be an inclusion of rings, then the integral closure of A inB is the union of all elements b ∈ B which are integral over A (i.e. they satisfy a monicpolynomial with coefficients in A). If A is a domain, then the integral closure A of A isthe integral closure of A in its quotient field Q(A). Clearly A ⊂ A ⊂ Q(A). If A = A, thenwe say that A is integrally closed or normal.

Exercise 16.2. Let A ⊂ B be an inclusion of rings. Show that the integral closure A ⊂ Bof A in B is a ring.

Exercise 16.3. Show that if A is factorial (i.e. it has a unique factorization) then A isintegrally closed.

Exercise 16.4. Show that the integral closure A ⊂ B is integrally closed in B.

Exercise 16.5. Show that the integral closure of Z ⊂ Q(i) is Z[i].

Exercise 16.6. Show that the integral closure of Z ⊂ Q(√

5) is (a+ b√

5)/2| 4|(a2−5b2).

Exercise 16.7. Let ξ be an n-th root of 1. Show that the integral closure of Z ⊂ Q(ξ) isZ[ξ].

Exercise 16.8. Suppose that Ai ⊂ B are normal in B, then show that A = ∩Ai ⊂ B isnormal in B.

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Lemma 16.9. Let A ⊂ B be an inclusion of domains and S a multiplicatively closed subsetof A. If A is the integral closure of A in B, then S−1A is the integral closure of S−1A inS−1B.

Proof. If b ∈ B is integral over A then it satisfies a monic polynomial xn+a1xn−1 +. . .+an ∈

A[x]. But then, for any s ∈ S, the element b/s ∈ S−1A satisfies the monic polynomialxn + (a1/s)x

n−1 + . . .+ (an/sn) ∈ S−1A[x] and so S−1A is contained in the integral closure

of S−1A in S−1B.Conversely, suppose that b/s is in the integral closure of S−1A in S−1B, then it satisfyes a

monic polynomial xn+(a1/s1)xn−1+. . .+(an/sn) ∈ S−1A[x]. But then bs1 · · · sn satisfies themonic polynomial xn+a1s1 · · · sns/s1x

n−1 + . . .+an(s1 · · · sns)n/sn ∈ A[x]. Thus bs1 · · · sn ∈A and hence b/s ∈ S−1A.

Definition 16.10. Let X be an affine variety, then X is normal if and only if A(X) isintegrally closed.

Lemma 16.11. Let X be an affine variety, then X is normal if and only if the local ringsOp are integrally closed rings for all points p ∈ X.

Proof. Suppose that X is normal and hence A = A(X) is integrally closed in Q = Q(A).Let m ⊂ A be the maximal ideal of p ∈ X then Op = S−1A where S = A \ m is thecorresponding multiplicative subset. Since X is normal A = A is integrally closed. ByLemma 16.9, Op = S−1A is integrally closed (in Q(S−1A) = Q(A)).

Conversely, suppose that the local rings Op are integrally closed rings for all points p ∈ X.Let b ∈ A ⊂ Q, then as A ⊂ Op ⊂ Q and Op = Op, it follows that b ∈ ∩p∈XOp = A (wherefor the last equality we used 1).

The advantage of the above lemma, is that the definition of normal can be made localand hence it applies to any variety.

Definition 16.12. A quasi-projective variety X is normal at a point p ∈ X if Op is anintegrally closed ring. X is normal if it is normal at every point p ∈ X.

Definition 16.13. Let X be an affine variety, then the normalization ν : Xν → X is definedby the inclusion of rings A(X)→ A(X). Note that Xν is normal and affine.

Lemma 16.14. Let Y be a normal affine variety and f : Y → X be a dominant morphismof affine varieties, then f factors uniquely through the normalization Xν → X.

Proof. Let A = A(X) and B = A(Y ), then f : Y → X is determined by the inclusionf ∗ : A→ B2 which induces an inclusion f ∗ : Q(A)→ Q(B). It follows immediately that wehave A ⊂ B = B and hence an induced map Y → Xν → X.

Lemma 16.15. Let X be an affine variety and ν : Xν → X its normalization. If U ⊂ Xis an affine open subset, then ν−1(U)→ U is the normalization of U .

Proof. By Lemma 16.14 there is a unique map Uν → Xν factoring Uν → U → X and hencea map Uν → ν−1(U). On the other hand since ν−1(U) is normal, there is a unique mapν−1(U)→ Uν . By the uniqueness of these maps, it follows that ν−1(U) ∼= Uν .

1ref2ref

39

Lemma 16.16. If X is a quasi-projective variety, then there exists a morphism ν : Xν → Xsuch that for any affine subset U ⊂ X, ν−1(U)→ U is the normalization of U (in particularν is uniquely determined by this property).

Proof. Let X = ∪Ui where Ui is affine. Let νi : Uνi → Ui be the corresponding normaliza-

tions, it suffices to show that there is a natural isomorphism between ν−1i (Uij) and ν−1

j (Uij).By Lemma 16.14

Next we compute an interesting example. Let X = V(y2 − x2(x + 1)) ⊂ C2. we wishto compute the normalization Xν . Let t = y/x ∈ Q(A) where A = A(X) = C[x, y]/(y2 −x2(x + 1)), Then y/x ∈ A since it satisfies the monic polynomial t2 − (x + 1) ∈ A[t]. Notethat t2 = (y/x)2 = x+1 and so x = t2−1 and y = tx = t(t2−1) and so there is an inclusionA ⊂ C[t] ⊂ Q(A) given by x → t2 − 1 and y → t(t2 − 1). Clearly C[t] is integrally closedand hence A = C[y/x] ∼= C[t].

Exercise 16.17. Let X = V(xy) ⊂ C2 compute the normalization ν : Xν → X.

Exercise 16.18. Let X = V(x2 − y3) ⊂ C2 compute the normalization ν : Xν → X.

Exercise 16.19. Let X = V(x2 − y4) ⊂ C2 compute the normalization ν : Xν → X.

Note that in all cases, we have that Xν is smooth. This is a general fact:

Theorem 16.20. Let X be a one dimensional quasi projective variety (a curve), then Xν

is smooth.

Corollary 16.21. Let X be a normal quasi-projective variety, then there is a closed subsetZ ⊂ X of codimension ≥ 2 such that X \ Z is smooth.

Proof. Exercise. This is not completely obvious. One approach is to mimic the proof ofTheorem 16.20. See http://www.math.utah.edu/ bertram/6030/Codim1.pdffor details.

Before proving Theorem 16.20, we make a brief detour.

Definition 16.22. A discrete valuation is a (non-zero) function ν : K∗ → Z where K isa field, such that

(1) ν(ab) = ν(a) + ν(b) for all a, b ∈ K∗, and(2) ν(a+ b) ≥ minν(a), ν(b).

We let

Aν = k ∈ K∗|ν(k) ≥ 0 ∪ 0be the corresponding DVR (discrete valuation ring).

Exercise 16.23. Show that if p is a prime number and ν : Q∗ → Z is defined by ν(a/b) =ν(a) − ν(b) where ν(a) is the maximum power of p dividing a, then ν : Q∗ → Z is a DVRand Avu = Z(p) = S−1Z where S = Z \ (p).

Exercise 16.24. For any c ∈ C and f/g ∈ C(x) let ν(f/g) be the order of the zero orpole of f/g at c (so that if c = 0, ν(xk) = k). Show that ν is a discrete valuation andAν = C[x](x−c).

Exercise 16.25. For any f/g ∈ C(x) let ν(f/g) = deg(g) − deg(f). Show that ν is adiscrete valuation and Aν = C[x−1]x−1. What happens if we let ν(f/g) = deg(f)− deg(g).

40

Exercise 16.26. Show that if ν is a discrete valuation , then ν(K∗) = dZ for some d > 0.If we let ν ′(k) = ν(k)/d, then show that ν ′(K∗) = Z, ν ′ is a discrete valuation and Aν′ = Aν.

Lemma 16.27. Let ν be a discrete valuation (such that ν(K∗) = Z), then m := k ∈K∗|ν(k) > 0 is a principal ideal in Aν. Thus m = (π) where ν(π) = 1. We say that π isthe uniformizing parameter.

Proof. Let a ∈ m and b ∈ Aν , then ν(ab) = ν(a) + ν(b) > 0 and so ab ∈ m. If a, a′ ∈ m,then ν(a+a′) ≥ minν(a), νa′ > 0. Thus m is an ideal in Aν . To check that it is principal,pick π ∈ m of minimal degree. Clearly ν(π) = 1 (as ν(K∗) = 1). For any m ∈ m wewrite m = π · b where b = m/π. Note that ν(m/π) = ν(m) − 1 ≥ 0 and so m/π ∈ Aν asrequired.

Exercise 16.28. Show that if I ⊂ Aν is any ideal, then I = mk = (πk) for some k ≥0. It follows that any DVR is Noetherian and in fact a UFD. Note that m is the uniqueprime/maximal ideal in Aν.

Lemma 16.29. Let (A,m) be a Noetherian local ring (so that A is a domain and m is theunique maximal ideal). If the Krull dimension of A is one (i.e. m is the unique prime ideal),

then for any ideal 0 6= I 6= A we have√I = m and there is a unique n ∈ N such that mn ⊂ I

but I 6⊂ mn−1.

Proof. We begin by observing that if a ∈ m, then A[a−1] = K(A). To this end, note that ifJ ⊂ A[a−1] is a proper prime ideal, then J = P [a−1] for some prime ideal P = J ∩ A ⊂ A.By our assumption P = m and since a ∈ m, 1 = a/a ∈ J which is impossible.

We will now prove that√I = m for any ideal 0 6= I ⊂ A. It suffices to show m ⊂

√I.

Pick m ∈ m and 0 6= a ∈√I \m, then (by what we have seen above) a−1 = b/mn for some

b ∈ A and n ≥ 0. Note that in fact n > 0 since a 6∈ m. But then mn = ab and so m ∈√I.

Finally, it suffices to show that mn ⊂ I for some n. This is where we use the fact thatA is Noetherian and so m = (m1, . . . ,mk). Since

√I = m, there are integers ni such that

mnii ∈ I, but then if n = 1− k +

∑ni, it is easy to see that mn ∈ I for any m ∈ m. Notice

in fact that m =∑miai and mn = (a1m1 + . . .+ akmk)

n. Expanding this out, we see thatevery term contains a power mni

i for some 1 ≤ i ≤ k and hence belongs to I.

We have the following important result.

Theorem 16.30. Let (A,m) be an integrally closed Notherian local ring of dimension 1,then A is a DVR.

Proof. We begin by proving that m is principal i.e. m = (π) for some π ∈ A. Pick any0 6= a ∈ m. By Lemma 16.29, there is an integer n such that mn ⊂ (a) but mn−1 6⊂ (a). Pickb ∈ mn−1 \ (a), then b/a ∈ K(A) \ A and (b/a)m ⊂ A (this is because bm ⊂ mn ⊂ (a)). Weclaim that however (b/a)m 6⊂ m. Grant this for the time being, then there is π ∈ m such that(a/b)π = u 6∈ m and hence u is a unit. But then for any m ∈ m, we have (a/b)m = c ∈ A(as (b/a)m ⊂ A) and so m = cπu−1 ∈ (π) as required.

To see the claim, suppose that (b/a)m ⊂ m, then we have an inclusion

A[a

b] → a · A[

a

b] ⊂ m[

a

b] ⊂ m.

Since A is Noetherian, it then follows that A[ab] is a finitely generated A module. But since

A is integrally closed, ab∈ A which is the required contradiction.3

3add refs41

Let 0 6= a ∈ A, if a is not a unit, then a ∈ m = (π) and we write a = πa1. If a1 isnot a unit, then we write a1 = πa2 and hence a = π2a2. We claim that after finitely manyiterations, we may write a = πnan where an is a unit. If this were not the case, then wewould have a sequence of ideals (a) ⊂ (a1) ⊂ (a2) ⊂ . . .. Since A is Noetherian, this sequenceis eventually constant so that (an) = (an+1) Since an = an+1π we have an+1 = an+1πb forsome b ∈ A. But then 1 = πb as A is a domain) and so π is a unit, which is impossible.Writing a = πnan where an is a unit, we let ν(a) = n.

We must check that ν is well defined: If a = πnan = πmb where m ≥ n, then an = πm−nb.Since π is not a unit, m− n = 0.

We leave the check that ν is a discrete valuation as an exercise.

Exercise 16.31. Check that ν is a discrete valuation.

Proof of Theorem 16.20. Replacing X by Xν we may assume that X is normal. Since Xis smooth on some open subset, replacing X by an affine open subset, we may assumethat X is affine and x ∈ X is the unique singular point. Note that C[x] is integrallyclosed and hence so is C[x]mxC. But then, by Theorem 16.30, C[x]mxC is a DVR. Let πbe the uniformizing parameter so that (π) = mxC[x]mxC. After possibly replacing X by asmaller open subset containing x, we may assume that π ∈ C[x] and mx = (π). But thendimmx/m

2x = 1 = dimX and so x ∈ X is smooth.

17. Conics

We begin by investigating conics in C2 i,.e. subsets of C2 defined by a polynomial ofdegree 2 say

q(x, y) = ax2 + by2 + 2cxy + 2dx+ 2ey + f = 0.

It is often convenient to view this in matrix form

q(x, y) =(x y 1

)a c dc b ed e f

xy1

=(x y 1

)M

xy1

.

Conics define familiar shapes in R2 such as parabolas (eg. y = x2), ellipses (eg. x2 +3y2 =2) and hyperbolas (eg. xy = 1). Any two of these shapes can be identified via a change ofcoordinates, eg. the ellipse 2(3x + y − 5)2 + 3(x − 2y − 7)2 = 3 can be identified with thecircle x2 + y2 = 3 by letting x→

√2(3x+ y − 5) and y →

√3(x− 2y − 7). In other words

we are allowed to act on (x, y) by an invertible linear transformation and to translate byarbitrary points in C2. Finally, we allow ourselves to rescale the constant term so that wecan identify the circle x2 + y2 = 3 with the unit circle x2 + y2 = 1.

It is then easy to see that after a linear transformation we may assume that the degreetwo term of q(x, y) is of the form

x2 + y2, x2, 0

depending on the rank of the matrix

(a cc b

)Suppose that we are in the rank 2 case so that

we may write

q(x, y) = x2 + y2 + 2dx+ 2ey + f = 0,

then, replacing x by (x−d) and y by y−e (and f by f −d2−e2) we obtain an expression ofthe form q(x, y) = x2 +y2 +f = 0. There are 2 cases to consider according to the rank of M .

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If the rank of M is 2 then we have the equation x2 + y2 = 0 and if the rank of M is 3, thenafter rescaling the constant term we have the equation x2 + y2 + 1 = 0 (note that since weare over the complex numbers, it does not matter if f is positive or negative; x2 +y2 +1 = 0is equivalent to x2 + y2− 1 = 0 via replacing (x, y) with (ix, iy). If the rank of M is 1, thenafter changing coordinates, we may assume that q(x, y) = x2 +2dx+2ey+f = 0. Replacingx by x+ d we may assume that q(x, y) = x2 + 2ey + f . If e 6= 0 (i.e. if the rank of M is 3),then replacing y by 2ey + f , we may assume that q(x, y) = x2 + y and otherwise if e = 0(so that the rank of M is 2 or 1) then we may assume that q(x, y) = x2 + 1 or q(x, y) = x2.All in all we have the following cases:

(1) Ellipse: x2 + y2 + 1 = 0(2) Two lines meeting at a point: x2 + y2 = 0 (note that this equation is equivalent to

(x+ iy)(x− iy) = 0).(3) Parabola: x2 + y = 0(4) Two parallel lines: x2 + 1 = 0(5) A double line x2 = 0.

Remark 17.1. Note that over the real numbers, the classification is different since x2+y2 =1 and x2 + y2 = −1 have different zero sets. In this case the classification depends on theranks of minors of M as well as their signature.

The above classification is somewhat cumbersome because we must treat the coordinates

x, y differently from the constant term (in particular we care about the rank of

(a cc b

)as

well as the rank of M . If we however consider the projective closure of these conics, i.e.conics in P2, then the situation is straightforward. A conic in P2 is defined by a homogeneouspolynomial of degree 2 say

q(x, y, z) = ax2 + by2 + 2cxy + 2dxz + 2eyz + fz2 = 0.

Two conics are equivalent if they are obtained by an invertible base change i.e. by a matrixA ∈ GL(3,C). If we replace (x, y, z) by coordinates (x, y, z)t = A(x′, y′, z′)t, then we obtaina new equation (

x y z)M

xyz

=(x′ y′ z′

)AtMA

x′y′z′

In other words, the base change replaces M by AtMA. Since M is symmetric, we may picka base change that diagonalizes M and so we may assume that q(x, y, z) = ax2 + by2 + fz2.If a 6= 0, then replacing x by

√ax, we may assume a = 1 (and similarly for b and f). Finally,

since we may permute the variables, we only have three cases:

(1) Rank M is 3: x2 + y2 + z2 = 0 an ellipse.(2) Rank M is 2: x2 + y2 = 0 two lines.(3) Rank M is 1: x2 = 0 a double line.

Notice that with this description, we have lumped together the affine Ellipse and Parabolacases: a parabola is just an ellipse which is tangent to the line at infinity; for example y = x2

in homogeneous coordinates becomes zy − x2 = 0. the line at infinity is defined by z = 0.The intersection point is [0 : 1 : 0] and in local coordinates (u = x/y, v = z/y) on the charty 6= 0 we have v = u2 which is tangent to the line at infinity v = 0. We have also lumped

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together two lines meeting at a point and two parallel lines (because two parallel lines aresimply projective lines meeting at infinity).

Exercise 17.2. Study the classification of conics over P2R.

Exercise 17.3. What are the possible quadrics in P3 (i.e. solutions of a homogeneousequation of degree 2). Show that any nonsingular quadric is isomorphic to V(xy−zw) ⊂ P3.

Exercise 17.4. Show that any nonsingular quadric in P3 is isomorphic to P1 × P1.

18. Smooth varieties

We begin with a few examples. Let y = x2 be a parabola, we wish to understand tangentlines to y = x2 at any given point (a, a2). From calculus, we know that the correct answerin parametric form is the line

t→ (a+ t, a2 + 2at) = (a+ a2) + t(1, 2a).

This reflects the fact that dy/dx = 2x and hence the slope at (a, a2) is 2a. If we evaluatex2 − y along the tangent line, we get

(a+ t)2 − (a2 + 2at) = t2

so that the induced equation vanishes to order at least 2 at the origin. In fact this property(vanishing to order at least two at the origin) characterizes the tangent lines. Considerfor simplicity the case a = 0 i.e. the point (0, 0). A line through the origin is givenparametrically by t→ (a1t, a2t) where a1, a2 are not both 0. We then restrict our equationx2 − y to this line we obtain

(a1t)2 − a2t

which vanishes to order 2 at the origin if and only if a2 = 0. From a slightly different pointof view, one can consider the gradient of f = x2 − y, given by ∇f = (2x,−1) so that thetangent line at (a, a2) is given by

0 = ∇f(a,a2) · (x− a, y − a2) = (2a,−1) · (x− a, y − a2) = 2a(x− a)− (y − a2).

As a second example, consider the sphere x2 + y2 + z2 = 1 and the tangent plane at(0, 0, 1). Since ∇(f)(0, 0, 1) = (0, 0, 2), the tangent plane is given by 2(z− 1) = 0. The linescontained in this plane are of the form

(0, 0, 1) + t(a, b, 0).

Once again, restricting the equation x2 +y2 +z2 = 1 to the line (0, 0, 1)+ t(a, b, 0) we obtain

t2(a2 + b2) = 0

so that t = 0 is a zero of order 2. On the other hand, restricting the equation x2 +y2 +z2 = 1to the line (0, 0, 1) + t(a, b, c) where c 6= 0 we obtain

t2(a2 + b2 + c2) + 2tc = 0

so that t = 0 is a zero of order 1. This leads us to the following definition.

Definition 18.1. Let a ∈ X = V(f1, . . . , fr) ⊂ Cn be a point on an affine variety (or analgebraic set) and l(t) = a + tv for some 0 6= v ∈ Cn be a line, then we say that a line l(t)is tangent to X at the point a ∈ X if l(0) = a and t = 0 is a zero of multiplicity at least 2for fi(l(t)) for all 1 ≤ i ≤ r. If the minimum multiplicity of t = 0 in fi(l(t)) is m+ 1, thenwe say that l(t) is tangent to order m at a ∈ X. The tangent space to X at a is the

44

subspace TaX ⊂ Cn given by the union of the translates through the origin of all the linestangent to X at a.

We consider next two more examples.

(1) X = V(x2 − y3) at (0, 0, 0). If l = t(a, b), then f(l(t)) = t2(a2 + b3t) has a zero oforder 2 at t = 0 for all a 6= 0 and a zero of order 3 if a = 0. therefore T(0,0)X = C2.Note that ∇(x2 − y3)(0,0) = (2x,−3y2)(0,0) = (0, 0).

(2) X = V(x2 + y2 + y3). If l = t(a, b), then f(l(t)) = t2(a2 + b2 + b3t) which has a zeroof order 2 at 0 if a 6= ±ib and a zero of order 3 if a = ±ib.

Proposition 18.2. The tangent space TaX does not depend on the choice of the basis ofthe ideal.

Proof. Assume a = (0, . . . , 0). Suppose that (f1, . . . , fr) = (g1, . . . , gs). It suffices to showthat if l(t) = tv is a line such that fi(tv) is divisible by t2 for i = 1, . . . , r, then gk(tv) isdivisible by t2 for any 1 ≤ k ≤ s. However note that gk ∈ (f1, . . . , fr) and so gk =

∑hifi

where h1, . . . , hr ∈ C[x1, . . . , xn]. But then

gk(tv) =∑

hi(tv)fi(tv)

is divisible by t2 since each fi(tv) is divisible by t2.

Proposition 18.3. Let X = V(f1, . . . , fr) be an affine variety and p ∈ X, then a linel = p+ tv is tangent to X at p if and only if

l ⊂ V((∇f1)p(x− p), . . . , (∇fr)p(x− p))where (x− p) = (x1 − p1, . . . , xr − pr) and so (∇fj)p(x− p) =

∑ ∂fk∂xi

(p) · (xi − pi).

Proof. Assume that p = (0, . . . , 0), then

fi −∇fi(p) · x ∈ (x1, . . . , xn)2

and so t2 divides fi(tv) if and only if t2 divides ∇fi(p) · tv = t(∇fi(p) ·v) which is equivalentto ∇fi(p) · v = 0.

Definition 18.4. Let p ∈ X be a point on an affine variety, then we say that p ∈ X issmooth if dimTpX = dimpX. If p ∈ X is not smooth, then dimTpX > dimpX and wesay that p ∈ X is a singular point.

Remark 18.5. Note that we have to check that the definition is independent of the particularembedding of X ⊂ CN and we have not yet shown that for non smooth points we have theinequality dimTpX > dimpX (or equivalently that dimTpX ≥ dimpX holds for any p ∈ X).

Proposition 18.6. Let X = V(f1, . . . , fr) ⊂ Cn be a d-dimensional affine variety, thenSing(X) is a proper closed subvariety defined by the vanishing of the (n − d) × (n − d)minors of the matrix

[∇f1, . . . ,∇fr].

Proof. For any p ∈ X, we have seen that the tangent space is defined by

TpX = V(∇f1(p) · (x− p), . . . ,∇fr(p) · (x− p)).Writing this in matrix form, we obtain that TpX is determined by the equation

[∇f1(p), · · · ,∇fr(p)]t · (x− p) = 045

where [∇f1(p), · · · ,∇fr(p)]t denotes the r×nmatrix with entries ([∇f1(p), · · · ,∇fr(p)]t)i,j =∂fi∂xj

(p) and x− p = (x1− p1, . . . xr− pr)t is a column vector. By linear algebra, we have that

dim(TpX) = dimCn − rank([∇f1(p), · · · ,∇fr(p)])and so dim(TpX) > dimX if and only if rank([∇f1(p), · · · ,∇fr(p)]) < n − d. By linearalgebra, this last condition is equivalent to showing that all (n− d)× (n− d) minors vanish.

We must now show that there is at least one non singular point on X. If X = V (f) is ahypersurface (i.e. X is defined by an irreducible polynomial f), and every point p ∈ X issingular, then we would have that ∂f/∂xi vanishes at every point of X and so that f divides∂f/∂xi. But this is impossible for degree reasons. To conclude the general case, it sufficesto observe that every variety contains an open subset that is isomorphic to a hypersurface(see Proposition 21.3).

Exercise 18.7. Show that if p ∈ X ∩Y where X, Y are varieties not containing each other,then p ∈ X ∪ Y is a singular point.

19. Intrinsic definition of the tangent space

Let 0 ∈ Cn be the origin and M = (x1, . . . , xn) ⊂ C[x1, . . . , xn] the corresponding maximalideal, then we can identify the tangent space T0(Cn) with M/M2. To see this, consider themap

d : M → (Cn)∗ := linear functions on Cn

defined by d(f) = f1 where f = f1 + f2 + . . . and fi ∈ C[x1, . . . , xn]i are homogeneous ofdegree i. Note that df =

∑∂f/∂xi(0, . . . , 0) · xi = ∇f(0, . . . , 0) · x. It is also easy to see

that d is surjective, since if L =∑aixi ∈ (Cn)∗, then d(L) = L. Therefore we have shown

thatd : M/M2 → (Cn)∗

is an isomorphism. We claim that this isomorphism holds for any point of any affine variety.

Theorem 19.1. Let P ∈ X ⊂ Cn be a point on an affine variety and mP ⊂ A(X) be themaximal ideal of P ∈ X, then TP (X) ∼= (mp/m

2p)∗.

Proof. Recall that TP (X) ⊂ Cn is defined by the linear terms g1 of polynomials g ∈ I(X).(We may assume that the coordinates are chosen so that P = (0, . . . , 0).) If (Cn)∗ is the dualspace, then g1 ∈ (Cn)∗ are the linear equations defining TP (X) and TP (X) = V(g1g∈I(X)).Notice that mP = M/I(X). Consider the natural map

D = (i∗ d) : M → (Cn)∗ → (TP (X))∗

where d : M → (Cn)∗ was defined above and i∗ is the dual map to i : TP (X) → Cn

(cf. Exercise 19.2). It is easy to see that D is surjective (cf. Exercise 19.3). We musttherefore understand the kernel of D i.e. the set of functions f ∈M such that f1|TP (X) = 0.This condition is equivalent to saying that f1 ∈ (g1g∈I(X)) or that f1 =

∑aigi1 for some

gi ∈ I(X) and ai ∈ C. In other words f−∑aigi ∈M2 for some gi ∈ I(X) i.e. f ∈M2+I(X).

But then

(TP (X))∗ =M

M2 + I(X)=

mp

m2p

where we have used the fact that mp = M/I(X) and Exercise 19.4. 46

Exercise 19.2. Show that ((Cn)∗)∗ = Cn and show that if V is an infinite dimensionalvector space, then (V ∗)∗ 6= V .

Exercise 19.3. Given a linear map of vector spaces f : V → W , show that there is anaturally defined linear map of vector spaces f ∗ : W ∗ → V ∗ and that if f ∗ = 0, then f = 0.If V,W are finite dimensional, then show that f ∗ is injective (resp surjective) if and only iff is surjective (resp. injective).

Exercise 19.4. Let M, I ⊂ R be ideals. Show that M/(M2 + I) = (M/I)/(M/I)2.

20. Resolution of singularities

Recall that if X and Y are quasi-projective varieties, then X and Y are birational ifthere exists a rational map f : X 99K Y and open subsets U ⊂ X and V ⊂ Y suchthat f |U : U → V is an isomorphism. Equivalently, we have seen that two quasi projectivevarieties are birational if and only if their function fields are isomorphic. Birational geometryis the study of quasi-projective varieties up to birational isomorphism. One of the mostimportant results in birational geometry is Hironaka’s resolution of singularities.

Theorem 20.1. Let X be any quasi projective variety, then there exists a projective mor-phism µ : X ′ → X such that X ′ is smooth and if X0 = X \ Sing(X) and X ′0 = µ−1(X0),then µ|X′0 : X ′0 → X0 is an isomorphism. In particular µ is birational.

Remark 20.2. Hironaka’s theorem is a very deep result whose proof is beyond the scope ofthese notes. We will however illustrate it with several examples and prove it in dimension1.

It turns out that the resolution µ : X ′ → X can be constructed rather explicitely in termsof certain maps known as blow ups. We begin by considering the blow up of Cn at theorigin. Let

B = (x, y) ∈ Cn × Pn−1|x ∈ y ⊂ Cn × Pn−1,

and the morphism µ : B → Cn given by µ(x, y) = x. Notice that we are identifyingPn−1 with the set of lines through the origin in Cn. Note that B is a quasi-projectivevariety defined by equations xiyj = xjyi for all 1 ≤ i, j ≤ n. It is clear that B is aquasi-projective variety and µ is a projective morphism. If x 6= (0, . . . , 0) it follows thatµ−1(x) = (x, [x]) and E := µ−1((0, . . . , 0)) = (0, . . . , 0) × Pn−1. It follows easily sees thatµ : (B \E)→ Cn \ (0, . . . , 0) is an isomorphism. One should think of E as parametrizingthe tangent directions to Cn at the origin. We will now begin to investigate the effects ofblowing up on affine varieties.

Definition 20.3. Let p ∈ X ⊂ Cn be a point on an affine variety and µ : B → Cn be theblow up of Cn at p, then we define Bp(X) the blow up of X at p by taking the closure (inthe Zariski topology) of µ−1(X \ p). In other words

Bp(X) := µ−1(X \ p).

Note that µ−1(Bp \ p) is isomorphic to Bp \ p and µ−1(p) = E ∩ Bp(X). We nowillustrate the procedure with several examples.

(1) Lines: Let X = V(ax1 + bx2) ⊂ C2 be a line (so that a, b ∈ C are not both 0).Recall that X ⊂ C2 × P1 is defined by the equation x1y2 = x2y1 where x1, x2 arecoordinates on C2 and y1, y2 are homogeneous coordinates on P1. Recall also that

47

we may cover P1 by two affine charts U2 = y2 6= 0 ∼= C with coordinate t2 = y1/y2

and U1 = y1 6= 0 ∼= C with coordinate t1 = y2/y1. Focusing on the first chartX1 = X ∩ (C2 ×U1) is defined by x1t1 = x2 and E1 = E ∩C2 ×U1 is defined by theequation x1 = 0. The equation of the line then becomes

ax1 + bx2 = x1(a+ bt1).

This means that µ−1(X) has two irreducible components, namely

µ−1(X) = E ∪Bp(X)

where Bp(X)∩C2×U1 is defined by V(a+ bt1, x1t1−x2). Note that the intersectionBp(X) ∩E is given by letting x2 = 0 and hence is given by the point (0, 0,−a/b) Inother words this intersection recovers the slope of the original line.

(2) Cusp: Let X = V(x21 − x3

2) ⊂ C2 and B → C2 be the blow up at the origin (0, 0).We follow the above computation on C2 × U1 where we have x1t1 = x2 so that

x21 − x3

2 = x21(t21 − x2).

This means thatµ−1(X) = 2E +Bp(X)

where Bp(X) intersects E at the point (0, 0, 0) ∈ C2×U1. In local coordinates x2, t1we have that Bp(X) = V(t21 − x2) is a parabola tangent to E. Note that we haveused ”2E” to record the fact that µ−1(X) vanishes to second order along E; in settheoretic terms we of course have µ−1(X) = E +Bp(X).

(3) Node: Let X = V(x21 − x3

2 − x22). Proceeding as above we have

x21 − x3

2 − x22 = x2

2(t21 − x2 − 1)

so thatµ−1(X) = 2E +Bp(X)

where Bp(X) intersects E at the points (0, 0,±1) ∈ C2 × U1. In local coordinatesx2, t1 − 1 (near (0, 1)) we have that Bp(X) = V(u(t1 − 1)− x2) where u = t1 + 1 isnon-zero on a neighborhood of (0, 1) and so Bp(X) intersects E transversely at thepoints (0, 0,±1).

(4) Consider a cone V(x21 + x2

2 − x23) ⊂ C3 and blow up the origin p = (0, 0, 0). We now

have three affine charts P2 = U1 ∪ U2 ∪ U3 where Ui = yi 6= 0. Computing on U1

we let t2 = y2/y1 and t3 = y3/y1 so that

Bp(C3) ∩ C3 × U1 = V(x2 − t2x1, x3 − t3x1)

and then µ−1(X) is defined by

x21 − x2

2 − x23 = x2

1(1 + t22 + t23)

so that µ−1(X) = 2E+Bp(X) where Bp(X) is smooth. A little reflection shows thatBp(X)∩E = V(x2

1 + x22 − x2

3) ⊂ P2 is a smooth circle and the map Bp(X)→ X is aresolution of singularities as in Hironaka’s theorem.

(5) This should not lead one to expect that one blow up always resolves a singularity.For example if X = V(y2 − x2y + y4) ⊂ C2, then letting y = tx we obtain

y2 − x2y + y4 = x2(t2 − xt+ t4x2) = x2t(t− x+ t3x2)

and so one sees that Bp(X) is defined locally by V(t(t − x + t3x2)) which is clearlysingular at q = (0, 0) (in fact it corresponds to two curves meeting transversely). To

48

resolve the singularities of X it is therefore necessary to perform a second blow upBq(Bp(X))→ X.

(6) It is easy to check that if X = V(f) ⊂ Cn is a hypersurface and B → Cn is theblow up of a point p ∈ X, then Bp(X) ∩ E = V(fd) ⊂ Pn−1 is the projectivehypersurface defined by the leading term of the Taylor expansion of f at p. Forexample if p = (0, . . . , 0) is the origin, the f − fd ∈ (x1, . . . , xn)d+1 (i.e. f = fd+higher order terms).

Exercise 20.4. Define the blow up of Cn+r along a subspace Cn as follows. Pick coordinatesx0, . . . , xn+r such that Cn = V(xn+1, . . . , xn+r). Then consider the subset X ⊂ Cn+r × Pr−1

defined by the equations xn+iyj − xn+jyi for 1 ≤ i, j ≤ r. Use this definition to define theblow up of a smooth n+ r-dimensional variety along a smooth n-dimensional subvariety.

Exercise 20.5. Show that the blow up of Pn+r along a linear subspace L ∼= Pn is isomorphicto the graph of the corresponding projection Pn+r 99K Pr.4

21. Smooth projective varieties and generic smoothness

Theorem 21.1. Let X be a smooth projective variety of dimension d, then X is isomorphicto a subvariety of P2d+1.

Proof. By assumption, X ⊂ Pn for some n > 0. If n > 2d + 1, then the secant varietyS(X) ⊂ Pn given by the closure of the union of all lines through two points of X hasdimension ≤ 2d + 1 and the tangent variety T (X) ⊂ Pn given by the union of all tangentlines has dimension ≤ 2d and hence we may pick a point p ∈ Pn \ S(X) ∪ T (X). Considerthe projection map φ : Pn 99K Pn−1. Since p 6∈ X, then φ|X : X → Pn−1 is a morphism.Since p 6∈ S(X), φ|X is bijective and since p 6∈ T (X), the differential of φ|X has maximalrank at each point x ∈ X and so X → φ(X) induces an isomorphism of tangent spaces.5

Exercise 21.2. Show that the secant variety S(X) ⊂ Pn is a variety of dimension ≤ 2d+ 1and the tangent variety T (X) ⊂ Pn is a variety of dimension ≤ 2d.

Proposition 21.3. Every projective variety is birational to a hypersurface.

Proof. Consider an embedding X ⊂ Pn. Suppose that dimX ≤ n − 2. Let p ∈ Pn \ X.Consider the projection map φ : Pn 99K Pn−1. If φ|X is not birational, then for any x ∈ X,the line px through p and x intersects X at least in one other point, say y. Consider the mapπ : X ×X × P1 → S(X) ⊂ Pn, then by what we observed above dimπ−1(p) ≥ d = dimX.It follows that the set B ⊂ Pn of points p such that the corresponding projection φ|X is notbirational, has dimension ≤ dim(X ×X × P1)− d = d+ 1. Therefore we may pick a pointp ∈ Pn \ B and obtain a birational map φ|X : X → X1 ⊂ Pn−1. Repeating this processn− d− 1 times yields the proposition.

Theorem 21.4 (Bertini’s Theorem). Let X ⊂ Pn be a smooth projective variety and Pn be

the dual projective variety parametrizing hyperplanes H ⊂ Pn. Let W ⊂ Pn be the subset ofplanes such that H ∩X is smooth. Then W is a non-empty open subvariety of Pn.

4other examples. WU?5Remark on why this is an isom.

49

Proof. Consider the subset B ⊂ X × Pn consisting of pairs (x,H) such that x is a singularpoint of H ∩ x. Since X is smooth, it is easy to see that for any x ∈ X, H ∩ x is singularif and only if Tx(X) ⊂ H and therefore the fibers of p : B → X are projective spaces ofdimension n− d− 1 where d = dimX (it is one condition to vanish at x and d conditions to

contain Tx(X)). Therefore, the dimension of B is n−d− 1 +d = n− 1 and so if q : B → Pnis the second projection, then q(B) ⊂ Pn is a proper subvariety and hence W = Pn \ q(B)

is a non-empty open subvariety of Pn.

Exercise 21.5. Prove a version of Bertini’s Theorem for varieties with isolated singulari-ties.

Theorem 21.6 (Generic smoothness). Let f : X → Y be a morphism of complex projectivevarieties, then there exists a non-empty open subset X0 ⊂ X such that f |X0 is smooth. Ifmoreover X is smooth, then there exists a non empty open subset Y 0 ⊂ Y such that forevery y ∈ Y 0, the fiber Xy = f−1(y) is smooth.

Proof. Assume for simplicity that Y = C. Let U ⊂ X be an affine open subset, then f isgiven by an element F ∈ OX(U) and since F is not constant, the differential dF : Tx(X)→C = Ty(Y ) is surjective on an open subset.

We must now show that if X is smooth, then for all but finitely many y ∈ C the differentialTx(X) → C = Ty(Y ) is surjective. By what we have seen above, that there is an opensubset X0 ⊂ X where this holds. Let X1 be the complement of this open subset. For eachirreducible component of X1, there is an open subset X0

1 such that Tx(X01 ) → Ty(Y ) is

surjective for every x ∈ X01 . Since Tx(X

01 ) ⊂ Tx(X) it follows that Tx(X)→ Ty(Y ) is onto.

By Noetherian induction, there is an open subset Y 0 ⊂ Y such that Tx(X)→ Ty(Y ) is ontofor all x ∈ f−1(Y 0).

22. Cubic surfaces

In this section we will study cubic surfaces i.e. smooth surfaces X ⊂ P3 defined by ahomogeneous equation of degree 3 sat f(x, y, z, w) ⊂ C[x, y, z, w]. We start we the followingobvious remark.

Lemma 22.1. Let H ⊂ P3 be a hyperplane, then H intersects X in one of the following:

(1) an irreducible cubic (degree 3) curve,(2) the union of a line and a smooth conic,(3) the union of three lines.

Proof. Since X is smooth, H can not be contained in X (otherwise X = H ∪Y where Y is aconic and X is singular along H∩Y , see Exercise 18.7). Therefore f |H 6= 0 is a homogeneouspolynomial of degree 3. We must exclude the possibility that X ∩H contains a double line2L. We may choose local coordinates so that H = V(t) and L = V(t, z). We then have thatz2 divides f |H(x, y, z) = f(x, y, z, 0) and so

f(x, y, z, w) = A(x, y, z)z2 +B(x, y, z, w)w

where A,B are homogeneous of degree 1 and 2 respectively. Since

∇f = (Ax · z2 +Bx · w,Ay · z2 +By · w,Az · z2 + A · 2z +Bz · w,Bw · w +B),

then ∇f = 0 and f = 0 if z = w = 0 and B(x, y, 0, 0) = 0. (Note that B(x, y, 0, 0) hasdegree 2 and hence there are two such solutions counted with multiplicity.) Therefore Xhas a singular point which is impossible.

50

Lemma 22.2. If X contains three lines through a point, then the lines are coplanar.

Proof. Suppose that x ∈ L1, L2, L3 ⊂ X, then Li = Tx(Li) ⊂ Tx(X).

Proposition 22.3. X contains at least one line L.

Proof. There is a very elementary but rather lengthy proof of this fact in [Reid]. Herewe follow the proof of [Beauville] which is less elementary but much shorter. The set ofall cubic threefolds (including the singular ones) is just PH0(OP3(3)) ∼= P19. We considerG the space of all lines in P3. Note that a line in P3 is determined by a pair of distinctpoints X = [x0 : x1 : x2 : x3], Y = [y0 : y1 : y2 : y3] ∈ P3, but of course different pointsmay determine the same line. It turns out that the line is uniquely determined by thepoint [λ12 : λ13 : λ14 : λ23 : λ24 : λ34] ∈ P5 where λij = xiyj − xjyi. Note that X, Y arelinearly independent and hence at least one 2 × 2 minor is non-zero. One way to think ofthis is to use exterior algebra. The plane is determined by X ∧ Y (up to non-zero scalar).So we write X =

∑xiei, Y =

∑yiei and X ∧ Y =

∑xiyjei ∧ ej. The coefficient of

e1 ∧ e2 is then x1y2 − x2y1 = λ12. Note that the six minors satisfy the following relationλ12λ34 − λ13λ24 + λ14λ23 = 0 (this is an easy check). In fact G is defined by this equation(this is also an easy check, since λ12λ34− λ13λ24 + λ14λ23 = 0 is irreducible, it suffices to seethat dimG = 4, but this follows easily since X = [1 : 0 : x2 : x3] and Y = [0 : 1 : y2 : y3]determine an open subset of G which is isomorphic to C4.) We now consider the incidencecorrespondence and projections

Z = (L, S)|L ⊂ S ⊂ G× P19, p : Z → G, q : Z → P19.

Fix a line L, then in appropriate coordinates x, y, z, w on P3, we may assume that L =V(z, w) and hence X contains L if and only if f ∈ (z, w) i.e. if and only if the coefficientsof x3, x2y, xy2, y3 are all 0. this shows that the co-dimension of each fiber of p is 4 andhence dimZ = dim(G × P19) − 4 = 19. We also claim that Z is non empty. To see this,simply exibit a cubic containing a line! Consider now the projection q. If the propositionfails, then q(Z) ⊂ P19 is a proper closed subset and thus each fiber of Z → q(Z) is positivedimensional.6 It suffices then to exhibit one cubic with finilely many lines. This requires alittle preparation, and we will prove it later (see Lemma 22.10).

Exercise 22.4. Show that the open subset λ1,2 6= 0 is isomorphic to C4.

Exercise 22.5. Find the equation of a smooth cubic containing a line 9say V(z, w)).

Proposition 22.6. Let L be a line on X, then there are 5 pairs of lines Li, L′i with 1 ≤ i ≤ 5

such that L,Li, L′i are coplanar for 1 ≤ i ≤ 5 and (Li ∪ L′i) ∩ (Lj ∪ L′j) = ∅ for any

1 ≤ i < j ≤ 5.

Proof. Consider a plane Pt ⊃ L, then as we have seen X ∩ Pt = L ∪ Ct where Ct is a planeconic. We must show that there are exactly five planes for which Ct is a singular conic (andhence the union of two lines). The fact that (Li ∪ L′i) ∩ (Lj ∪ L′j) = ∅ is then immediatefrom Lemma 22.2. Suppose in fact that (without loss of generality) L1 ∩ L2 6= ∅, then byconstruction L1 and L2 must meet at a point of L. But then L1, L2, L are coplanar and soL2 = L′1 a contradiction.

We now choose coordinates such that L = V(z, w) and so f ∈ I(L) = (z, w), and so

f = Ax2 + 2Bxy + Cy2 + 2Dx+ 2Ey + F

6explain some details such as semicontinuity of fiber dimension.51

where A,B,C,D,E, F are homogeneous of degrees 1, 2, 3 in C[z, w]. As we have seen inSection 17, this conic is singular exactly when the

∆(z, w) = det

A B DB C ED E F

= ACF + 2BDE − CD2 −B2F − AE2 = 0.

A plane P containing L is given by P[s:t] = V(sz − tw). If s 6= 0, then we let λ = t/s andPλ = V(z − λw). Restricting to Pλ we obtain A|P = A(z, w)|P = A(λw,w) = A(λ, 1)w andsimilarly for B|P , C|P , D|P , E|P and F |P . Since f ∈ (z, w), we have f |P = w · g where

g = A(λ, 1)x2 + 2B(λ, 1)xy + C(λ, 1)y2 + 2D(λ, 1)xw + 2E(λ, 1)yw + F (λ, 1)w2.

Since ∆(λ, 1) is a polynomial of degree 5 in λ, it follows that ∆(λ, 1) = 0 has five solutions,and we aim to show that these 5 solutions are distinct or equivalently that there are nomultiple solution.

Pick any such root, we aim to show that this is not a repeated root. Without loss ofgenerality we may assume that this root is λ = 0 and we aim to find a contradiction. Inthis case P = V(z) and P ∩X contains three lines (including the line L = V(z, w). Thereare two cases depending on if the three lines contain a point or not. In this case we canassume that the coordinates have been chosen so that the remaining lines are given by thefollowing equations:

(1) L1 = V(z, x), L2 = V(z, x− z),(2) L1 = V(z, x), L2 = V(z, y).

Suppose that we are in case (2), then f |P = xyw and so f = xyw+zq where q ∈ C[x, y, z, w]is homogeneous of degree 2. But then z divides A, 2B−w,C,D,E, F so that B = w/2 + zband A = za, C = zc,D = zd, E = ze and hence

∆(z, w) = z2(acfz + 2(w + bz)de− cd2z − (wb+ zb2)f − ae2z)− w2F/4.

Since X is smooth, we have that z2 does not divide F (for otherwise (0, 0, 1, 0) is a point ofmultiplicity ≥ 2 for f = Ax2 +2Bxy+Cy2 +2Dx+2Ey+F i.e. X is singular at (0, 0, 1, 0)).But then z2 does not divide ∆(z, w) and so λ = 0 is not a repeated root.

Case (1) is similar and we leave it as an exercise.

Exercise 22.7. Prove case (1).

Proposition 22.8. X is birational to P2.

Proof. Let L,M be two disjoint lines (eg L1 and L2) and define rational maps

φ : S 99K L1 × L2, ψ : L1 × L2 99K S

as follows: For any point P ∈ P3 \ (L1 ∪L2) there exists a unique line LP containing P andintersecting L1 and L2 (this line is the intersection of the two planes containing p and Lifor i = 1, 2). For any P ∈ S \ (L1 ∪ L2), we then define

φ(P ) = (LP ∩ L1, LP ∩ L2).

Suppose now that Pi ∈ Li, then consider the line P1, P2 and note that if P1, P2 is notcontained in X, then P1, P2 = P1 ∪ P2 ∪Q and so we define

ψ(P1, P2) = Q.

It is easy to see that φ and ψ are inverses (on appropriate open subsets) and hence X isbirational to P1 × P1 (which in turn is clearly birational to P2).

52

Remark 22.9. It is not hard to push these arguments a little further and to check that asmooth cubic surface contains exactly 27 lines (see [Reid] for the details.)

Finally we must check that a smooth cubic threefold contains finitely many lines.

Lemma 22.10. Let X be a smooth cubic threefold, then X contains only finitely many lines.

Proof. Suppose that X contains a line (otherwise the claim is clear), then we may pick aplane P such that X ∩P = L∪L1 ∪L′1 is the union of 3 lines. If M is another line, then Mintersects one of the lines L,L1, L

′1. Without loss of generality we may assume that infinitely

many lines intersect L. But this is a contradiction as we have seen that exactly 10 linesintersect L.

23. Rational and unirational varieties

Definition 23.1. A projective variety X is rational if it is birational to Pn. EquivalentlyX is rational if C(X) ∼= C(x1, . . . , xn). A variety X is unirational if there is a dominantmorphism Pn 99K X.

In general, it is very hard to tell if a variety is rational. For example it is clear thatrational varieties are unirational but the converse is far from obvious. It is known that

Theorem 23.2. If X is unirational and dimX ≤ 2, then X is rational.

We can rephrase this result as follows: Let K ⊂ C(x, y) be a field of transcendence degree2, then K ∼= C(u, t). Note that we are not claiming that K = C(x, y), as we could forexample have the inclusion K = C(x2, y2) ⊂ C(x, y) which is not an equality. The proof ofthis result goes beyond the scope of these notes, however we mention the intuition behindthe dimX = 1 case. In this case we must show that if P1 → X is a dominant map, thenX ∼= P1. Let g(X) be the topological genus of the Riemann surface X, then we have toshow that g(X) = 0. This follows from the well known Hurwitz formulaZ which says

Theorem 23.3. Let f : X → Y be a dominant map of degree d of projective varieties ofdimension 1, then

2(g(X)− 2) = d(2g(Y )− 2) +∑

(ri − 1),

where ri denotes the ramification index.

Note that there exist open subsets Y0 ⊂ Y and X0 = f−1(Y0) ⊂ X such that f0 =f |X0 : X0 → Y0 is a smooth map (the rank of the differential is maximal at each pointof X0). It follows that X0 → Y0 is an isomorphism locally in the analytic topology (thismeans that for any point x ∈ X0, there are analytic neighborhoods x0 ∈ B ⊂ X0 andy0 = f(x0) ∈ B′ ⊂ Y0 such that f |B : B → B′ is an isomorphism). If however x0 ∈ X \X0,then we can choose analytic neighborhoods x0 ∈ B ⊂ X and y0 = f(x0) ∈ B′ ⊂ Y andanalytic coordinates z on B and w on B′ such that w = f(z) = zr for some r ≥ 1. We saythat the ramification index at x0 is r (note that r = 1 precisely when f is smooth, so if weaccidentally include a smooth point in X \X0, then the above formula is not affected). Thequantity 2g− 2 = v− e+ f can be computed from any triangulation where v is the numberof vertices, e the number of edges and f the number of faces. If we choose a triangulationof Y such that each point of Y \ Y0 is a vertex, then the inverse image of this triangulationvia f gives a triangulation with df faces, de edges and dv −

∑(ri − 1) vertices. Therefore

2−2g(X) = df−de+dv−∑

(ri−1) = d(f−e+v)−∑

(ri−1) = d(2g(Y )−2)−∑

(ri−1).53

Applying the Hurwitz formula to a dominant morphism P1 → X of degree d gives

−2 = 2g(P1)− 2 = d(2g(X)− 2) +∑

(ri − 1).

Since∑

(ri − 1) ≥ 0, we must have 2g(X)− 2 < 0 and so g(X) = 0.The proof of the two dimensional case is much harder. First of all it is necessary to

generalize the concept of the genus of a curve (Riemann surface) to a complex surface (4-manifold). There are several natural options. First of all, we note that g(X) = dimH0(Ω1

X).Here Ω1

X = T∨X is the cotangent bundle and H0(Ω1X) denotes the space of global holomorphic

1-forms. Recall that in local coordinates, a holomorphic 1-form may be written as f(z)dz, soif X = ∪Ui is an open cover and zi is a local coordinate on Ui, then an element ω ∈ H0(Ω1

X)can be represented locally by

ωi = fidzi ∈ Γ(Ω1Ui

)which patch together on Ui,j = Ui ∩ Uj. This means that if zi = gi,j(zj) on Ui,j, then

dzi =∂gi,j∂zj

dzj. For example, if X = P1, then there are two charts: U1 and U2 with local

coordinates z and w respectively such that z = 1/w on U1,2 = U1∩U2. Suppose that f(z)dzand g(w)dw represent an element of H0(Ω1

P1), then dz = − 1w2dw and so on U1,2 we must

have f(z)dz = g(1/z)−1z2 dz. Since g(w) is a polynomial, g(1/z) 1

z2 only involves powers za

with a ≤ −2 while f(z) only involves positive powers of z. Thus f(z) = 0 and g(w) = 0.This shows that H0(Ω1

P1) = 0 and hence dimH0(Ω1P1) = g(P1).

Returning to our goal of generalizing the genus to higher dimensional varieties, there arenow two natural choices.

Definition 23.4. Let X be a smooth projective variety, then the irregularity of X is

q(X) = dimH0(Ω1X).

A second natural choice involves the canonical line bundle which is just the top exteriorpower of the cotangent bundle. More precisely, if X is a smooth projective variety ofdimension n, then let ωX = ∧nΩ1

X = ΩnX so that in local coordinates z1, . . . , zn, any section

of ωX can be locally written as

f(z1, . . . , zn)dz1 ∧ . . . ∧ dzn,where f is a regular function. If Ui ⊂ X is an open cover and zi1, . . . , zin are localcoordinates, then on Ui,j we can write zik = gi,jk (zj1, . . . , z

jd). It is easy to see that the

transition functions are now represented by the determinant of the matrix whose (k, l)-thentry is ∂gi,jk /∂z

jl . Similarly, we can consider the line bundles ω⊗mX for any m > 0 whose

sections can be locally written as

f(z1, . . . , zn)(dz1 ∧ . . . ∧ dzn)⊗m,

and have transition functions det(∂gi,jk /∂zjl )m. We then define

Definition 23.5. Let X be a smooth projective variety and m > 0 a positive integer, thenthe m-th plurigenera of X is the number

Pm(X) = dimH0(ω⊗mX ).

Here H0(ω⊗mX ) denotes the space of global sections of ω⊗mX described above, that patch togetherto give a global section.

We are now ready to state Castelnuovo’s Criterion.54

Theorem 23.6. Let X be a projective surface with q(X) = P2(X) = 0, then X is rational.

The proof of this result is non-trivial and we do not recall it here. Note however thatthe irregularity and plurigenera are birational invariants. This means that if X 99K Y is abirational map of smooth projective varieties, then q(X) = q(X ′) and Pm(X) = Pm(X ′) forall m > 0. In particular q(X) = P2(X) = 0 is a necessary condition for rationality.

Note that Theorem 23.3 in dimension 2 follows immediately from Castelnuovo’s criterion.In fact, if f : P2 99K X is a dominant map of surfaces, then it is easy to see that P2(X) =q(X) = 0. The reason is that if ω ∈ H0(Ω1

X) (or ω ∈ H0(ω⊗2X )), then there is a naturally

defined element f ∗ω ∈ H0(Ω1P2) (or f ∗ω ∈ H0(ω⊗2

X )). If V ⊂ X and U = f−1(V ) ⊂ P2 areopen subsets such that f |U : V → U is smooth, then locally in the analytic topology U and Vare isomorphic and ω and f ∗ω are identified via this isomorphism. Since q(P2) = P2(P2) = 0it follows that f ∗ω = 0 and hence that q(X) = P2(X) = 0. But then by Castelnuovo’sCriterion X is birational to P2 (we do not claim that f is a birational map).

It turns out that in general it is very hard to determine if an unirational variety is rational,however Clemens and Griffiths proved the following amazing result.

Theorem 23.7. Let X ⊂ P4 be a cubic threefold i.e. a smooth hypersurface of degree 3,then X is unirational but not rational. In particular C(X) ⊂ C(x1, x2, x3) but C(X) is notisomorphic to C(t1, t2, t3).

The proof of this result is beautiful but very non-trivial. In what follows we will onlyexplain why a smooth cubic threefold is unirational.

Proposition 23.8. Let X ⊂ P4 be a cubic threefold, then X is unirational.

Proof. Claim 1. We may find a line L ⊂ X.To see this, simply consider a general hyperplane P3 ∼= H ⊂ P4. Then H ∩X ⊂ H ∼= P3

is a smooth cubic surface (by Bertini’s Theorem), and we have seen that a cubic surfacecontains a line (and in fact 27 lines). Just pick one of these.

Consider now the rational map P4 99K P2 defined by projecting from L. This map is amorphism on P4 \ L and it becomes a morphism BLP4 → P2 after blowing up P4 along theline L. Note that the exceptional divisor is isomorphic to L× P2.

Claim 2. Consider The blow up BLX ⊂ BLP4. Then, the fibers of the induced mapBLX → P2 are quadrics.

Since any plane Πx through L (corresponding to a point x ∈ P2) intersects X along Land a residual quadric curve say Cx, we have that the fibers of φ are quadrics φ−1(x) ∼= Cx,and the claim follows. Note that there is a non-empty open subset U ⊂ P2 such that forany x ∈ U , we have Cx is smooth and hence Cx ∼= P1.

Claim 3. Consider E = BLX ∩ L × P2 the exceptional divisor lying over L. Then theinduced map E → P2 has generic degree 2 and the fibers of E → L are quadrics.

Since E = L × P2 ∩ BLX and, the fiber of E over x ∈ P2 is given by L ∩ Cx and henceconsists of two points. In other words E 99K P2 is generically of degree 2.

Consider now the fiber product Y = BLX ×P2 E, then the fibers of BLX ×P2 E → E7 arejust the rational curves Cx and Y → E has a rational section. But then Y is birational toE × P1 and hence X is uniruled.

Exercise 23.9. Show that H0(Ω1Pn) = 0 and H0(ω⊗kPn ) = 0.

7to be defined55

Exercise 23.10. Let X0 ⊂ X be a big open subset of a smooth projective variety. Showthat H0(Ω1

X) ∼= H0(Ω1X0

) and H0(ω⊗mX ) ∼= H0(ω⊗mX0) for all m > 0.

Exercise 23.11. Let f : X → Y be a generically finite morphism. Show that there is anaturally defined homomorphism H0(Ω1

Y ) → H0(Ω1X) and H0(ω⊗mY ) → H0(ω⊗mX ) for any

m > 0.

Exercise 23.12. Let f : X → Y be a birational morphism, show that H0(ω⊗kX ) ∼= H0(ω⊗kY )and H0(Ω1

X) ∼= H0(Ω1Y ). Deduce that Pm(X) and q(X) are birational invariants. (Hint.

Use the fact if f : X → Y is a birational morphism, then there is a big open subset Y0 ⊂ Ysuch that f : f−1(Y0)→ Y0 is an isomorphism).

Exercise 23.13. Let C1, C2 be curves of genus g1, g2, compute Pm(C1×C2) and h0(Ω1C1×C2

).

24. Sheaves

Definition 24.1. If X is a topological space, then a presheaf (resp. a sheaf) of abeliangroups F on X is given by assigning:

(1) an abelian group F(U) to any open subset U ⊂ X, and(2) for any inclusion of open subsets V ⊂ U ⊂ X a “restriction” homomorphism

ρUV : F(U)→ F(V ),

such that conditions (1-3) (resp. conditions (1-5)) below are satisfied by F and ρ:

(1) F(∅) = 0,(2) ρU,U = idU ,(3) if W ⊂ V ⊂ U ⊂ X are open subsets, then ρUW = ρVW ρUV ,(4) if U ⊂ X is an open subset and

⋃i Vi is an open cover of U and s ∈ F(U) satisfies

ρU,Vi(s) = 0 for all i, then s = 0, and(5) if U ⊂ X is an open subset and

⋃i Vi is an open cover of U and if si ∈ F(Vi)

satisfy ρViVi,j(si) = ρVj ,Vi,j(sj) for all i, j then there is an element s ∈ F(u) such thatρU,Vi = si for all i.

Note that as usual Vi,j denotes Vi ∩ Vj and in what follows, we will denote ρUV simplyby |V . The above definitions seems at first very abstract but in practice there are manyexamples of a sheaf that you are familiar with.

(1) The sheaf of regular functions OX . Let X be a quasi-projective variety, then OX(U)defines a sheaf of rings on X. (The hardest axiom to check is (5) which says thatif f is a function which is regular when restricted to each open subset of an opencover, then f is in fact regular).

(2) The sheaf of continuous functions on a topological space.(3) The sheaf of analytic functions on an analytic variety.(4) Given an abelian group A and a topological space X, one can define the constant

presheaf by letting A(∅) = 0 and A(U) = A for any open subset ∅ 6= U ⊂ X. It iseasy to see that this is not a sheaf in general as (5) may fail.

(5) Given an abelian group A and a topological space X, one can define the constantsheaf on X as follows: If U is an open subset, assuming that its connected compo-nents Ui are also open, then let A(U) =

∏i∈I A(Ui) =

∏i∈I A. In other words we let

A(U) = A if U is an irreducible open subset and otherwise we let A(U) be a productof one copy of A for each irreducible component of A.

56

Notation 24.2. We call the elements of the abelian groups associated to X by the sheaf“sections” of F , i.e. an element s ∈ F(X) is called a global section of F and an elements ∈ F(U) is called a local section on the open set U . We often denote the set of globalsections using the notation Γ(X,F) := F(X) and the set of local sections on U using thenotation Γ(U,F) := F(U).

By now the reader must expect that each time we introduce new objects we also introducea notion of maps (and isomorphisms) between these objects. In this case we have

Definition 24.3. Let F and G be sheaves on a topological space X, then a morphismof sheaves φ : F → G is a collection of homomorphisms φU : F(U) → G(U) for anyopen subsets U ⊂ X which are compatible with inclusion of open subsets V ⊂ U ⊂ X (i.e.ρUV φU = ρUV φV where, by abuse of notation, ρUV denotes both restrictions F(U)→ F(V )and G(U) → G(V )). An isomorphism of sheaves is a morphism of sheaves φ : F → Gsuch that there is a morphism of sheaves ψ : G → F such that φ ψ and ψ φ are theidentity morphism on F and G.

Just as in the case of groups or rings, we can talk about kernels, cokernels and images ofa morphism.

Definition 24.4. Let φ : F → G be a morphism of pre-sheaves on a topological space X,then we have:

(1) kernel presheaf: ker(φ)(U) = ker(φU : F(U)→ G(U)),(2) cokernel presheaf: coker(φ)(U) = coker(φU : F(U)→ G(U)) = G(U)/φU(G(U)),(3) image sheaf: im(φ)(U) = im(φU : F(U)→ G(U)) = φU(F(U)).

Exercise 24.5. Show by example that image and cokernel presheaves may not be sheaves.

Exercise 24.6. Show that the kernel presheaf is alway a sheaf.

It is much more convenient to work with sheaves. Luckily it is always possible to associatea sheaf to any presheaf in a natural way.

Definition 24.7. Let F be a presheaf on a topological space X, then we define the sheaf F+

on X in the following two steps: firstly, for any open subset U ⊂ X an element s ∈ F+(U)is given by an open cover U = ∪Ui and a collection of elements si ∈ F(Ui) such thatsi|Uij

= sj|Uijfor all i, j; next we identify two elements of F+(U) if there is an open cover

Vj of U such that si|Vj∩Ui= 0 for all i, j.

We say that F+ is the shefification of F .

Exercise 24.8. Show that the first step in the above definition ensures property (5) in thedefinition of a sheaf and the second step ensures property (4).

Note that the above definition is quite cumbersome (but natural). There is an alternativemore efficient way to proceed. Intuitevely, we are interested in the local behaviour of sheavesaround a point P ∈ X. For example, we say that s ∈ F(X) is zero locally near a point Pif there is an open subset P ∈ U ⊂ X such that s|U = 0. Therefore it is natural to considerthe following definition.

Definition 24.9. Let F be a presheaf on a topological space X, then for any point P ∈ Xthe stalk of F at P is given by

FP = < U, s > |U ⊂ X is open, s ∈ F(U)/ ∼57

where < U, s >∼< U ′, s′ > if and only if there is an open subset P ∈ V ⊂ U ∩ U ′ such thats|V = t|V . Given s ∈ F(U) we then have elements sP =< s, U >∈ FP for any P ∈ U . wewill say that sP is the germ of s at P .

With this definition we can give the following alternative definition for the sheaf F+

associated to a presheaf F .

Definition 24.10. For any open subset U ⊂ X let F ′(U) be the set of functions froms : U → ∪P∈UFP such that

(1) for any P ∈ U , s(P ) ∈ FP , and(2) for any P ∈ U there exists an open subset P ⊂ V ⊂ U and an element t ∈ F(V )

such that s(Q) = tQ for all Q ∈ V .

Exercise 24.11. Show that F+ = F ′.

Exercise 24.12. Show that if F is a presheaf on X, then F+P = F ′P for any P ∈ X.

Exercise 24.13. Show that there is a natural morphism φ : F → F+ satisfying the followinguniversal property: For any morphism of pre-sheaves ψ : F → G, if G is a sheaf, then thereis a unique morphism of sheaves ψ+ : F+ → G such that ψ+ φ = ψ.

Exercise 24.14. If φ : F → G is a morphism of sheaves then show that there is a naturalinduced map φp : FP → GP .

Exercise 24.15. If φ : F → G is a morphism of sheaves then φ is an isomorphism if andonly if the induced map of stalks φp : FP → GP is an isomorphism for all P ∈ X.

Definition 24.16. If φ : F → G is a morphism of sheaves, then we define

(1) kernel sheaf: ker(φ) is just the kernel presheaf of φ (as observed above this isautomatically a sheaf).

(2) image sheaf: If we denote the corresponding image presheaf by im′(φ) ⊂ G, thenwe define the image sheaf by im(φ) = im′(φ)+ ⊂ G+ = G. (In other words the imagesheaf is the sheaf naturally associated to the image presheaf.)

(3) cokernel sheaf: If we denote the corresponding cokernel presheaf by coker′(φ), thenwe define the cokernel sheaf by coker(φ) = coker′(φ)+. (In other words the cokernelsheaf is the sheaf naturally associated to the cokernel presheaf.)

Definition 24.17. If φ : F → G is a morphism of sheaves, then we that φ is injective ifker(φ) = 0, φ is surjective if im(φ) = G.

Exercise 24.18. Check that φ : F → G is injective (resp. surj.) if and only if φP is injective(resp. surjective) for all P ∈ U .

Definition 24.19. Consider a sequence of morphism of sheaves φi : F i → F i+1. We saythat this sequence is exact if ker(φ) = im(φi−1). We will be particularly interested in shortexact sequences i.e. in exact sequences of the form

0→ F ′ → F → F ′ → 0.

This is equivalent to requiring that F ′ → F is injective, F → F ′′ is surjective and thatker(F → F ′′) = im(F ′ → F).

58

Exercise 24.20. Show that a sequence of sheaves

0→ F ′ → F → F ′ → 0

is exact if and only if the corresponding sequence of abelian groups

0→ F ′P → FP → F ′P → 0

given on stalks is exact for every point P ∈ X.

Given a continuous map f : X → Y of topological spaces, we can define several usefuloperations determined by f for sheaves on X and Y . Here is one such operation.

Definition 24.21. If F is a sheaf on X, we define the direct image presheaf f∗F on Yto be given by f∗F(V ) = F(f−1(V )) for any open set V in Y .

Exercise 24.22. Show that the direct image presheaf of a sheaf is automatically a sheaf.

Exercise 24.23. Let f : X → Y be a morphism of quasi-projective varieties, and let OXand OY be the sheaves of regular functions on X and Y respectively. What do you expect thenatural induced morphism of sheaves f ] : OY → f∗OX to look like? Compute this morphismfor some of your favorite such examples f .

25. Sheaves of modules

We begin by recalling some definitions and exercises from earlier sections on sodules toremind you about some of the central constructions.

Definition 25.1. Let R be a domain, then the fraction field of R is the set

Q(R) = (f, g)|f ∈ R, g ∈ R∗/ ∼, (f, g) ∼ (h, k), iff fk = gh.

Addition is defined by (f, g) + (h, k) = (fk + gh, gk) and multiplication is defined by(f, g)(h, k) = (fh, gk). It will be convenient to denote the equivalence class of (f, g) simplyby f/g.

We saw that we can generalize this construction, allowing us to “invert” specified elementsof any ring A provided they together form a multiplicatively closed set, a process calledlocalization that is essential to many basic constructions in algebraic geometry.

Definition 25.2. Let A be a commutative ring with identity and let S ⊆ A such that itcontains the identity and is multiplicatively closed, meaning that a, b ∈ S =⇒ ab ∈ S.The localization of A at S (or the ring of fraction of A with respect to S) is the set

S−1A = (f, g)|f ∈ A, g ∈ S/ ∼where the equivalence relation ∼ is defined by

(f, g) ∼ (h, k) ⇐⇒ ∃x ∈ S such that x(fk − gh) = 0.

Addition is defined by (f, g) + (h, k) = (fk + gh, gk) and multiplication is defined by(f, g)(h, k) = (fh, gk). It will be convenient to denote the equivalence class of (f, g) simplyby f/g.

If you haven’t yet done some of the exercises about localization in the earlier sections, werecall them here.

Exercise 25.3. Prove that ∼ is indeed an equivalence relation.59

Exercise 25.4. There is a natural map π : A → S−1A sending a 7→ a/1. Prove thatS−1A together with this map satisfies the following universal property: given any morphismu : A→ B of commutative rings (sending 1 7→ 1) where u(d) is a unit ∀d ∈ S, there existsa unique morphism v : S−1A→ B such that u = v π.

Exercise 25.5. Prove that π in the previous exercise is one-to-one if and only if S containsno zero-divisors.

Exercise 25.6. Prove that S−1A = 0 if and only if 0 ∈ S.

If R is a domain, then choosing S = R \ 0 produces the fraction field Q(R).

Exercise 25.7. Given any element f ∈ A, we define Af = S−1f A where Sf = fn|n ≥ 0

is the set of nonnegative powers of f . Under what condition is Af = 0? Prove that Af =R[x]/(xf − 1).

Let P ⊆ A be a prime ideal. Recall that S = A \ P contains the identity and is multi-plicatively closed. We call the ring S−1A the localization of A at the prime ideal P .We now recall the definition of an A-module.

Definition 25.8. Let A be a ring, (M,+) an abelian group and A×M →M an operation(called scalar multiplication and denoted by (a,m)→ a·m), then M is a (left) A-moduleif the following hold for all a, a′ ∈ A and m,m′ ∈M :

(1) a · (m+m′) = a ·m+ a ·m′,(2) (a+ a′) ·m = a ·m+ a′ ·m,(3) (aa′) ·m = a · (a′ ·m), and(4) 1 ·m = m.

Remind yourself the obvious basic definitions in the theory of modules. For example,an A-submodule of an A-module M is simply a subgroup of M that is itself closed underscalar multiplication. Morphisms of A-modules are group homomorphisms that respect theA-module structure.

Exercise 25.9. Go back and review the following notions in the theory of modules:

(1) Quotient of an A-module by an A-submodule,(2) kernel of a morphism of A-modules,(3) cokernel of a morphism of A-modules,(4) image of a morphism of A-modules,

Exercise 25.10. Let n be a positive integer and let A be a commutative ring with identity.Show that the cartesian product An is an A-module under componentwise vector addtion andcomponentwise scalar multiplication. We call this the free A-module of rank n. We sayan A-module M is a free module of rank n if it is isomorphism to An as A-modules forsome positive integer n.

Exercise 25.11. Let I ⊂ A be an ideal, show that I and A/I are A-modules and that wehave a short exact sequence of A-modules

0→ I → A→ A/I → 0.

Definition 25.12. Let A be a commutative ring with identity and let S ⊆ A be a multi-plicatively closed subset containing the identity. Let M be an A-module. The localization

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of M at S (or the module of fraction of M with respect to S) is the set

S−1M = (f,m)|f ∈ S, m ∈M/ ∼where the equivalence relation ∼ is defined by

(f,m) ∼ (g, n) ⇐⇒ ∃x ∈ S such that x(fn− gm) = 0.

This set becomes a S−1A module under the following operations: addition is defined by(f,m) + (g, n) = (gm + fn, fg) and scalar multiplication is defined by (g/h)(f,m) =(gm, hf). It will be convenient to denote the equivalence class of (d,m) simply by m/d.

Exercise 25.13. Given an element f ∈ A, what does the Af -module Mf = S−1f M look like?

Exercise 25.14. Let M be an A-module and P a prime ideal of A. If AP = (A \ P )−1A isthe localization of A at P , then show that MP = (A \ P )−1M is an AP -module.

We now introduce sheaves of OX-modules. These form the building blocks for many ofthe central algebraic objects of study in algebraic geometry.

Definition 25.15. Let X be a quasi-projective variety and F be a sheaf on X, then Fis a sheaf of OX-modules (or just OX-module for short) if for any open subset U ⊂X, the group F(U) is a OX(U)-module and the module structures are compatible with therestriction morphisms (so that if V ⊂ U is an open subset, s ∈ F(U) and t ∈ OX(U), then(t · s)|V = t|V · s|V ).

In particular if X is an affine variety with coordinate ring A = A(X), then Γ(X,OX) =OX(X) = A and so M := F(X) is an A-module. It turns out that there is a natural way ofinverting this procedure.

Definition 25.16. Let X be an affine variety with ring of regular functions A = A(X) and

M an A-module, then there is a natural sheaf M on X associated to M such that for any

point P ∈ X, the stalk MP of M at P ∈ X is MP = MmPthe localization of M at the

maximal ideal mP of P ∈ X. For any open subset U ⊂ X, this is defined by M(U) is theset of functions s : U →

∏P∈U MmP

such that for every P ∈ U , s(P ) ∈MmP, and such that

for any P ∈ U there is an open subset P ∈ V ⊂ U and elements m ∈ M and a ∈ A suchthat for any Q ∈ V we have s(Q) = m/a ∈MmQ

.

Exercise 25.17. Check that M is a sheaf of OX-modules such that the stalks MP of M at

P ∈ X is MP = MmPthe localization of M at the maximal ideal mP of P ∈ X.

Proposition 25.18. With the notation above, if f ∈ A and Xf is the corresponding prin-

cipal open subset, then Γ(Xf , M) = Mf and in particular Γ(X, M) = M .

Exercise 25.19. Prove Proposition 25.18

Exercise 25.20. Define the following notions for sheaves of OX-modules:

(1) submodule of an OX-modules,(2) morphisms of OX-modules,(3) quotients of an OX-modules by an submodule,(4) kernel of a morphism of OX-modules,(5) image of a morphism of OX-modules,

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Exercise 25.21. Let X be a quasi-projective variety. How can we view⊕n

i=1OX as anOX-module?

Definition 25.22. A sheaf of OX-modules F is said to be free of rank n if F is isomorphicto⊕n

i=1OX for some positive integer n as sheaves of OX-modules. There is a much lessrestrictive but related condition: a sheaf of OX-modules F is said to be locally free if thereexists an open covering

⋃i Ui of X on which, for every open set Ui in the covering, the

restriction F|Uiis a free OX |Ui

-module.

Exercise 25.23. Show that if X is connected and F is a locally free OX-module on X, thenF has the same rank everywhere.

Exercise 25.24. Define the tensor product F ⊗OXG of two OX-modules to be the sheaf

associated to the presheaf U 7→ F(U)⊗OX(U)G(U) (see Section ??). We often write F⊗OXG

as F ⊗ G and s⊗ t as st without confusion.

Exercise 25.25. Prove that the tensor product of two locally free OX-modules of rank n isagain a locally free OX-module of rank n.

Exercise 25.26. Let F and G be locally free rank 1 OX-modules on X. Show that thesheaf of homomorphisms Hom(F ,OX) (defined by U 7→ HomOX |U(F|U ,OX |U)) satisfiesF ⊗Hom(F ,OX) ∼= OX .

Definition 25.27. A sheaf F on a quasi-projective variety X is quasi-coherent if there is a

cover X =⋃i Ui of affine open subsets such that F|Ui

= Mi where Mi is the OX(Ui) = A(Ui)

module Γ(Ui, M). If moreover each Mi is a finitely generated Ai module, we say that F iscoherent.

In what follows we will be interested in coherent sheaves. Typical examples of coherentsheaves are the structure sheaf OX and any ideal sheaf of OX modules (i.e. any subsheaf ofOX-modules J ⊂ OX).

Exercise 25.28. Prove that any ideal sheaf is coherent.

Exercise 25.29. Find an example of a sheaf which is not quasi-coherent.

Exercise 25.30. Prove that if X is an affine variety and F is a quasi-coherent OX-moduleon X, then X can be covered by a finite collection of principle open subsets Xfi such that for

each i, F|Xfiis isomorphic to Mi for some S−1

fiΓ(X,OX)-module Mi (where Sfi = fni |n ≥

0 as in Exercise 25.7).

Lemma 25.31. Let X be an affine variety, let f ∈ Γ(X,OX), let Xf be the correspondingprinciple open subset of X, and let F be a quasi-coherent sheaf on X. If s is a global sectionof F and s|Xf

= 0, then there is some positive integer n making fns = 0 everywhere on X.

Proof. By Exercise 25.30, there is a finite cover of principal open subsets Xfi over each of

which F restricts to Mi for some S−1fi

Γ(X,OX)-module Mi. Let si be the restriction of s to

each of these subsets, meaning si ∈ Γ(Xfi , Mi) = Mi for each i. Consider now the restrictionof si to Xf ∩Xfi = Xffi . By Proposition 25.18, si|Xffi

is an element of the localized module

(Mi)f , so looks like an element mi/fn for some mi ∈ Mi and some non-negative integer n.

Since si|Xffi= 0 by assumption, fnsi = 0 in Mi. Since we are only using a finite collection

of principal open subsets Xfi in our cover, we can take the maximum integer n appearingas i ranges among these subsets.

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Exercise 25.32. Under the same assumptions as the above lemma, prove that given anysection t ∈ F(Xf ) there is some positive integer n making fnt extend to a global section ofF on X.

Consider a short exact sequence of sheaves

0→ F ′ → F → F ′′ → 0.

Evaluating this over X (i.e. taking Γ(X, . . .)) we obtain a sequence

0→ Γ(X,F ′)→ Γ(X,F)→ Γ(X,F ′′)→ 0.

Theorem 25.33. The above sequence is always exact on the left (so that Γ(X,F ′) →Γ(X,F) is injective and ker(Γ(X,F)→ Γ(X,F ′′)) = im(Γ(X,F ′)→ Γ(X,F))) and if X isaffine and F ′ is quasi-coherent, then it is exact on the right (so that Γ(X,F) → Γ(X,F ′′)is surjective).

Proof. (c.f. [Hartshorne, Proposition 5.6]) Let φ : F ′ → F be the first map and ψ : F → F ′′be the second map in the exact sequence of sheaves 0 → F ′ → F → F ′′ → 0 given. Wehave that φ is an injection, ψ is a surjection, and Ker(ψ) = Im(φ). We now consider themaps

φ(X) : Γ(X,F ′)→ Γ(X,F)

andψ(X) : Γ(X,F)→ Γ(X,F ′′).

Since φ is an injection, we have automatically that φ(X) is injective by definition. Fur-thermore, from Exercise 24.20, we know that for every point P ∈ X, the induced sequenceon stalks 0 → F ′P → FP → F ′′P → 0 is exact. We leave it as an exercise to show (usingsimply the exactness of this sequence on stalks, the definition of a stalk, and the definitionof a sheaf) that Ker(ψ(U)) = Im(φ(U)) for any open set U ⊂ X. We are left to show thatwhenever X is affine and F ′ is quasi-coherent, then ψ(X) is surjective.

Let s ∈ Γ(X,F ′′) be any global section of F ′′. Since ψ is surjective on sheaves, given anypoint P ∈ X there is a principal open set Xf containing P such that s|Xf

has a preimaget ∈ F(Xf ) (why?). We will first show that the exists an integer n so that the elementfns ∈ Γ(X,F ′′) has a some preimage in Γ(X,F).

To this end, let

X =N⋃i=1

Xgi

be a finite covering of X using principle open subsets Xgi such that on each Xgi , therestriction of s has a preimage ti ∈ F(Xgi) (remind yourself why such a finite cover exists).Notice now that on each open set Xf ∩ Xgi = Xfgi we have two elements t|Xfgi

and tieach mapping to s through the map ψ(Xfgi). Then t − ti is in the kernel of ψ(Xfgi), somust be in the image of φ(Xfgi). Use this fact, the Lemma above with the fact that F ′ isquasi-coherent, the gluing property of sheaves, and some ingenuity, to finish showing thatthe integer n desired above exists.

We now show the desired surjectivity of ψ(X). Cover X with a finite cover⋃rj=1 Xfj of

principal open sets Xfj such that s|Xfjhas a preimage in F(Xfj). We know that for each j,

we can find an integer nj such so that the element fnj s ∈ Γ(X,F ′′) has a some preimage in

Γ(X,F). Let N = maxjnj. Then the element fNj s ∈ Γ(X,F ′′) also has a some preimage63

tj ∈ Γ(X,F) for each j. Since X =⋃rj=1 Xfj is a covering, we must have A = (f1, . . . , fr),

so in particular we can write 1 =∑r

j=1 ajfj. Convince yourself that t =∑r

j=1 ajtj is thepreimage we are looking for.

26. Bezout’s Theorem

Let C ⊂ C2 be a curve in the affine plane, given by a polynomial f(x, y) ∈ C[x, y]. Wecan write the polynomial f as a finite sum of forms of increasing degree, i.e.

f(x, y) =d∑

i=m

fi(x, y)

where each term fi(x, y) is a sum only of monomials of degree i. Obvously, the degree ofthe polynomial f(x, y) is the nonnegative integer d. We define the multiplicity of f at 0 tobe the nonnegative integer m. Given any point p = (a, b) ∈ C2, we can use a simple affinechange of coordinates T (u, v) = (u+ a, v + b) to define a notion of multiplicity for f at thepoint p.

Definition 26.1. We define the multiplicity of f at p = (a, b), denoted mpf , to be themultiplicity of f(x+ a, y + b) at 0.

Exercise 26.2. Show that we could equivalently define mpf to be the degree of the firstnonzero term appearing in the Taylor expansion of f centered at p.

Exercise 26.3. Show that mpf = 1 if and only if p is a nonsingular point of the curve C.

Exercise 26.4. Show that mpf is always less than or equal to the degree of f .

Exercise 26.5. Let G : C2 → C2 be a map defined coordinatewise by polynomials that sendsp ∈ C2 to Q ∈ C2. Show that mQ(f G) is always greater than or equal to mpf . Whatconditions are necessary for equality to be attained? Are these conditions sufficient?

We quickly recall the construction of the local ring at a point P .

Definition 26.6. Let X ⊂ Cn be an affine variety and let p ∈ X. Define the local ring ofX at p, denoted Op, to be the subring of the ring of rational function on X that are definedat p. Setting mp for the maximal ideal at p, it is not hard to see that the set S = O(X) \mp

is a multiplicative set and Op = S−1O(X).

Exercise 26.7. Characterize the value of a function f ∈ Op at p using the quotient mapOp → Op/mp.

Exercise 26.8. Prove that mp is the unique maximal ideal in Op.

Theorem 26.9. Let C ⊂ C2 be an irreducible curve defined by f ⊂ C[x, y]. Then for anypositive integer n sufficiently large,

mPf = dimC(mnp/m

n−1p

).

Proof. (c.f. [Fulton, Section 3.2]) First, notice that the sequence

0→ mnp/m

n+1p → Op/mn+1 → O/mn → 0

is a short exact sequence. It is therefore enough to show that the dimension (as a complexvector space) ofOp/mn looks like n·mpf+s for every n and some constant s (why?). Using an

64

affine change of coordinates, we can assume that p is the origin, and so mn = (x, y)nOp andOp/mn ∼= C[x, y]/((x, y)n, f). Notice that for any polynomial g ∈ (x, y)n−mpf , fg ∈ (x, y)n.Use this to show that there is a short exact sequence

0→ C[x, y]/(x, y)n−mpf → C[x, y]/(x, y)n → C[x, y]/((x, y)n, f)→ 0.

Compute the desired dimension of the third term of this sequence to be

n ·mpf −mpf(mpf − 1)

2

for all n ≥ mpf .

We say curves C,D ⊂ C2 intersect properly at p ∈ C2 if they have no common compo-nents passing through p. For two such curves, we now formalize an important multiplicityassociated to a point p ∈ C ∩D in the intersection.

Definition 26.10 (Definition-Theorem). Let C,D ⊂ C2 be curves defined respectively bypolynomials f, g ∈ C[x, y] and let p ∈ C2 be a point, and denote O = OC2. There exists aunique intersection multiplicity Ip(f, g) satisfying Fulton’s [Fulton] seven axioms:

(1) Ip(f, g) ∈ Z≥0 when C and D intersect properly at p. Ip(f, g) =∞ otherwise;(2) Ip(f, g) = 0 if and only if p /∈ C ∩D;(3) for any affine change of coordinates T : C2 → C2, Ip(f, g) = IT−1(p)(f T, g T );(4) Ip(f, g) = Ip(g, f);(5) Ip(f, g) ≥ mpf ·mpg with equality if and only if C and D have no tangent lines at p

in common;(6) If f =

∏i f

rii and g =

∏j g

sii then Ip(f, g) =

∑i,j risjIp(fi, gj);

(7) Ip(f, g) = Ip(f, g + af) for any a ∈ C[x, y];

given by

Ip(f, g) = dimC (Op/(f, g))

Proof. Proof of uniqueness: Assuming it exists, there is an inductive procedure for calcu-lating the value of Ip(f, g) using the axioms above. Using (1), (2), and (3), we may assumeit is finite and nonzero, and that p is the origin. Assume that Ip(f, g) = k > 0 and thatany intersection multiplicity values less than k can always be calculated. Let r and s bethe respective nonnegative degrees of the single variable polynomials f(x, 0) and g(x, 0),assuming r ≤ s without loss of generality.

If r = 0, then f(x, y) = yh(x, y) for some h ∈ C[x, y]. By (6),

Ip(f, g) = Ip(y, g) + Ip(h, g).

In this case, use the axioms to show that Ip(y, g) is equal to the multiplicity of x in g(x, 0),which must be positive since g(p) = 0 by assumption. Thus Ip(h, g) < k and we havecomputed Ip(f, g) by induction. If r ≥ 0 we can reduce to the case r = 0 in a finite numberof steps using (7) and (4). We leave this as an exercise.

Proof of existence: Show that the definition

Ip(f, g) = dimC (Op/(f, g))

satisfies the axioms. This is some difficult algebra. Try as many axioms as you can, andthen read through to the proof of Bezout’s theorem below, where we provide another gooddefinition for Ip(f, g) which should be easier to show satisfies these axioms.

65

Exercise 26.11. Show that any two parallel lines in C2 do not intersect. Homogenize theequations for those lines, and count the number of resulting intersections in P2.

Exercise 26.12. How many points of intersection are there between the line V(x− a) andthe curve V(x2+y2−1) in C2? What happens at a = 1? What happens when you homogenizeand study the intersection in P2?

Theorem 26.13 (Bezout’s Theorem). Let C,D ⊂ P2 be projective plane curves defined byhomogeneous polynomials f, g ∈ C[x, y, z] of degrees m and n respectively. Assume C andD share no common components. Then∑

p

Ip(f, g) = mn.

The value Ip(f, g) always makes sense since it is defined locally. We can interpret f and gas dehomogonized polynomials in two variables in affine coordinates, using one of the threebasic affine open sets of the projective plane containing p.

Definition 26.14. Let R be a ring, and let f =∑n

i=0 aixi and g =

∑mj=0 bjx

j be polynomials

in a polynomial ring R[x] of one variable. The resultant Res(f, g;x) is defined to be thedeterminant of the Sylvester matrix which, by example in the case m = 5 and n = 3, is thedeterminant of the (m+ n)× (m+ n) matrix:∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

a5 a4 a3 a2 a1 a0 0 00 a5 a4 a3 a2 a1 a0 00 0 a5 a4 a3 a2 a1 a0

b3 b2 b1 b0 0 0 0 00 b3 b2 b1 b0 0 0 00 0 b3 b2 b1 b0 0 00 0 0 b3 b2 b1 b0 00 0 0 0 b3 b2 b1 b0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣Exercise 26.15. Find the resultant of x−a and x−b for a, b ∈ C. Show that Res(f, g;x) = 0if and only if a = b.

Exercise 26.16. Find the resultant of x − a and (x − b)(x − c) for a, b, c ∈ C. Show thatRes(f, g;x) = 0 if and only if a = b or a = c.

Exercise 26.17. If f, g ∈ C[x1, . . . , xn], show that Res(f, g;xi) ∈ C[x1, . . . , xi−1, xi+1, . . . , xn].

Exercise 26.18. Let A be the Sylvester matrix of f and g. Assume f(a) = g(a) = 0 forsome a. Show that the vector x = (am+n−1, am+n−2, . . . , a1, a0)T satisfies Ax = 0. Concludethat Res(f, g;x) = 0.

Exercise 26.19. Assume f(a) = g(a) = 0 for some a. Prove that fg

= pq

for some poly-

nomials p and q, each of which is exactly one degree lower. Use this to show that f and gshare a root if and only if the resultant is zero.

Exercise 26.20. Let f, g ∈ C[x, y, z] be homogeneous polynomials of degrees m and n,respectively, with f(0, 0, 1) 6= 0 and g(0, 0, 1) 6= 0. Show their resultant Res(f, g; z) (thinkingof x and y as constants) is homogeneous of degree mn in x and y.

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Exercise 26.21. Let h ∈ C[x, y] be a nonzero homogeneous polynomial. Prove that h canbe written in the form

h(x, y) = c

t∏i=1

(six− riy)mi

where c is a nonzero constant and [ri : si] are distinct points of P1. Show that

V(h) = [r1 : s1], . . . , [rt : st] ⊂ P1.

Proof. Proof of Bezout’s Theorem: Using an affine transformation of C3, choose coor-dinates of P2 so that [0 : 0 : 1] is not in C or D and, furthermore, is not colinear withany two other points of C ∩ D. Let p ∈ C ∩ D be described using projective coordinates[u : v : w]. Prove that another good definition for Ip(f, g) is the exponent of (vx − uy)in the factorization of Res(f, g; z) given in the above exercise (show it satisfies the sevenaxioms). Using this and the previous two exercises, you should be able to deduce Bezout’sTheorem.

27. Riemann-Roch Theorem for Curves

Let C be a smooth, projective curve over C of topological genus g.

Definition 27.1. A divisor D on C is a finite linear combination of points pi ∈ C,

D =∑i

ni[pi],

with integer coefficients ni ∈ Z. Under the assumption that only a finite number of coeffi-cients may be non-zero, we may also write

D =∑p∈C

np[p]

for convenience. We say that D is an effective divisor (and write D ≥ 0) if all the ni ≥ 0.We say the degree of D is the integer degD =

∑i ni. For two divisors D =

∑p∈C np[p] and

D′ =∑

p∈Cmp[p], we write D ≥ D′ if and only if np ≥ mp for all p ∈ C. Given a rationalfunction f on C, we can always write

div(f) =∑p∈C

ordp(f)[p]

where

ordp(f) =

a if f has a zero of multiplicty a at p

−a if f has a pole of order a at p

0 otherwise.

Exercise 27.2. Show that every rational function f on C has only a finite number of zerosand poles, so that div(f) is a well-defined divisor.

Exercise 27.3. Show that the set of divisors on C form a free abelian group.

Exercise 27.4. Show that div(fg) = div(f) + div(g). Show that div(f−1) = −div(f).

Exercise 27.5. Show that for any rational function f on C, deg div(f) = 067

Exercise 27.6. Let f be a nonzero rational function on C. Show that div(f) is effective ifand only if f is a constant function.

Exercise 27.7. Let f and g be nonzero rational functions on C. Show that div(f) = div(g)if and only if f = kg for some nonzero constant k ∈ C.

Definition 27.8. Let D =∑p∈c

np[p] be a divisor on C and consider the sheaf of regular

functions OC on C. Define OC(D) to be the sheaf of functions on C who’s zeros and polesare bounded locally by D, meaning that on an open set U ⊂ C,

OC(D)(U) = f ∈ C(U)| ordP (f) ≥ np for all p ∈ U

Exercise 27.9. Prove that OC(D) is always a locally free sheaf of rank 1 (i.e. the sheaf ofsections of a line bundle).

Exercise 27.10. Show that the set L(D) := Γ(C,OC(D)) may be characterized as the setof all global rational function on C such that div(f)+D is effective (or f is zero). Convinceyourself of the following facts:

(1) If D ≥ D′ then L(D) ⊇ L(D′) and dimC(L(D)/L(D′)) ≤ deg(D −D′).(2) L(0) = C and L(D) = 0 whenever degD < 0.

Exercise 27.11. Show that L(D) is always a finite-dimensional complex vector space overC. Denote l(D) := dimC L(D).

Our goal is to better understand the vector spaces L(D), their dimension, and how theycompare for different choices of D. One fact to notice is the following.

Exercise 27.12. Show that if D and D′ are divisors on C with D −D′ = div(f) for somerational function f , then multiplication by f induces an isomorphism between L(D) andL(D′).

We would like to work in a sense modulo this isomorphism, motivating the followingimportant definition.

Definition 27.13. Two divisors D and D′ on C are said to be linearly equivalent if D−D′ =div(f) for some rational function f , in which case we write D ∼ D′.

Exercise 27.14. Show that:

(1) linear equivalence is indeed an equivalence relation on the group of divisors on C;(2) D ∼ 0 if and only if D = div(f) for some f ;(3) if D ∼ D′ then degD = degD′;(4) if D ∼ D′ and E ∼ E ′ then D + E ∼ D′ + E ′.

Essentially, the fundamental collection of objects we want to study is the set of divisorson C modulo rational equivalence.

Exercise 27.15. Denote the set of divisors on C modulo rational equivalence by Div(C)/ ∼.Show it is a group.

Exercise 27.16. Show that if D ∼ D′ then OC(D) ∼= OC(D′) as sheaves. Denote by Pic(C)the set of isomorphism classes of locally free rank one sheaves on C. Show that the mappingD 7→ OC(D) induces a one-to-one correspondence between Div(C)/ ∼ and Pic(C).

68

Exercise 27.17. Prove that Pic(C) is a group under tensor product. Is the correspondencein the previous question a group homomorphism? This uses the exercises 25.24, 25.25, and25.26.

One important locally free rank one sheaf on C is the sheaf of sections of the cotangentbundle, the cotangent sheaf ΩC . Since C has dimension one, the cotangent sheaf is it’s owntop exterior power, so we will denote it by ΩC = ωC . We denote the divisor corresponding toωC under the correspondence between Div(C)/ ∼ and Pic(C) by KC , the canonical divisor.We are now ready to state the Riemann-Roch Theorem, which relates the dimension ofL(D) to the degree of D and genus g of C using KC .

Theorem 27.18 (Riemann-Roch Theorem).

l(D)− l(KC −D) = degD + 1− g

We first note a couple of facts about the sheaves OC provided to us by a famous and usefultheorem called Serre duality. For proofs of these facts, we refer the reader to [Hartshorne,Chapter 4]

Theorem 27.19.

(1) H1(C,OC) ' H0(C, ωC), and

(2) H1(C,OC(D)) ' H0(C, ωC ⊗Hom(OC(D),OC)).

Notice, in particular, that since the topological genus g of our curve C can be defined asthe dimension of H0(C, ωC), this theorem tells us that it is also the dimension of H1(C,OC).Notice also that the second isomorphism tells us that the left hand side of the Riemann-RochTheorem is nothing more than χ(OC(D)).

Proof. Proof of Riemann-Roch Theorem We need to show that

χ(OC(D)) = degD + 1− g

where

χ(OC(D)) = dimH0(C,OC(D))− dimH1(C,OC(D)).

As a quick exercise, show that this is true in the case D = 0. Now, let D be any divisorand p ∈ C be any point. To finish the proof, it is enough to show that the formula is truefor D if and only if it is true for D + [p] (why?).

Consider the skyscraper sheaf C(p) at p, that is, the sheaf given by

U 7→

C if p ∈ U0 otherwise.

Prove that the following sequence is exact:

0→ OC(−[p])→ OC → C(p)→ 0.

If you tensor the entire sequence with OC(D + [p]), what is the result? Prove that it isthe short exact sequence

0→ OC(D)→ OC(D + [p])→ C(p)→ 0.69

By Lemma ??, this gives

χ(OC(D + [p])) = χ(OC(D)) + χ(C(p))

χ(OC(D + [p])) = χ(OC(D)) + 1

but deg(D + [p]) = degD + 1, so the formula is true for D + [p] if and only if it is true forD.

28. Tensor products

Definition 28.1. Let R be a ring and M,N be R modules, then the tensor productM ⊗R N is the module generated by

m⊗ n|m ∈M, n ∈M/ ∼where ∼ is the equivalence relation induced by

(am+ a′m′)⊗ (bn+ bn′) = ab(m⊗ n) + a′b(m′ ⊗ n) + ab′(m⊗ n′) + a′b′(m′ ⊗ n′)for all m,m′ ∈M , n, n′ ∈ N and a, a′, b, b′ ∈ R. (In other words M ⊗RN is the quotient ofthe free module M ×N by the relation ∼).

Exercise 28.2. Show that the above equivalence relation is equivalent to requiring that

(1) (m,n) + (m′, n) ∼ (m+m′, n) for m,m′ ∈M and n ∈ N ,(2) (m,n) + (m,n′) ∼ (m,n+ n′) for m ∈M and n, n′ ∈ N ,(3) r(m,n) ∼ (rm, n) ∼ (m, rn) for r ∈ R, m ∈M and n ∈ N .

We have a natural bilinear map

π : M ×N →M ⊗R N, π(m× n) = m⊗ nby which we mean a map of sets such that

π((am+ a′m′), (bn+ bn′)) = abπ(m,n) + a′bπ(m′, n) + ab′π(m,n′) + a′b′π(m′, n′).

Exercise 28.3. Show that the set of R module homomorphisms φ : M ⊗RN → Q is in oneto one correspondence with the set of bilinear maps ψ : M ×R N → Q.

Exercise 28.4. We say that an element v ∈ M ⊗R N is a simple tensor if v = m ⊗ n forsome m ∈M and n ∈ N . Show that not all tensors are simple.

Exercise 28.5. Show that M ⊗R R ∼= R⊗RM and M ⊗R N ∼= N ⊗RM .

Exercise 28.6. Show that (⊕ni=1Mi)⊗ (⊕nj=1Nj∼= ⊕i,jMi ⊗Nj.

Exercise 28.7. Show that if M,N,P are R-modules, then (M⊗RN)⊗RP ∼= M⊗R(N⊗RP ).

Exercise 28.8. If φ : M → M ′ and ψ : N → N ′ are homomorphisms of R-modules, thenthere is a natural induced homomorphism Φ : M ⊗N →M ′⊗N ′ such that the The inducedmaps M →M ⊗N →M ′ ⊗N ′ →M ′ and N →M ⊗N →M ′ ⊗N ′ → N ′ coincide with φand ψ.

Exercise 28.9. Show that the tensor product is right exact so that if

0→M ′ →M →M ′′ → 0

is a short exact sequence of R modules and N is an R module, then

M ′ ⊗N →M ⊗N →M ′′ ⊗N → 0

is exact.70

Exercise 28.10. If R, S are rings, φ : R→ S is a homomorphism and M is an S module,then M ⊗S R is an R module.

Exercise 28.11. Let R be a ring, S a multiplicative subset of R and M an R module. Showthat S−1R⊗RM ∼= S−1M .

Definition 28.12. We say that an R module M is flat if given any injective homomorphismN ′ → N of R-modules, then the induced homomorphism M ⊗R N ′ → M ⊗R N is alsoinjective.

Exercise 28.13. Show that M is a flat R-module if and only if the tensor product by M isexact so that if

0→M ′ →M →M ′′ → 0

is a short exact sequence of R modules and N is an R module, then

0→M ⊗N ′ →M ⊗N →M ⊗N ′′ → 0

is a short exact sequence.

Lemma 28.14. If S is a multiplicative subset of a ring R, then S−1R is flat.

Proof. Let φ : M ′ → M be an injective homomorphism, then we must show that φ :S−1M ′ → S−1M is injective (see Exercise 28.11). Suppose that φ(m′/s) = 0, then φ(m′)t =0 for some t ∈ S. But then φ(tm′) = 0 and so tm′ = 0 (as φ is injective). But then m′/s = 0in S−1M ′ as required.

29. Fiber products

We will next introduce the notion of fibered products. Unluckily, in order to define thisnotion we need to consider a more general setting.

Definition 29.1. A ringed space (X,OX) is a topological space X together with a sheaf ofrings OX on X. A morphism of ringed spaces (f, f ]) : (X,OX) → (Y,OY ) is a continuousmap f : X → Y and a homomorphism f ] : OY → f∗OX of sheaves of rings on Y . We saythat the ringed space (X,OX) is a locally ringed space if each stalk OX,P is a local ringand a morphism of ringed spaces (f, f ]) : (X,OX) → (Y,OY ) is a morphism of locally

ringed spaces if for any P ∈ X, the map f ]P : OY,f(P ) → OX,P is a local homomorphismof local rings.

Exercise 29.2. Explain how the map f ]P : OY,f(P ) → OX,P is defined and show that it is alocal homomorphism of local rings (i.e. (f ])−1mP = mf(p) where mP denotes the maximalideal of OX,P and mf(p) denotes the maximal ideal of OY,f(P )).

Definition 29.3. let A be a ring, then the corresponding affine scheme (Spec(A),O) isdefined as follows:

(1) The spectrum of A is the set Spec(A) given by the union of all prime ideals ofA. The topology on Spec(A) is defined by letting the closed sets be of the formV(a) = P ⊃ a be the union of all prime ideals containing an ideal a ⊂ A.

(2) The sheaf of rings O on Spec(A) is defined by

O(U) = s : U →∏P∈U

AP

71

where U ⊂ Spec(A) is an open subset and s are functions such that s(P ) ∈ AP forany P ∈ U and for any P ∈ U there exists a neighborhood P ∈ V ⊂ U such thats(Q) = a/f ∈ AQ for all Q ∈ V where a, f ∈ A and f(Q) 6= 0 for all Q ∈ V .

Exercise 29.4. Show that the sets V(a) define a topology on Spec(A).

Exercise 29.5. Let A = C[x], then show that the points of Spec(A) correspond to themaximal ideals ma = (x− a) ⊂ C[x] and the zero ideal. Show that the zero ideal is the onlynon-closed point. What is its closure.

Exercise 29.6. What are the points (and their closures) of X = Spec(C[x, y]).

Exercise 29.7. Show that any affine scheme is a locally ringed space.

Exercise 29.8. Show that if X and Y are affine schemes corresponding to C-algebras A,B, then there is a natural bijection between the C algebra homomorphisms φ : B → A andmorphisms of locally ringed spaces f : X → Y .

Exercise 29.9. Let (X,OX) be a scheme. Show that there is a natural morphism (X,OX)→Spec(Z).

Exercise 29.10. Let a ⊂ A be an ideal, X = Spec(A) and Y = Spec(A/a). Show thatthe surjective homomorphism a : A → A/a induces a closed immersion Y → X (i.e. amorphism of schemes which is a homeomorphism on the topological space Y and such thatthe induced map of sheaves on X, f ] : OX → f∗OY is surjective).

Note that different ideals a, a′ ⊂ A may define the same topological space but havedifferent associated scheme structures. For example (xn) ⊂ C[x] defines a scheme whosetopological space is just the origin in C, but has scheme structure given by O = C[x]/(xn).We think of the corresponding scheme as the n-th infinitesimal neighborhood of the origin.

For another example, consider the ideals (x, y), (x2, y2), (x2, y2, xy) ⊂ C[x, y] definingdifferent scheme structures on (0, 0) ⊂ C2.

Finally consider (x2, xy) ⊂ C[x, y]. The topological space corresponding to this schemeis given by the y-axis, however there is a non-reduced structure at the origin correspondingto a non-zero tangent direction parallel to x-axis.

Definition 29.11. A scheme (X,OX) is a locally ringed space such that X = ∪Ui where(Ui,OX |Ui

) are affine schemes.

Definition 29.12. If (X,OX) and (Y,OY ) are schemes and U ⊂ X and V ⊂ Y areopen subsets such that (U,OX |U) and (Y,OY |V ) are isomorphic, then there exists a scheme(Z,OZ) and open subsets Zi such that (Z1,OZ |Z1) ∼= (X,OX) and (Z2,OZ |Z2) ∼= (Y,OY ).we say that (Z,OZ) is obtained by glueing the schemes (X,OX) and (Y,OY ) along(U,OX |U) ∼= (Y,OY |V ).

Exercise 29.13. Let (X,OX) = (Y,OY ) = Spec(C[x]) and U ∼= V = Spec(C[x, x−1]). Showthat the scheme Z obtained by glueing (X,OX), (Y,OY ) along U ∼= V is isomorphic to theaffine line with the origin replaced by a double point.

Definition 29.14. Let f : X → S and g : Y → S be morphisms of affine varieties so thatA(X) = A and A(Y ) = B are rings and f, g correspond to homomorphisms φ : R→ A andψ : R → B. Define the fiber product X ×S Y to be the affine variety corresponding tothe ring A ⊗R B and morphisms α : X ×S Y → X and β : X ×S Y → Y determined by

72

the homomorphisms A → A ⊗R B and B → A ⊗R B which send a → a ⊗ 1 and b → 1 ⊗ brespectively.

Lemma 29.15. Let Z be an affine variety and a : Z → X, b : Z → Y be S-morphisms (i.e.morphisms such that the f a = g b), then there exists a unique morphism c : Z → X×S Ysuch that α c = a and β c = β.

Proof. Let T = A(Z), then the morphisms a, b correspond to homomorphisms A → T andA⊗R B → B. These homomorphisms are compatible with the S-module structure T → Aand T → B and hence we obtain a natural homomorphism T → A⊗R B.

One can define the fiber product of two morphisms of quasi-projective varieties f : X → Sand g : Y → S as follows.

Definition 29.16. The fiber product of X, Y over S is a quasi-projective variety Z withmorphisms α : Z → X and β : Z → Y (compatible with the structure maps X → S andY → S so that f α = g β) such that for any quasi-projective variety W with morphismsa : W → X and b : W → Y (compatible with the structure maps X → S and Y → S sothat f a = g b), then there exists a unique morphism c : W → Z such that α c = β c.We denote the fiber product by z = X ×S Y .

One can use the above lemma to show the following.

Theorem 29.17. Let f : X → S and g : Y → S be morphisms of quasi-projective varieties,then the fiber product X ×S Z exists.

Proof. [Hartshorne, II.3.3].

We can use fibered products to give a definition of fibers of a morphism.

Definition 29.18. let f : X → Y be a morphism of schemes and y ∈ Y a point (corre-sponding to a morphism Spec(C)→ Y ), then the fiber of f over y is Xy := X ×Y y.

We make the following observation. Even if we assume that X is a quasi-projective variety,it may happen that the fiber Xy is not a quasi-projective variety. Consider for example (cf.[Hartshorne, III.3.3.1]) X = Spec(C[x, y, t]/(ty − x2)) and Y = Spec(C[t]). Let f : X → Ybe the morphism induced by the inclusion C[t] → C[x, y, t]/(ty − x2), then for any a ∈ C,the fiber Xa is determined by X = Spec(C[x, y]/(ay− x2)). For a 6= 0, this is a plane curveof degree 2 in C2, but for a = 0, we obtain a double line.

30. Monomial orders

Recall that a monomial in C[x1, · · · , xn] is simply an expression of the form xa11 · · ·xann = xa

where a = (a1, . . . , an) ∈ Zn≥0. A monomial ordering > on C[x1, · · · , xn] is a relation onthe set of all monomials such that

(1) ≥ is a total order, i.e. for any monomials xa, xb and xc (a, b, c ∈ Zn≥0)

– if xa < xb and xb < xc, then xa < xc, and– for any two monomials xa and xb then one and only one of the following is true:

xa < xb, xb < xa or xa = xb.(2) for any monomials xa, xb and xc (a, b, c ∈ Zn≥0), if xa < xb, then xa+c < xb+c (the

order respect multiplication), and73

(3) (Well ordering) any decreasing sequence of elements xa1 > xa2 > . . . is finite (orequivalently, for any non-empty subset of monomials xaii∈I there is a smallestelemet say xa0).

Note that we can equivalently talk about the corresponding order on Zn≥0. We will switchfrom one to the other often without any further comment. It is easy to see that the orderingon C[x] given by degree:

xa > xb if and only if a > b

is a monomial order. However, the relation C[x, y] given by degree

xayb > xcyd if and only if a+ b > c+ d

is not a monomial ordering. It is however possible to give many monomial orderings onC[x1, · · · , xn]. For example we have

Definition 30.1. Lexicographic order: Given a, b ∈ Zn≥0, we say a >lex b if the firstnon-zero entry in a− b is ≥ 0. Thus

x31x2 >lex x

21x

72, x2

1x32x3 >lex x

21x

32x

24.

Notice that this is just the usual alphabetical order, eg. a = x1, b = x2 etc then

artic > apt, bbb > burp.

Graded lexicographic order: Given a, b ∈ Zn≥0, we say a >grlex b if either |a| > |b| or|a| = |b| and a >lex b. Thus

x31x2 >grlex x

21x

72, x2

1x32x

24 >grlex x

21x

32x3, .

Exercise 30.2. Show that >grlex and >lex are monomial orders.

Exercise 30.3. Show that the order defined by xayb > xcyd if and only if a+b√

2 > c+d√

2is a monomial order.

The notion of ordering is extremely useful when we wish to run any recursive procedure.For example the division algorithm and the Euclidean algorithm. Let’s begin by recallingthese in the standard setting.

Definition 30.4. Division algorithm: Let a, b ∈ N then there exist q, r ∈ N such thata = bq + r where 0 ≤ r < b.

Recall that the greatest common divisor (or GCD) of two integers a, b ∈ N is aninteger c = gcd(a, b) such that

(1) c divides a and b (written c|a, c|b),(2) for any integer d dividing a and b, then d divides c.

Exercise 30.5. Show that (a, b)− (c) as ideals in Z where c = gcd(a, b).

Exercise 30.6. Show that c = gcd(a, b) is the smallest integer in (a, b) ∩ N.

Definition 30.7. Euclidean algorithm:Let a ≤ b ∈ N then let a = bq+r where q, r ∈ Z≥0

and 0 ≤ r < b. Let a = a0, b = b0 and r = r0. We then proceed inductively as follows.Assume that ai, bi, ri have been defined for 0 ≤ i ≤ n. If rn = 0, then bn = gcd(a, b)otherwise let an+1 = bn, bn+1 = rn and write an+1 = bn+1qn+1 + rn+1 where 0 ≤ rn+1 < bn.Since ri > ri + 1 > . . . ≥ 0 the procedure terminates after finitely many steps.

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Exercise 30.8. Use the Euclidean Algorithm to find the GCD of 66 and 42.

Once we have fixed a monomial ordering > we can also perform long division on polynomials.For example you are familiar with long division of polynomials in one variable: Givenf, g ∈ C[x] there exist q, r ∈ C[x] such that

f = gq + r, deg r < deg g.

here of course we use the monomial ordering induced by degree.

Exercise 30.9. Divide x4 + 3x2 by x2 + x+ 1.

We can also use long division to find the GCD of two polynomials in one variable.

Exercise 30.10. Show that gcd(x4 − 1, x3 − 3x+ x2 − 3) = x2 + 1.

Exercise 30.11. Show that if f, g ∈ C[x], then (f, g) = (h) where h = gcd(f, g). thus h isany non-zero polynomial of minimal degree in (f, g)

Turning now to polynomials in several variables, let’s use >lex to divide x3y + xy3 byx+ y2. We have LT (x3y + xy3) = x3y and LT (x+ y2) = x. Now x3y = (x2y) · x and so wehave

x3y + xy3 = (x2y) · (x+ y) + xy3 − x2y2

Since LT (xy3 − x2y2) = x2y2 = xy2LT (x+ y), we next write

xy3 − x2y2 = −xy2(x+ y) + 2xy3 = (x+ y)(xy2 + 2y3)− 2y4.

Putting all of this together we obtain

x3y + xy3 = (x+ y)(x2y + xy2 + 2y3)− 2y4.

Therefore the remainder is 2y4 and note that x+y > y4. Unluckily, it is not always possibleto assume that if we divide f by g the remainder will satisfy g > r. Consider in fact

f = x2 + y2, g = xy

wrt >lex. Since xy does not divide any term in f , it follows that r = f but g < r (notr < g). Therefore when dividing f by g we can only require that the leading term of g doesnot divide any term of r:

Definition 30.12. If > is a monomial order on C[x1, . . . , xn], then for any f, g ∈ C[x1, . . . , xn]there exist q, r ∈ C[x1, . . . , xn] such that

f = gq + r,

where LT (g) does not divide any monomial in r. Similarly if we divide f by g1, . . . , gt thenwe write

f =∑

giqi + r, qi ∈ C[x1, . . . , xn]

where LT (gi) does not divide any monomial in r (for 1 ≤ i ≤ n).

Exercise 30.13. Use >grlex to divide x3y + xy3 by x + y2. Notice that you get a differentanswer.

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31. Monomial ideals

Definition 31.1. A monomial ideal I ⊂ C[x1, . . . , xn] is an ideal generated by monomials.

For example (x2, y3) is a monomial ideal, but (x2 + y3) is not a monomial ideal. Noticehowever that (x2 + y3, x2 − y3) is a monomial ideal since (x2 + y3, x2 − y3) = (x2, y3).

Exercise 31.2. Show that I ⊂ C[x1, . . . , xn] is a monomial ideal if and only if for any f ∈ Ithen all the monomials of f are in I. It follows that two monomial ideals I, J ⊂ C[x1, . . . , xn]are equal if and only if they contain the same monomials.

Next we will reprove the Hilbert’s basis theorem in the context of monomiaml ideals.

Theorem 31.3 (Dicksons Lemma). Let A ⊂ Zn≥0 and I = (xαα∈A) ⊂ C[x1, . . . , xn] thecorresponding ideal, then there exists a finite subset B ⊂ A such that I = (xαα∈B) ⊂C[x1, . . . , xn]

Proof. We proceed by induction on n. If n = 1, then it suffices to consider the smallestnon-zero element a ∈ A. We claim that I = (xa). Consider in fact

We will now show how to associate a monomial ideal to any ideal I.

Definition 31.4. Let I ⊂ C[x1, . . . , xn] be an ideal, then let LT (I) = (LT (f)f∈I).

Note that LT (I) is a monomial ideal. Note that it may happen that I = (f1, . . . , fr) butLT (I) 6= (LT (f1), . . . , LT (fr)).

Exercise 31.5. Let f = x and g = x + y, and consider >lex. Show that LT (f, g) 6=(LT (f), LT (g)).

We will now use Dickson’s Lemma to reprove the Hilbert’s basis theorem:

Theorem 31.6 (Hilbert’s basis theorem). Let I ⊂ C[x1, . . . , xn] be an ideal, then there existf1, . . . , fr ∈ I such that I = (f1, . . . , fr).

Proof. [CLS, Theorem 4].

Definition 31.7. If f1, . . . , fr ∈ I are such that (LT (f1), . . . , LT (fr)) = LT (I)) then wesay that f1, . . . , fr is a Groebner basis of I.

It is an immediate consequence of Dickinson’s Lemma that every ideal I ⊂ C[x1, . . . , xn]has a Groebner basis. We also observe that a Groebner basis is a basis of I so that(f1, . . . , fr) = I. To see this, pick f ∈ I and divide it by f1, . . . , fr to get f =

∑qifi + r

where LT (r) is not divisible by LT (fi). However r ∈ I so that LT (r) ∈ LT (I). But then....

Lemma 31.8. Let G = (g1, . . . , gr) be a Groebner basis for an ideal I ⊂ C[x1, . . . , xn]and f ∈ C[x1, . . . , xn], then there exists a unique polynomial r ∈ C[x1, . . . , xn] such thatr ∈ f + I, and no term of r is divisible by LT (gi).

Proof. The existence of such a polynomial r was observed avbove: simply divide f byg1, . . . , gr to get f =

∑aigi + r with a1, . . . , ar ∈ C[x1, . . . , xn].

To see uniqueness, suppose that we have two such expressions r ∈ f + I and r′ ∈ f + Isuch that no term of r, r′ is divisible by LT (gi), then r − r′ ∈ I and so some term of r − r′is divisible by LT (gi). This is impossible and so r − r′ = 0.

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Definition 31.9. g1, . . . , gr is a minimal Groebner basis of an ideal I ⊂ C[x1, . . . , xn]if it is a Groebner basis such that for any 1 ≤ i ≤ r we have LC(gi) = 1 and

LT (gi) 6∈ (LT (g1), . . . , LT (gi−1, LT (gi+1), . . . , LT (gr)).

Definition 31.10. g1, . . . , gr is a minimal Groebner basis of an ideal I ⊂ C[x1, . . . , xn]if it is a Groebner basis such that for any 1 ≤ i ≤ r we have

LT (gi) 6∈ (LT (g1), . . . , LT (gi−1, LT (gi+1), . . . , LT (gr)).

Definition 31.11. g1, . . . , gr is a reduced Groebner basis of an ideal I ⊂ C[x1, . . . , xn]if it is a Groebner basis such that for any 1 ≤ i ≤ r we have LC(gi) = 1 every monomial ofgi is not in (LT (g1), . . . , LT (gi−1, LT (gi+1), . . . , LT (gr)).

Proposition 31.12. Reduced Groebner basis always exist and are unique.

Proof.

32. Solutions and hints for the exercises

(1) Exercise 1.24. Suppose that Cn = X ∪ Y where X, Y are proper closed subsets. Wemay pick polynomials p, q ∈ C[x1, . . . , xn] such that p ∈ I(X), q ∈ I(Y ). But thenpq ∈ I(X ∪ Y ) = I(Cn), but this is impossible as any non-constant polynomial has a(infinitely many if n > 1) solution. v

(2) Exercise 1.40. Suppose that X is compact in the Euclidean topology. Supposethat there exists a sequence xk such that no subsequence has a limit in X. wemay assume that all xk are distinct and we pick open neighborhoods Ui such thatUi ∩ xi|i ∈ N = xi. Let U = X \ xi|i ∈ N. U is open in X since no subsequencehas a limit in X (check). Then U ∪i∈NUi is an open cover of X. Since X is compact,it admits a finite open subcover, but this is clearly impossible.

(3) Exercise 2.11. Hint: Consider the projection V(xy = 1)→ C.(4) Exercise 3.31. Hint: Consider the inclusion Z → Q.(5) Exercise 9.7.(6) Exercise 29.8. See [Hartshorne, II.2.3].(7) Exercise 12.18. Hint: Use the diagonal trick.(8) Exercise 12.25. Hint: Use the d-uple embedding.

References

[Allcock] D. Allcock: Hilberts nullstellensatz, available at http://www.ma.utexas.edu/users/allcock/expos/nullstellensatz3.pdf[CLS][Beauville] A. Beauville: Complex Algebraic Surfaces London Mathematical Society student text 34[Fulton] W. Fulton: Algebraic Curves[Hartshorne] R. Hartshorne: Algebraic Geometry[Reid] M. Reid, Undergraduate Algebraic Geometry, LMS Student Texts 12 , C.U.P., Cambridge

1988. http://homepages.warwick.ac.uk/staff/Miles.Reid/MA4A5/UAG.pdf[SKKT00] K. Smith, L. Kahanpaa, P. Kekalainen, W. Treves: An invitation to Algebraic Geometry,

Springer-Verlag, New York, 2000. xii+155.

Department of Mathematics, University of Utah, Salt Lake City, UT 84112, USAE-mail address: hacon@math.utah.edu

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