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APPENDIX A

Basics from Algebraic Geometry

In this appendix we gather together some notions and results from algebraicgeometry which have been used in the text. We concentrate on affine algebraicgeometry which simplifies a lot the notational part and makes the subject mucheasier to access in a first attempt. In the second appendix, we discuss the relationbetween the Zariski topology and theC-topology. With its help we are able to use

certain compactness arguments replacing the corresponding results from projectivegeometry.The appendix assumes a basic knowledge in commutative algebra. Although

we give complete proofs for almost all statements they are mostly rather short.This was done on purpose. For advanced readers we only wanted to recall brieflythe basic facts, while beginners are going to find a more detailed study of alge-braic geometry is necessary. We recommend the text books [Har77], [Mum99],[Mum95], [Sha94a,Sha94b] and the literature cited below. As a substitute wehave presented many examples which should make the new ideas clear and withwhich one can check the results. In addition, a number of exercises are included.The reader is advised to look at them carefully; some of them will be used in theproofs.

99
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100 A. BASICS FROM ALGEBRAIC GEOMETRY

Contents

1. Affine Varieties 1011.1. Regular functions 1011.2. Zero sets and Zariski topology 1031.3. Hilberts Nullstellensatz 1051.4. Affine varieties 1071.5. Special open sets 1091.6. Decomposition into irreducible components 1111.7. Rational functions and local rings 1132. Morphisms 1152.1. Morphisms and comorphisms 1152.2. Images, preimagees and fibers 1172.3. Dominant morphisms 119

2.4. Products 1202.5. Fiber products 1223. Dimension 1223.1. Definitions 1223.2. Finite morphisms 1243.3. Krullsprincipal ideal theorem 1273.4. Decomposition Theorem and dimension formula 1303.5. Constructible sets 1323.6. Degree of a morphism 1334. Tangent Spaces and Differentials 1344.1. Zariski tangent space 1344.2. Tangent spaces of subvarieties 1364.3. Epsilonization 1364.4. Nonsingular varieties 1374.5. Associated graded algebras 1384.6. Vector fields and tangent bundle 1424.7. Differential of a morphism 1454.8. Tangent spaces of fibers 1464.9. Morphisms of maximal rank 1475. Normal Varieties and Divisors 1495.1. Normality 1495.2. Integral closure and normalization 1515.3. Discrete valuation rings and smoothness 1535.4. ZariskisMain Theorem 1555.5. Divisors 158
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1. AFFINE VARIETIES 101

1. Affine Varieties

1.1. Regular functions. Our base field is the field C of complex numbers.Every polynomialp C[x1, . . . , xn] can be regarded as aC-valued function onCnin the usual way:

a= (a1, . . . , an) p(a) =p(a1, . . . , an).These functions will be called regular. More generally, let V be aC-vector spaceof dimension dim V =n < .

Definition 1.1. AC-valued function f: V Cis called regular iff is givenby a polynomial p C[x1, . . . , xn] with respect to one and hence all bases ofV.This means that for a given basis v1, . . . , vn ofV we have

f(a1v1+ + anvn) =p(a1, . . . , an)

for a suitable polynomialp. The algebra of regular functions onVwill be denotedbyO(V).

By our definition, every choice of a basis (v1, v2, . . . , vn) ofVdefines an isomor-

phismC[x1, . . . , xn] O(V) by identifyingxi with the ith coordinate function on

Vdefined by the basis, i.e.,

xi(a1v1+ a2v2+ + anvn) :=ai.Another way to express this is by remarking that the linear functions on V areregular and thus the dual space V := Hom(V,C) is a subspace ofO(V). Soif (v1, v2, . . . , vn) is a basis of V and (x1, x2, . . . , xn) the dual basis of V

then

O(V) = C[x1, x2, . . . , xn] and the linear functions xiare algebraically independent.

Example 1.2. Denote by Mn = Mn(C) the complex n n-matrices so thatO(Mn) = C[xij | 1 i, j n]. Consider det(tEn X) as a polynomial inC[t, xij , i , j = 1, . . . , n] where X := (xij ). Developing this as a polynomial in twe find

det(tEn X) =tn s1tn1 + s2tn2 + (1)nsnwith polynomialsskin the variablesxij. This defines regular functionssk O(Mn)which are homogeneous of degree k. Of course, we have s1(A) = tr(A) = a11+ + ann and sn(A) = det(A) for any matrix A Mn.

The same construction applies to End(V) for any finite dimensional vectorspaceV and defines regular function sk O(End(V)).

Example 1.3. Consider the vector space Pn of unitary polynomials of degreen:

Pn := {tn a1tn1 + a2tn2 + (1)nan| a1, , an C} Cn.There is a regular functionD O(Pn), thediscriminant, with the following prop-erty: D(p) = 0 for ap Pn if and only ifp has a multiple root.

Proof. Expandingn

i=1(tyi) =tn1(y)tn1+ +(1)nn(y) we see thatthe polynomials j (y) are the elementary symmetric polynomials in n variables

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y1, . . . , yn, i.e.

k

=k

(y) =k

(y1

, . . . , yn

) := i1

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1.2. Zero sets and Zariski topology. We now define the basic object ofalgebraic geometry, namely the zero set of regular functions. Let V be a finite

dimensional vector space.Definition 1.5. Iff O(V) then we define the zero set off by

V(f) := {v V| f(v) = 0} =f1(0).More generally, the zero set off1, f2, . . . , f s O(V) or of a subset S O(V) isdefined by

V(f1, f2, . . . , f s) :=s

i=1

V(fi) = {v V| f1(v) = =fs(v) = 0}

or

V(S) := {v V| f(v) = 0 for all f S}.

Remark 1.6. The following properties of zero sets follow immediately fromthe definition.

(1) LetS O(V) and denote by a := (S) O(V) the ideal generated by S.ThenV(S) = V(a).

(2) IfS T O(V) thenV(S) V(T).(3) For any family (Si)iIof subset Si O(V) we have

iI

V(Si) = V(iI

Si).

Example 1.7. (1) SLn(C) = V(det 1) Mn(C).(2) On(C) = V(

n=1 xixj ij| 1 i j n).

(3) Iff=f(x, y) C

[x, y] is a non-constant polynomial in 2 variables, thenV(f) C is called a plane curve. In order to visualize a plane curve, weusually draw a real picture R2.

Lemma 1.1. LetVbe a finite dimensional vector space and leta, b be ideals inO(V) and(ai| i I) a family of ideals ofO(V).

(1) Ifa b thenV(a) V(b).(2)

iI V(ai) = V(

iIai).(3)V(a) V(b) = V(a b) = V(a b).(4)V(0) =V andV(1) = .

Proof. Properties (1) and (2) follow from Remark 1, and property (4) is easy.So we are left with property (3). Since a

a

b

a

b, it follows from (1) that

V(a) V(a b) V(a b). So it remains to show thatV(a b) V(a) V(b). Ifv Vdoes not belong to V(a)V(b) then there are functionsf aand h bsuchthatf(v) = 0 =h(v). Sincef h a band (f h)(v) = 0 we see thatv / V(a b),and the claim follows.

Definition1.8. The lemma shows that the subsets V(a) wherearuns throughthe ideals ofO(V) form the closed sets of topology on V which is called Zariskitopology. From now on all topological terms like open, closed, neighborhood,continuous, etc. will refer to the Zariski topology.

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Example 1.9. (1) Thenilpotent coneN Mn consisting of all nilpotentmatrices is closed and is a cone, i.e. stable under multiplication with

scalars. E.g. for n = 2 we haveN= V(x11+ x22, x11x22 x12x21) M2 .

(2) The subset M(r)n Mn of matrices of rank r are closed cones.

(3) The set of polynomials f Pn with a multiple root is closed (see Exam-ple1.3).

(4) The closed subsets ofC are the finite sets together with C. So the non-empty open sets ofC are the cofinite sets.

Exercise 1.3. A regular function f O(V ) is called homogeneous of degree d iff(tv) = tdf(v) for all t Cand all v V.

(1) A polynomialf C[x1, . . . , xn] is homogeneous of degree d as a regular functiononCn if and only if all monomials occurring in fhave degreed.

(2) Assume that the ideala O(V) is generated by homogeneous functions. Thenthe zeros setV(a) V is a cone.

(3) Conversely, ifX V is a cone, then the ideal I(X) can be generated by homo-geneous functions.

Exercise 1.4. Show that the subset A:={(n, m) C2 | n, mZ and mn0}is Zariski-dense inC2.

Definition 1.10. LetX Vbe a closed subset. A regular function onX isdefined to be the restriction of a regular function on V:

O(X) := {f|X| f O(V)}.The kernel of the (surjective) restriction map res:O(V) O(X) is called theideal ofX:

I(X) := {f O(V) | f(x) = 0 for all x X}.Thus we have a canonical isomorphismO(V)/I(X) O(X).

Remark 1.11. Every finite dimensional C-vector space V carries a naturaltopology given by the choice of a norm or a hermitian scalar product. We willcall it the C-topology. Since all polynomials are continuous with respect to theC-topology we see that the C-topology is finer than the Zariski topology.

Exercise 1.5. Show that every non-empty open set in Cn is dense in theC-topology.(Hint: Reduce to the casen = 1 where the claim follows from Example1.9(4).)

Remark 1.12. In the Zariski topology the finite sets are closed. This followsfrom the fact that for any two different points v, w

V one can find a regular

function f O(V) such that f(v) = 0 and f(w)= 0. (One says that the regularfunctions separate the points.) But the Zariski topology is not Hausdorff (see thefollowing exercise).

Exercise 1.6. LetU, U Cn be two non-empty open sets. Then UU is non-empty,too. In particular, the Zariski topology is not Hausdorff.

Exercise 1.7. Consider a polynomial f C[x0, x1, . . . , xn] of the form f = x0p(x1, . . . , xn), and let X= V(f) be its zero set. Then I(X) = (f) and O(X) C[x1, . . . , xn].
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1.3. Hilberts Nullstellensatz. The famous Nullstellensatz of Hilbertap-pears in many different forms which are all more or less equivalent. We will give

some of them which are suitable for our purposes.Definition 1.13. If a is an ideal of an arbitrary ring R, its radical

a is

defined by a:= {r R | rm a for some m >0}.

The ideal a is called perfect ifa=a. The ringR is called reduced if

(0) = (0),

or equivalently, ifR contains no nonzero nilpotent elements.

It is easy to see that

a is an ideal and that

a=

a. Moreover,

a= Rimplies thata= R.

Theorem 1.2 (Hilberts Nullstellensatz). Let a O(V) be an ideal andX :=

V(a)

V its zero set. Then

I(X) =I(V(a)) = a.A first consequence is that every strict ideal has a non-empty zero set, because

X= V(a) = implies thata= I(X) = O(V) and so a= O(V).Corollary 1.3. For every ideala = O(V) we haveV(a) = .Let mC[x1, . . . , xn] be a maximal ideal and a= (a1, . . . , an) V(m) which

exists by the previous corollary. Then m (x1 a1, . . . , xn an) and so these twoare equal.

Corollary 1.4. Every maximal idealm ofC[x1, . . . , xn] is of the form

m= (x1

a1, . . . , xn

an).

Another way to express this is by using the evaluation map ev v :O(V)C(see Exercise1.1).

Corollary 1.5. Every maximal ideal ofO(V) equals the kernel of the evalu-ation map evv :O(V) C at a suitablev V.

Exercise 1.8. If X V is a closed subset and m O(X) a maximal ideal thenO(X)/m= C. Moreover, m = ker(evx : f f(x)) for a suitable x X.

Proof of Theorem 1.2. We first prove Corollary1.4 which implies Corol-lary1.5as we have seen above, and also Corollary 1.3.

Letm C[x1, . . . , xn] be a maximal ideal and denote by K := C[x1, . . . , xn]/mits residue class field. Then K contains C and has a countable C-basis, becauseC[x1, . . . , xn] does. IfK=C and p K\ C then p is transcendental over C. Itfollows that the elements ( 1pa| a C) fromKform a non-countable set of linearlyindependent elements over C. This contradiction shows thatK= C. Thusxi +m=ai+ m for a suitable ai C(for i= 1, . . . , n), and so m= (x1 a1, . . . , xn an).This proves Corollary1.4.

To get the theorem, we use the so-called trick of Rabinowich. Let a C[x1, . . . , xn] be an ideal and assume that the polynomial f vanishes onV(a).Now consider the polynomial ring R := C[x0, x1, . . . , xn] in n+ 1 variables and
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the ideal b := (a, 1 x0f) generated by 1 x0f and the elements of a. Clearly,V(b) = and so 1 b. This means that we can find an equation of the form

i

hifi+ h(1 x0f) = 1

wherefi a and hi, h R. Now we substitute 1f for x0 and findi

hi(1

f, x1, . . . , xn)fi= 1.

Clearing denominators finally gives

i hifi = fm, i.e., fm a, and the claim

follows.

Corollary 1.6. For any ideala O(V) and its zero setX := V(a) we have

O(X) =

O(V)/

a.

Exercise 1.9. Let a Rbe an ideal of a (commutative) ring R. Then a is perfect ifand only if the residue class ring R/a has no nilpotent elements different from 0.

Example 1.14. Let f C[x1, . . . , xn] be an arbitrary polynomial and con-sider its decomposition into irreducible factors: f = pr11 p

r22 prss . Then

(f) =

(p1p2 ps) and so the ideal (f) is perfect if and only if the polynomial f itis square-free. In particular, if f C[x1, . . . , xn] is irreducible, thenO(V(f))C[x1, . . . , xn]/(f). A closed subset of the formV(f) is called a hypersurface.

Example 1.15. We haveO(SLn(C)) O(Mn)/(det 1) because the polyno-mial det 1 is irreducible. (In fact, if det 1 =f1 f2 then each factor fi is linearin the variables which occur. But ifxi0j0 occurs in f1 then all the variables xij0

and xi0j have to occur in f1, too, since det 1 does not contain products of theformxij0xij0 and xi0j xi0j. This implies that all variables occur in f1, hence f2 isa constant.)

Example 1.16. Consider the plane curve C :=V(y2 x3) which is calledNeils parabola. ThenO(C) C[x, y]/(y2 x3) C[t2, t3] C[t] where thesecond isomorphism is given by : x t3, y t2.

Proof. Clearly, y 2 x3 ker . For anyf C[x, y] we can write f=f0(x) +f1(x)y+ h(x, y)(y

2 x3). Iff ker then 0 = (f) = f0(t2) +f1(t2)t3 and sof0= f1= 0. This shows that ker = (y

2 x3), and the claim follows. Exercise 1.10. Let C

C2 be the plane curve defined by y

x2 = 0. Then I(C) =

(y x2) andO(C) is a polynomial ring in one variable.Exercise 1.11. Let D C2 be the zero set ofxy 1. ThenO(D) is not isomorphic

to a polynomial ring, but there is an isomorphismO(D) C[t, t1].Exercise 1.12. Consider the parametric curve

Y := {(t, t2, t3) C3 | t C}.ThenYis closed inC3. Find generators for the ideal I(Y) and show that O(Y) is isomor-phic to the polynomial ring in one variable.

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Another important consequence of the Nullstellensatz is a one-to-one corre-spondence between closed subsets ofCn and perfect ideals of the coordinate ring

C[x1, . . . , xn].Corollary 1.7. The map X I(X) defines a inclusion-reversing bijection

{X V closed} {a O(V) perfect ideal}whose inverse map is given by a V(a). Moreover, for any finitely generatedreducedC-algebraR there is a closed subsetXCn for somen such thatO(X)is isomorphic to R

Proof. The first part is clear sinceV(I(X)) = X and I(V(a)) =a for anyclosed subsetX Vand any ideal a O(V).

If R is a reduced and finitely generated C-Algebra, R = C[f1, . . . , f n], thenR C[x1, x2, . . . , xn]/a where a is the kernel of the homomorphism defined byxi fi. Since R is reduced we havea= a and soO(V(a)) C[x1, . . . , xn]/a R.

Exercise 1.13. LetX Vbe a closed subset and f O(X) a regular function suchthat f(x)= 0 for all x X. Then f is invertible inO(X), i.e. the C-valued functionx f(x)1 is regular on X.

Exercise 1.14. Every closed subsetX Cn is quasi-compact, i.e., every covering ofXby open sets contains a finite covering.

Exercise 1.15. Let X Cn be a closed subset. Assume that there are no non-constant invertible regular function on X. Then every non-constantf O(X) attains allvalues inC, i.e. f: X C is surjective.

1.4. Affine varieties. We have seen in the previous section that every closedsubset X V (or X Cn) is equipped with an algebra ofC-valued functions,namely the coordinate ringO(X). We first remark thatO(X) determines thetopology ofX. In fact, define for every ideal a O(X) the zero set inXby

VX(a) := {x X| f(x) = 0 for all f a}.Clearly, we haveVX(a) =V(a)X where a O(V) is an ideal which mapssurjectively onto a under the restriction map. This shows that the setsVX(a)are the closed sets of the topology on X induced by the Zariski topology of V.Moreover, the coordinate ringO(X) also determines the points of X since theyare in one-to-one correspondence with the maximal ideals ofO(X):

x X mx:= ker evx O(X)where evx :O(X) C is the evaluation mapf f(x). This leads to the followingdefinition of an abstract zero set, not embedded in a vector space.

Definition 1.17. A set Ztogether with aC-algebraO(Z) ofC-valued func-tions on Z is called an affine variety if there is a closed subset XCn for somen and a bijection : Z

Xwhich identifiesO(X) withO(Z), i.e., :O(X)O(Z) given by f f , is an isomorphism.

The functions fromO(Z) are called regular, and the algebraO(Z) is calledcoordinate ring ofZor algebra of regular functions onZ.

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The affine variety Z has a natural topology, the Zariski topology, the closedsets being the zero sets

VZ(a) := {z Z| f(z) = 0 for all f a}wherea runs through the ideals ofO(Z). If follows from what we said above thatthe bijection : Z

Xis a homeomorphism with respect to the Zariski topology.Clearly, every closed subset X V or X Cn together with its coordinate

ring O(X) is an affine variety. More generally, ifX is an affine variety and Y Xa closed subset, then Ytogether with the restrictionsO(Y) := {f|Y| f O(X)}is an affine variety, called a closed subvariety.

Less trivial examples are the following.

Example 1.18. LetMbe a finite set and define O(M) :=CM = Maps(M,C)to be the set of all

C-valued functions on M. Then (M,

O(M)) is an affine variety

and O(M) is isomorphic to a product of copies ofC. This follows immediately fromthe fact that any finite subset N Cn is closed and thatO(N) = Maps(N,C).

Example1.19. LetXbe a set and define thesymmetric product Symn(X) tobe the set of unordered n-tuples of elements from X, i.e.,

Symn(X) =X X X/ where (a1, a2, . . . , an) (b1, b2, . . . , bn) if and only if one is a permutation of theother.

In case X =C we defineO(Symn(C)) to be the symmetric polynomials in nvariables and claim that Symn(C) is an affine variety.

To see this consider the map

:Cn Cn, a= (a1, . . . , an) (1(a), 2(a), . . . , n(a))where1, . . . , n are the elementary symmetric polynomials (see Example 1.3). Itis easy to see that is surjective and that (a) = (b) if and only ifa b. Thus, defines a bijection : Symn(C)

Cn, and the pull-back of the regular functionsonCn are the symmetric polynomials: :C[x1, . . . , xn]

O(Symn(C)).Exercise 1.16. Let Z be an affine variety with coordinate ringO(Z). Then every

polynomial f O(Z)[t] with coefficients inO(Z) defines a function on the product ZCin the usual way:

f=m

k=0 fktk : (z, a)

m

k=0 fk(z)ak

C

Show that Z C together withO(Z)[t] is an affine variety.(Hint: For any ideala C[x1, . . . , xn] there is a canonical isomorphismC[x1, . . . , xn, t]/(a) (C[x1, . . . , xn]/a)[t].)

Exercise 1.17. For any affine variety Z there is a inclusion-reversing bijection

{A Z closed} {a O(Z) perfect ideal}given by A I(A) := {f O(Z) | f|A= 0} (cf. Corollary1.7).
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For the last example we start with a reduced and finitely generated C-algebraR. Denote by spec R the set of maximal ideals ofR:

spec R:= {m | m Ra maximal ideal}.We know from Hilberts Nullstellensatz (see Exercise 1.8) that R/m = C forall maximal ideals m spec R. This allows to identify the elements from R withC-valued functions on spec R: For f Rand m spec R we define

f(m) :=f+ m R/m= C.Proposition 1.8. LetR be a reduced and finitely generatedC-algebra. Then

the set of maximal idealsspec Rtogether with the algebraR considered as functionsonspec R is an affine variety.

Proof. We have already seen earlier that every such algebra R is isomorphicto the coordinate ring of a closed subset X

Cn. The claim then follows by

using the bijection X

spec O(X), x mx = kerevx, and remarking that forf O(X) and x Xwe have f(x) = evx(f) =f+mx, by definition.

Exercise 1.18. Denote byCn the n-tuples of complex numbers up to sign, i.e.,Cn :=Cn/ where (a1, . . . , an) (b1, . . . , bn) if ai =bi for all i. Then every polynomialin C[x21, x

22, . . . , x

2n] is a well-defined function on Cn. Show that Cn together with these

functions is an affine variety.(Hint: Consider the map : Cn Cn, (a1, . . . , an) (a21, . . . , a2n) and proceed like inExample1.19.)

Although every affine variety is isomorphic to a closed subset ofCn for asuitable n, there are many advantages to look at these objects and not only atclosed subsets. In fact, an affine variety can be identified with many different

closed subsets of this form (see the following Exercise1.19), and depending on thequestions we are asking one of them might be more useful than another. We willeven see in the following section that certain open subsets are affine varieties in anatural way.

On the other hand, whenever we want to prove some statements for an affinevarietyXwe can always assume that X= V(a) Cn so that the regular functionson Xappear as restrictions of polynomial functions. This will be helpful in thefuture and quite often simplify the arguments.

Exercise 1.19. Let Xbe an affine variety. Show that every choice of a generatingsystem f1, f2, . . . , f n O(X) of the algebraO(X) consisting of n elements defines anidentification ofXwith a closed subsetV(a) Cn.(Hint: Consider the map X

Cn given by x

(f1(x), f2(x), . . . , f n(x)).)

1.5. Special open sets. Let X be an affine variety and f O(X). Definethe open set Xf X by

Xf :=X\ VX(f) = {x X| f(x) = 0}.An open set of this form is called a special open set.

Lemma 1.9. The special open sets of an affine varietyX form a basis of thetopology.
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Proof. If U X is open and x U, then X\ U is closed and does notcontain x. Thus, there is a regular function f O(X) vanishing on X\ U suchthatf(x) = 0. This implies x Xf U.

Given a special open set Xf Xwe see that f(x)= 0 for all xXfand sothe function 1f is well-defined on Xf.

Proposition1.10. Denote byO(Xf)the algebra of functions onXfgeneratedby 1f and the restrictionsh|Xf of regular functionsh onX:

O(Xf) := C[ 1f

, {h|Xf| h O(X)}] = O(X)|Xf[1

f].

Then(Xf, O(Xf)) is an affine variety andO(Xf) O(X)[t]/(f t 1).Proof. LetX= V(a) Cn and define

X := V(a, f xn+1 1) Cn+1.It is easy to see that the projection pr: Cn+1 Cn onto the first n coordinatesinduces a bijective map X

Xfwhose inverse : Xf Xis given by(x1, . . . , xn) = (x1, . . . , xn, f(x1, . . . , xn)

1).

The following commutative diagram now shows that (O(X)) is generated by(xn+1) =

1f and the restrictions h|Xf (h O(X)).

X X closed

Cn+1

pr

Xf open X closed CnThis proves the first claim. For the second, we have to show that the canonicalhomomorphismO(X)[t]/(f t 1) O(X) is an isomorphism. In other words,every functionh =

mi=0 hit

i O(X)[t] which vanishes on Xis divisible byft1.Sincef|Xis invertible we first obtain

i hif

mi = 0 which implies

h= h tmm

i=0

hifmi =

m1i=0

hiti(1 fmitmi),

and the claim follows.

Example 1.20. The group GLn(C) is a special open set of Mn(C), henceGLn(C) is an affine variety with coordinate ringO(GLn(C)) =C[{xij| 1 i, jn}, 1det ]. In particular, C := GL1 = C \ {0} is an affine variety with coordinateringC[x, x1].

Exercise 1.20. Let R be an arbitrary C-algebra. For any element s R defineRs := R[x]/(s x 1).

(1) Describe the kernel of the canonical homomorphism : R Rs.(2) Prove the universal property:For any homomorphism : R A such that(s)

is invertible inA there is a unique homomorphism : Rs Asuch that= .

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(3) What happens ifs is a zero divisor and what ifs is invertible?

1.6. Decomposition into irreducible components. We start with a purelytopological notion.

Definition 1.21. A topological space T is called irreducible if it cannot bedecomposed in the form T = A B where A, B T are strict closed subsets.Equivalently, every non-empty open subset is dense.

Lemma1.11. LetX Cn be a closed subset. Then the following are equivalent:(i) Xis irreducible.

(ii) I(X) is a prime ideal.(iii)O(X) is a domain, i.e., has no zero-divisor.

Proof. (i)

(ii): If I(X) is not prime we can find two polynomials f, hC[x1, . . . , xn] \ I(X) such that f h I(X). This implies that X V(f h) =

V(f) V(h), but X is neither contained inV(f) nor inV(h). Thus X= (V(g) X) (V(h) X) is a decomposition into strict closed subsets, contradicting theassumption.

(ii)(iii): This is clear sinceO(X) = C[x1, . . . , xn]/I(X).(iii)(i): IfX=A B is a decomposition into strict closed subsets there are

non-zero functions f, h O(X) such that f|A = 0 and h|B = 0. But then f hvanishes on all ofXand so f h= 0. This contradicts the assumption thatO(X)does not contain zero-divisor.

Example 1.22. Let f C[x1, . . . , xn]. Then the hypersurfaceV(f) is irre-ducible if and only iffis a power of an irreducible polynomial. This follows from

Example1.14and the fact that the ideal (f) is prime if and only iff is irreducible.More generally, iff=pr11 p

r22 prss is the primary decomposition, then

V(f) = V(p1) V(p2) V (pn)where each V(pi) is irreducible, and this decomposition isirredundant, i.e., no termcan be dropped.

Theorem 1.12. Every affine varietyX is a finite union of irreducible closedsubsetsXi:

(4) X=X1 X2 Xs.If this decomposition is irredundant, then the Xis are the maximal irreducible

subsets ofXand are therefore uniquely determined.The unique irredundant decomposition of an affine variety Xin the form (4) is

calledirreducible decompositionand theXis are called theirreducible components.For the proof of the theorem above we first recall that a C-algebraR is called

Noetherian if the following equivalent conditions hold:

(i) Every ideal ofR is finitely generated.(ii) Every strictly ascending chain of ideals ofR terminates.

(iii) Every non-empty set of ideals ofR contains maximal elements.

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(The easy proofs are left to the reader; for the equivalence of (ii) and (iii) one hasto use Zorns Lemma.)

The famous Basissatz of Hilbert

implies that every finitely generated C-algebra is Noetherian (see [Art91, Chap. 12, Theorem 5.18]). In particular, thisholds for the coordinate ringO(X) of any affine variety X. Using the inclusionreversing bijection between closed subsets of Xand perfect ideals ofO(X) (seeCorollary1.7and Exercise1.17) we get the following result.

Proposition1.13. LetXbe an affine variety. Then

(1) Every closed subsetA Xis of the formVX(f1, f2, . . . , f r).(2) Every strictly descending chain of closed subsets ofX terminates.(3) Every non-empty set of closed subsets ofX contains minimal elements.

Remark 1.23. It is easy to see that for an arbitrary topological space T theproperties (2) and (3) from the previous proposition are equivalent. If they hold

thenT is called Noetherian.

Proof of Theorem 1.12. We first show that such a decomposition exists.Consider the following set

M := {A X| A closed and not a finite union of irreducible closed subsets}.IfM = then it contains a minimal element A0. SinceA0is not irreducible, we canfind strict closed subset B , B A0 such that A0 =B B. But then B, B / Mand so both are finite unions of irreducible closed subsets. Hence A0 is a finiteunion of irreducible closed subsets, too, contradicting the assumption.

To show the uniqueness let X=X1 X2 Xs where allXi are irreducibleclosed subsets and assume that the decomposition is irredundant. Then, clearly,

every Xi is maximal. Let Y X be a maximal irreducible closed subset. ThenY = (Y X1) (Y X2) (Y Xs) and soY =Y Xj for some j . It followsthatY Xj and so Y =Xj because of maximality.

Remark1.24. The algebraic counterpart to the decomposition into irreduciblecomponents is the following statement about radical ideals in finitely generatedalgebrasR:Every radical ideala R is a finite intersection of prime ideals:

a= p1 p2 ps.If this intersection is irredundant then the pis are the minimal prime ideals con-taininga. (The easy proof is left to the reader.)

Example1.25. Consider the two hypersurfacesH1:= V(xyz),H2:= V(xzy

2

) inC3

and their intersection X :=H1 H2. ThenX= V(y, z) C where C := {(t, t2, t3) | t C},

and this is the irreducible decomposition.In fact, it is obvious that thex-axis V(y, z) is closed and irreducible and belongs

toX, and the same holds for the curveC(see Exercise1.12). If (a,b,c) X\V(y, z)then either b or c is= 0. But then b= 0 because ab= c. Hence a= cb1 and sob2 =ac = c2b1 which implies thatc2 =b3. Thusb = (cb1)2 and c = (cb1)3, i.e.(a,b,c) C.
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Another way to see this is by looking at the coordinate ring:

C[x,y ,z]/(xy

z,xz

y2)

C[x, y]/(x2y

y2).

Now (x2y y2) = y(x y2) = (y) (x y2) and the ideals (y) and (x y2)are obviously prime with residue class ring isomorphic to a polynomial ring inone variable. This shows again that Xhas two irreducible components, both withcoordinate ring isomorphic toC[t].

Exercise 1.21. The closed subvariety X :=V(x2 yz,xz x) C3 has threeirreducible components. Describe the corresponding prime ideals inC[x,y,z].

Example 1.26. The group O2 :={A M2| AAt = E} has two irreduciblecomponents, namely SO2 := O2 SL2 and

0 11 0

SO2, and the two components

are disjoint.

In fact, the condition AAt = E for A = a bc d implies that ab = dc.Since det

a bb a

= a2 +b2 we see that SO2 ={

a bb a

| a2 +b2 = 1} is

irreducible as well as

0 11 0

SO2= {

a bb a

| a2 + b2 = 1} and the claim follows

immediately.

Exercise 1.22. LetG GLn be a closed subgroup.(1) The irreducible components ofG are disjoint.(2) The irreducible component ofG containing Eis a normal subgroup G0.(3) The irreducible components are the cosets ofG0.

Exercise 1.23. LetX= Xibe the decomposition into irreducible components andletUi Xi be open subsets. Then U :=

Ui is open inX. It is dense in Xif and only if

all Ui are non-empty.

1.7. Rational functions and local rings. IfX is an irreducible affine va-riety thenO(X) is a domain by Lemma1.11. Therefore, we can form the field offractions ofO(X) which is called the field of rational functions on Xand will bedenoted by C(X). Clearly, ifX = Cn then C(X) = C(x1, x2, . . . , xn), the ratio-nal function field. An irreducible affine variety X is called rational if its field ofrational functionsC(X) is isomorphic to a rational function field.

A rational function f C(X) can be regarded as a function defined almosteverywhere on X. In fact, we say that f is defined inx

X if there are p, q

O(X) such that f= pq and q(x) = 0.Exercise 1.24. Iff C(C2) = C(x, y) is defined in C2 \ {(0, 0)}then fis regular.Exercise 1.25. Let fbe a rational function on the irreducible affine variety X and

denote by Def(f) Xthe set of points where f is defined.(1) The set Def(f) is open in X.(2) If Def(f) = Xthenfis regular onX. (Hint: Look at the ideal of denominators

a:= {p O(X) | p f O(X)}.)
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Example 1.27. Consider the irreducible plane curve C :=V(y2 x3) C2and put x:= x|C and y :=y |C. Then the rational function f := yx C(C) is notdefined in (0, 0). The interesting point here is that fhas a continuous extensionto all ofCwith value 0 at (0, 0), even in theC-topology.

Proof. There is an isomorphismO(C) C[t2, t3] (see Example1.16) whichmaps x to t2 and y to t3, and so f = yx is mapped to t. Since t / C[t2, t3] thefirst claim follows from Exercise1.25(2) above. The second part is easy becausethe mapC C: t (t2, t3) is a homeomorphism even in theC-topology.

Assume that Xis irreducible and let x X. DefineOX,x := {f C(X) | f is defined in x}.

It is easy to see thatOX,x is the localization ofO(X) at the maximal ideal mx.(For the definition of the localization of a ring at a prime ideal and, more generally,

for the construction of rings of fractions we refer to [Eis95, I.2.1].) This examplemotivates the following definition.

Definition 1.28. Let X be an affine variety and x Xan arbitrary point.Then the localizationO(X)mx of the coordinate ring O(X) at the maximal ideal inxis called thelocal ring ofXatx. It will be denoted by OX,x, its unique maximalideal by mX,x. ,

We will see later that the local ring ofX at x completely determines X in aneighborhood ofx (see Proposition2.4(3)).

Exercise 1.26. IfXis irreducible, thenO(X) = xX OX,x.Exercise 1.27. Let Xbe an affine variety, x

Xa point and X

X the union

of irreducible components of Xpassing through x. Then the restriction map induces anatural isomorphismOX,x OX,x.

Exercise 1.28. LetR be an algebra and : R RSthe canonical map r r1 whereRSis the localization at a multiplicatively closed subset 1 S R (0 / S).

(1) Ifa Ris an ideal and aS:=RS(a) RS then1((a)) = 1(aS) = {b R | sb a for somes S}.

Moreover, (R/a)S RS/aSwhere Sis the image ofS in R/a.

(Hint: For the second assertion use the universal property of the localization.)(2) Ifm R is a maximal ideal and S := R \ m, then mS is the maximal ideal of

RSand the natural maps R/mk RS/mkSare isomorphisms for all k 1.

(Hint: The image S in R/mk consists of invertible elements.)

Exercise 1.29. Let p < q be positive integers with no common divisor and defineCp,q :={(tp, tq)| t C} C2. Then Cp,q is a closed irreducible plane curve which isrational, i.e.C(Cp,q) C(t). Moreover, O(Cp,q) is a polynomial ring if and only ifp = 1.

Exercise 1.30. Consider the elliptic curveE:= V(y2 x(x2 1)) C2. Show thatE is not rational, i.e. that C(E) is not isomorphic to C(t). (Hint: IfC(E) = C(t) thenthere are rational functions f(t), h(t) which satisfy the equation f(t)2 =h(t)(h(t)2 1).)
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2. MORPHISMS 115

2. Morphisms

2.1. Morphisms and comorphisms. In the previous sections we have de-

fined and discussed the main objects of algebraic geometry, the affine varieties. Nowwe have to introduce the regular maps between affine varieties which should becompatible with the concept of regular functions.

Definition 2.1. Let X, Y be affine varieties. A map : X Y is calledregular ora morphismif the pull-back of a regular function on Y is regular onX:

f O(X) for all f O(Y).Thus we obtain a homomorphism :O(Y) O(X) ofC-algebras given by(f) :=f , which is called comorphismof.

A morphism is called an isomorphism if is bijective and the inverse map1 is also a morphism. If, in addition,Y =Xthen us called an automorphism.

Exercise 2.1. Letg GLnbe an invertible matrix. Then left multiplication A gA,right multiplication A Ag and conjugation A gAg1 are automorphisms of Mn.

Example 2.2. A map = (f1, f2, . . . , f m) :Cn Cm is regular if and only ifthe componentsfiare polynomials in C[x1, . . . , xn]. This is clear, since(yj) =fjwherey1, y2, . . . , ym are the coordinate functions on Cm.

More generally, let Xbe an affine variety and a = (f1, . . . , f m) : X Cm amap.Thenis a morphism if and only if the componentsfj are regular functionsonX. (This is clear since fj =

(yj).)

If a morphism = (f1, f2, . . . , f m): Cn Cm maps a closed subset XCn into a closed subset Y

Cm then the induced map : X

Y is clearly

a morphism, just by definition. This holds in a slightly more general setting, asclaimed in the next exercise.

Exercise 2.2. Let : X Y be a morphism. If X X and Y Y are closedsubvarieties such that(X) Y then the induced map : X Y,x (x), is againa morphism. The same holds ifX andY are special open sets.

These examples have the following converse which will be useful in many ap-plications.

Lemma2.1. LetX Cn andY Cm be closed subvarieties and let : X Ybe a morphism. Then there are polynomials f1, . . . , f m C[x1, . . . , xn] such thatthe following diagram commutes:

Cn :=(f1,...,fm) Cm

X Y

Proof. Lety1, . . . , ym denote the coordinate functions onCm. Put yj :=yj |Yand define fj :=

(yj) O(X), j = 1, . . . , m. SinceO(X) =C[x1, . . . , xn]/I(X)we can find representatives fj C[x1, . . . , xn], i.e. fj = fj +I(X). We claim

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that the morphism := (f1, . . . , f m) :Cn Cm satisfies the requirements of thelemma. In fact, let a X Cn and set (a) =:b = (b1, . . . , bm). Then

bj =yj(b) = yj (b) = yj ((a)) =(yj)(a) = fj(a) =fj(a),and so(a) = (a).

Example 2.3. The morphism t (t2, t3) from C C2 induces a bijectivemorphismC C := V(y2 x3) which is not an isomorphism (see Example1.16).

Similarly there is a morphism :C D:= V(y2 x2 x3) given byt (t2 1, t(t21)). This time is surjective, but not injective since(1) =(1) = (0, 0).

Exercise 2.3. (1) Every morphismC C is constant.(2) Describe all morphismsC C.(3) Every non-constant morphismC C is surjective.(4) An injective morphismC C is an isomorphism, and the same holds for mor-

phismsC

C.

Exercise 2.4. The graph of a morphism. Let :Cn Cm be a morphism anddefine

:= {(a, (a)) Cn+m}.which is called the graph of the morphism . Show that is closed in Cn+m, thatthe projection prCn :C

n+m Cn induces an isomorphism p : Cn and that =prCm p1.

Proposition 2.2. Let X, Y be affine varieties. The map induces abijection

Mor(X, Y) AlgC(O(Y), O(X)).

between the morphisms fromX to Yand the algebra homomorphism fromO(Y)to

O(X).

Remark 2.4. The mathematical term used in the situation above is that of acontravariant functor from the category of affine varieties and morphisms to thecategory of finitely generated reduced C-algebras and homomorphism, given byX O(X) and . In particular, we have (IdX) = IdO(X) and ( ) = whenever the expressions make sense. The proposition above then saysthat this functor is an equivalence, the inverse functor being Rspec R definedin Proposition1.8. It will be helpful to keep this functorial point of view in mindalthough it will not play an important role in the following.

Proof. (a) If1= 2 then, for all f O(Y) and all x X, we get

f(1(x)) =1(f)(x) =

2(f)(x) =f(2(x)).

Hence,1(x) =2(x) since the regular functions separate the points (Remark1.12).(b) Let :O(Y) O(X) be an algebra homomorphism. We want to con-

struct a morphism : X Y such that = . For this we can assume thatY Cm is a closed subvariety. Let yj := yj |Y be the restrictions of the coor-dinate functions on Cm and define fj := (yj) O(X). Then we get a mor-phism := (f1, . . . , f m) : X Cm such that (yj) = fj (see Example 2.2). Ifh= h(y1, . . . , ym) I(Y) then

h(f1, . . . , f m) =h((y1), . . . , (ym)) =(h(y1, . . . ,ym)) = 0
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2. MORPHISMS 117

because h(y1, . . . ,ym) = h|Y = 0 by assumption. Therefore h((a)) = 0 for alla Xand allh I(Y) and so (X) Y. This shows that induces a morphism : X Y such that

(yj ) =

(yj ) =fj =(yj), and so

=.

Example 2.5. LetXbe an affine variety, Va finite dimensional vector spaceand : X Va morphism. The linear functions onVform a subspace V O(V)which generatesO(V). Therefore, the induced linear map |V : V O(X)completely determines , and we get a bijection

Mor(X, V) Hom(V, O(X)) |V .

The second assertion follows from Proposition2.2and the well-known Substi-tution Principle for polynomials rings (see [Art91,Chap. 10, Proposition 3.4]).

Exercise 2.5. Show that for an affine variety Xthe morphismsX C correspondbijectively to the invertible functions onX.

Exercise 2.6. Let X, Y be affine varieties and : X Y, : Y X morphismssuch that = IdX . Then(X) Y is closed and : X (X) is an isomorphism.

2.2. Images, preimagees and fibers. It is easy to see that morphisms arecontinuous. In fact, the Zariski topology is the finest topology such that regularfunctions are continuous, and since morphisms are defined by the condition thatthe pull-back of a regular function is again regular, it immediately follows thatmorphisms are continuous. We will get this result again from the next propositionwhere we describe images and preimages of closed subsets under morphisms.

Proposition2.3. Let : X Ybe a morphism of affine varieties.(1) If B :=

VY(S)

Y is the closed subset defined by S

O(Y) then

1(B) = VX((S)). In particular, is continuous.(2) LetA :=V(a) X be the closed subset defined by the ideal a O(X).

Then the closure of the image(A) is defined by1(a) O(Y):(A) = VY(1(a)).

Proof. For x Xwe havex 1(B) (x) B f((x)) = 0 for allf S,

and this is equivalent to (f)(x) = 0 for all f S, hence to x VX((S)),proving the first claim.

For the second claim, let f O(Y). Then

f|(A)= 0 f|(A)= 0

(f)|A = 0

(f) I(A) =a

The latter is equivalent to the condition that a power of f belongs to 1(a).

Thus the zero set of1(a) equals the closed set (A).

Exercise 2.7. If1, 2 : X Yare two morphisms, then the kernel of coincidenceker(1, 2) := {x X| 1(x) = 2(x)} X

is closed in X

Exercise 2.8. Let : X Ybe a morphism of affine varieties.
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(1) IfX is irreducible, then (X) is irreducible.(2) Every irreducible component ofXis mapped into an irreducible component of

Y.

Exercise 2.9. Let : Cn Cm be a morphism, = (f1, f2, . . . , f m) where fiC[x1, x2, . . . , xn], and letX :=(Cn) be the closure of the image of. Then

I(X) = (y1 f1, y2 f2, . . . , ym fm) C[y1, y2, . . . , ym]where both sides are considered as subsets ofC[x1, . . . , xn, y1, . . . , ym]. SoI(X) is obtainedfrom the ideal (y1 f1, . . . , ym fm) byeliminating the variablesx1, . . . , xn.

Exercise 2.10. Let : X Xbe an automorphism andY Xa closed subset such

that(Y) Y. Then (Y) =Y and|Y: Y Yis an automorphism, too.A special case of preimagees are the fibers of a morphism : X Y. Let

y Y. Then

1

(y) := {x X| (x) =y}is called the fiber of y Y. By the proposition above, the fiber ofy is a closedsubvariety ofXdefined by(my):

1(y) = VX((my)).Of course, the fiber of a point y Ycan be empty. In algebraic terms this meansthat(my) generates the unit ideal (1) = O(X).

Exercise 2.11. Let : X Ybe a morphism of affine varieties. IfU Yis a specialopen set then so is 1(U).

Exercise 2.12. Describe the fibers of the morphism : M2 M2, A A2. (Hint:Use the fact that (gAg1) = g(A)g1 forg GL2.)

Definition2.6. Let : X Ybe a morphism of affine varieties and considerthe fiberF :=1(y) of a point y (X) Y. Then the fiber F is calledreducedif(my ) generates a perfect ideal inO(X), i.e. if

O(X) (my) = O(X) (my).The fiberF is called reduced in the pointx Fif this holds in the local ring OX,x,i.e.

OX,x (my) = OX,x (my).Example 2.7. Look again at the morphism :C C :=V(y2 x3)C2,

t (t2, t3). Then is the injectionO(C) C[t2, t3] C[t] and soC[t] (m(0,0)) = (t2, t3) (t2, t3) = (t).

Thus the zero fiber 1(0) is not reduced. On the other hand, all other fibersare reduced since induces an isomorphism ofC with the special open set C\{(0, 0)}(=Cx= Cy), where the inverse map is given by (a, b) ba .

Exercise 2.13. Show that all fibers of the morphism :C D:= V(y2 x2 x3) C2, t (t2 1, t(t2 1)), are reduced and that induces an isomorphismC \ {1, 1} D \ {(0, 0)}.

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2. MORPHISMS 119

Exercise 2.14. Consider the following morphism : SL2 C3,(

a bc d

) := (ab,ad,cd).

(1) The image of is a closed hypersurface H C3

.(2) The fibers of are the left cosets of the subgroup T := {

t

t1

| t C}.

(3) All fibers are reduced.

(Hint: Show that the left multiplication with some g SL2 induces an automorphism gofH and isomorphisms 1(y)

1(g(y)) for all yH. This implies that it sufficesto study just one fiber, e.g. 1((E)).)

Exercise 2.15. Consider the morphism :C2 C2 given by(x, y) := (x,xy).(1) (C2) = C2 \ {(0, y) | y= 0}which is not locally closed.(2) What happens with the lines parallel to thex-axis or parallel to they -axis?(3) 1(0) = y -axis. Is this fiber reduced?

(4) induces an isomorphismC2

\ y-axis

C2

\ y-axis

2.3. Dominant morphisms. Let : X Ybe a morphism of affine vari-eties,xa point ofXand y := (x) its image in Y. Then(my) mx, and soinduces a local homomorphism

x :OY,y OX,x.(A homomorphism between local rings is called local if it maps the maximal idealinto the maximal ideal.)

The next proposition tells us that, in a neighborhood of a point x X, amorphism is uniquely determined by the local homomorphism

x.

Proposition2.4. (1) Let , : X Y be two morphisms and x Xa point such that(x) =(x) andx =

x. Then and coincide on

every irreducible component ofXwhich containsx.(2) If x X, y Y and if :OY,y OX,x is a local homomorphism,

then there is a special open sets X X containing x and a morphism : X Y such thatx = .

(3) Ifx X, y Y and :OY,y OX,x an isomorphism, then there existspecial open sets X X and Y Y containing x and y, respectively,and an isomorphism : X

Y such thatx= .Proof.

(1) Let R be a finitely generated reduced C-algebra and m R amaximal ideal. The canonical map : R Rmis injective if and only ifm containsall minimal prime ideals of R. (In fact, ker ={r R| sr = 0 for some sR \ m}.)

Denote by X X the union of irreducible components passing through xand by Y Y the union of irreducible components passing through (x). Then(X) Y, because the image of an irreducible component ofXis contained in anirreducible component ofY(see Exercise2.8). Thus we obtain a morphism : XY with the following commutative diagram ofC-algebras and homomorphisms
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which shows that is completely determined by x:

O(Y)

O(Y)

OY ,(x)

=OY,(x) x

O(X) O(X) OX,x = OX,x(2) We can assume that all irreducible components ofXpass through x and allirreducible components ofYpass through y . Then

O(Y) OY,y OX,x O(X).Leth1, . . . , hm O(Y) be a set of generators and put gj :=(hj ). Then we can findan element t O(X) \ mx such that gj O(X)t for all j, i.e. (O(Y)) O(X)t.Hence there is a morphism : Xt Y such that =|O(X)t , and sox = .

(3) By (2) we can assume that there is a morphism : X Ysuch that

x=,i.e.(O(Y)) O(X). Letf1, . . . , f n O(X) be generators. Then fi= (hi)(s) wherehi O(Y) ands O(Y)\my . This implies that(O(Y)s) = O(X)twhere t= (s).Thus induces an isomorphismO(Y)s O(X)t, and the claim follows.

Definition2.8. LetX, Ybe irreducible affine varieties. A morphism :XY is called dominant if the image is dense in Y , i.e. (X) =Y. This is equivalentto the condition that :O(Y) O(X) is injective (see Proposition2.3(2)).

It follows that every dominant morphism :X Y induces a finitely gener-ated field extension :C(Y) C(X). If this is a finite field extension of degreed:= [C(X) : C(Y)] we will say that is amorphism of finite degreed. Ifd = 1, i.e.if induces an isomorphism C(Y)

C(X) thenis called abirational morphism.

Exercise 2.16. Let :C Cbe a non-constant morphism. Then has finite degreed, and there is a non-empty open set U C such that #1(x) = d for all x U.

There is a similar result as the second part of Proposition 2.4saying that affinevarieties with isomorphic function fields are locally isomorphic. The proof is similaras the proof above and will be left to the reader.

Proposition2.5. LetXandYbe irreducible affine varieties and let :C(Y) C(X) be an isomorphism. Then there exist special open setsX X andY Yand an isomorphism : X

Y such that= .

2.4. Products. Iffis a function onXandh a function onYthen we denotebyf h theC-valued function on the product X Y defined by (f h)(x, y) :=f(x) h(y).

Proposition2.6. The productX Yof two affine varieties together with thealgebra

O(X Y) := C[f h | f O(X), h O(Y)]ofC-valued functions is an affine variety. Moreover, the canonical homomorphismO(X) O(Y) O(X Y), f h f h, is an isomorphism.

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2. MORPHISMS 121

Proof. LetX Cn andY Cm be closed subvarieties. ThenXY Cn+mis closed, namely equal to the zero set V(I(X) I(Y)). So it remains to show thatO(X Y) = C[x1, . . . , xn, y1, . . . , ym]/I(X Y) is generated by the products f hand that f h O(X Y) for f O(X) and h O(Y). But this is clear sincexi = xi|XY =xi|X 1 and yj =yj |XY = 1 yj |Y, and f|X h|Y = (f h)|XY forf C[x1, . . . , xn] and h C[y1, . . . , ym].

For the last claim, we only have to show that the mapO(X) O(Y) O(X Y), f h f h, is injective. For this, let (fi| i I) be a basis ofO(Y).Then every elements O(X)O(Y) can b e uniquely written ass = finite sifi.Ifs is in the kernel of the map, then

si(x)fi(y) = 0 for all (x, y) X Y and

so, for every fixed xX, si(x)fi is the zero function on Y. This implies thatsi(x) = 0 for all x Xand so si= 0 for all i. Thuss = 0 proving the claim.

Example 2.9. (1) The two projections prX: X Y X, (x, y) x,and pr

Y

: X

Y

Y, (x, y)

y, are morphisms with comorphismsprX(f) =f 1 and prY(h) = 1 h.

(2) If : X X and :Y Y are morphisms, then so is : X Y X Y, (x, y) ((x), (y)).

(3) Diagonal: : X XX,x (x, x) is a morphism, (X) XXis aclosed subset defined by the set {f1 1 f| f O(X)}, andX (X)is an isomorphism whose inverse is induced by the projection prX.

(4) Graph: Let : X Ybe a morphism. Then() := {(x, (x)) | x X} X Y

is a closed subset. Moreover, the projection prX induces an isomorphism

p : ()

X and = prY

p1.

(5) Matrix multiplication: The composition of linear maps

: Hom(U, V) Hom(V, W) Hom(U, W), (A, B) B Ais a morphism. Choosing coordinates we find (zij ) =

kyikxkj .

Exercise 2.17. Show that the ideal of the diagonal (X) X Xis generated bythe function f 1 1 f, f O(X) (see Example2.9(3)).

Lemma 2.7. The projectionprX: X YX is an open morphism, i.e. theimage of an open set underprX is open.

Proof. It suffices to show that the image of a special open set U := (XY)g isopen. Writingg = fi

hiwith linearly independenthione gets prX(U) = i Xfiand the claim follows. Proposition2.8. IfX, Y are irreducible affine varieties thenX Y is irre-

ducible.

Proof. Assume that X Y =A B with closed subsets A, B. DefineXA := {x X| {x} Y A} and XB := {x X| {x} Y B}

SinceYis irreducible we see thatX=XAXB. Now we claim thatXAand XBareboth closed inXand so one of them equals X, sayXA = X. But thenA = X Y

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and we are done. To prove the claim we remark that X\ XA = prX(X Y\ A)which is open by Lemma2.7 above.

Corollary 2.9. IfX=i Xi andY =jYj are the irreducible decomposi-tions ofX andY, thenX Y =i,jXi Yj is the irreducible decomposition ofthe product.

2.5. Fiber products. Let X,Y ,S be affine varieties and let : X S, : Y S two morphisms. Then

XSY := {(x, y) X Y| (x) =(y)} X Yis a closed subset. In fact, it is the inverse image ( )1(S) of the diagonalS S Swhich is a closed subset (Example2.9(3)). We have the commutativediagram

XZY q

Yp

X

Swhere the morphisms p and q are induced by the projections X Y X andX Y Y . This affine variety is called the fiber product ofX, Y over S. It hasthe following universal property which defines it up to unique isomorphisms.

Proposition2.10. If : Z Xand: Z Yare two morphisms such that = , then there is a unique morphism (, ) : Z XSY such thatp (, ) = andq (, ) =:

Z (,)

XSYp

q Y

X S

Proof. Clearly, the morphism z ((z), (z)) X Y has its image inXZYand satisfies the conditions. It is also obvious that it is unique.

Exercise 2.18. Show thatO(XS Y) (O(X)O(S)O(Y))red where Rred :=R/(0).

3. Dimension

3.1. Definitions. Ifk is a field andAak-algebra then a seta1, a2, . . . , an Aof elements fromA are calledalgebraically independent overk if they do not satisfya non-trivial polynomial equation F(a1, a2, . . . , an) = 0 where F k[x1, . . . , xn].Equivalently, the canonical homomorphism ofk-algebrask[x1, . . . , xn] Adefinedbyxi ai is injective.

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3. DIMENSION 123

In order to define the dimension of a variety we will need the concept oftran-scendence degreetdegkKof a field extension K/k. It is defined to be the maximal

number of algebraically independent elements inK. We refer to[Art91,Chap. 13,Sect. 8] for the basic properties of transcendental extensions.

Definition 3.1. LetXbe an irreducible affine variety and C(X) its field ofrational functions. Then the dimension ofXis defined by

dim X:= tdegCC(X).

IfXis reducible and X=

Xi the irreducible decomposition (see1.6) then

dim X:= maxi

dim Xi.

Finally, we define the local dimension ofXin a point x X=

Xi to be

dimx X:= maxXix dim Xi.

Exercise 3.1. Let Xbe an affine variety.

(1) dim X is the maximal number of algebraically independent elements inO(X).(2) Assume that O(X) is generated by relements. Then dim X r, and if dim X=r

thenX Cr.Exercise 3.2. The functionx dimx Xis upper semi-continuous on X. (This means

that for all R the set{x X| dimx X < } is open in X.)Examples 3.2. (1) We have dimCn =n. More generally, ifV is a com-

plex vector space of dimension n then dim X=n.(2) If U

X is a special open subset which is dense in X, then dim U =

dim X.(3) For affine varietiesX, Y we have dim X Y= dim X+ dim Y.

Lemma 3.1. Let f C[x1, . . . , xn] be a non-constant polynomial and X :=V(f) Cn its zero set. Thendim X=n 1.

Proof. We can assume that fis irreducible and that the variable xn occursin f. Denote by xi O(X) = C[x1, . . . , xn]/(f) the restrictions of the coordi-nate functions xi. ThenC(X) =C(x1,x2, . . . ,xn). Since f(x1,x2, . . . ,xn) = 0 wesee that xn C(X) is algebraic over the subfield C(x1,x2, . . . ,xn1). Therefore,tdegC(X) = tdegC(x1,x2, . . . ,xn1) n1. On the other hand, the composition

C[x1, . . . , xn1]

C[x1, . . . , xn] res

O(X)is injective, since the kernel is the intersection (f) C[x1, . . . , xn1] which is zero.Thus, tdegC(X) n 1, and the claim follows.

The first part of the proof above, namely that dim V(f)< n= dimCn has thefollowing generalization.

Lemma 3.2. If X is irreducible and Y X a proper closed subset then wehavedim Y

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Proof. We can assume that Y is irreducible. Ifh1, . . . , hm O(Y) are alge-braically independent where m = dim Y, and hi = hi|Y for h1, . . . , hm O(X)then h1, . . . , hm are algebraically independent, too, and so dim X dim Y. Ifdim Y = dim X then every f O(X) is algebraic over C(h1, . . . , hm). Choosef O(X) in the kernel of the restriction map, i.e. f|Y = 0. Then f satisfies anequation of the form

fk +p1fk1 + +pk1f+pk = 0

where pj C(h1, . . . , hm) and k is minimal. Multiplying this equation with asuitableq C[h1, . . . , hm] we can assume thatpj C[h1, . . . , hm]. But this impliesthatpk|Y= 0. Thus pk = 0 and we end up with a contradiction.

Example 3.3. We have dim X = 0 if and only if X is finite, and this isequivalent to dimC

O(X) 0, and so b= f(a1, . . . , as) =

jhj (a1, . . . , as)fj(a1, . . . , as) ab.

The next proposition is a reformulation of our main Theorem 4.7.For later usewe will prove it in this slightly more general form.
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4. TANGENT SPACES AND DIFFERENTIALS 141

Proposition4.11. LetR be a finitely generatedC-algebra and letm Rbe amaximal ideal. Thendimgrm R= dim Rm. Moreover, dimCm/m

2 = dim Rm if and

only ifgrm R is a polynomial ring. If this holds, thenRm is a domain.Proof. Inverting an element from R\mdoes not change grm R(Lemma4.9(2)).

Therefore we can assume that all minimal primes ofRare contained inm. In partic-ular, we have dim Rm = dim R= maxidim R/pi where p1, . . . ,pr are the minimalprime ideals. Moreover, every element from R \m is a non-zero divisor.

Now consider the C-algebra R = R[t,mt1] R[t, t1] introduced above. Itfollows from Lemma4.8that R has the following two properties:

(i) R/Rt grm R, by (2).

(ii) dimR/Rt= dim R, by (5) and (6).

Hence, dim grm R= dim Rm, proving the first claim.

Assume now that dimCm/m2 = dim Rm =: n. Then we obtain a surjective

homomorphism :C[y1, . . . , yn] grm R

by sendingy1, . . . , yn to aC-basis ofm/m2. But every proper residue class ring ofC[y1, . . . , yn] has dimension < n, and so the homomorphism is an isomorphism.

On the other hand, if grm R is a polynomial ring then dim Rm = dimgrm R=dimCm/m

2. Moreover,

j>0mj = (0) by Lemma4.10,because every element from

R \ m is a non-zero divisor, and so R is a domain by Lemma4.9(1). Corollary 4.12. IfX is an affine variety, thenXsing Xis a closed subset

whose complement is dense inX.

Proof. LetX= i Xiis the decomposition ofXinto irreducible components.A point x Xi is a singular point ofXif and only if it is either a singular pointofXi or it belongs to two different irreducible components. Thus

Xsing=

i

(Xi)sing

j=k

Xj Xk,

and the claim follows easily.

Let us denote by OX,x the mx-adic completion of the local ringOX,x. It isdefined to be the inverse limit

OX,x := lim

O(X)/mkx.(We refer to[Eis95, I.7.1 and I.7.2] for more details and some basic properties.)

Sincemkx= {0} we have a natural embeddingOX,x OX,x .IfX= Cn andx = 0 then the completion coincides with the algebra of formal

power series in n variables:

OCn,0= C[[x1, . . . , xn]].The next result is an easy consequence of Theorem 4.7 above.

Corollary 4.13. The point x X is nonsingular if and only if OX,x isisomorphic to the algebra of formal power series indimx Xvariables.
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Remark 4.7. A famous result of Auslander-Buchsbaum states that the localringOX,x in a nonsingular point of a variety X is always a unique factorizationdomain. For a proof we refer to [Mat89,20, Theorem 20.3].

4.6. Vector fields and tangent bundle. LetXbe an affine variety. DenotebyT X :=

xXTxXthe disjoint union of the tangent spaces and byp :T X X

the natural projection, TxX x. We call T X the tangent bundle ofX. Wewill see later that T Xhas a natural structure of an affine variety and that p is amorphism.

A section : X T Xofp, i.e. p = IdX, is a collection (x)xXof tangentvectors x TxX. It is usually called a vector fieldand can be considered as anoperator on regular functions f O(X):

(f)(x) :=xf forx X.Definition

4.8.

Analgebraic (or regular) vector field onXis a section: XT Xwith the property that f O(X) for all f O(X). The space of algebraicvector fields is denoted by Vec(X).(In the following, we will mostly talk about vector fields and omit the termalgebraic or regular whenever it is clear from the context.)

Thus a vector fieldcan be considered as a linear map :O(X) O(X), andso Vec(X) is a subvector space of End(O(X)). More generally, the vector fieldsform a module overO(X) where the product f for f O(X) is defined in theobvious way: (f )x :=f(x)x.

Example 4.9. LetX=V be aC-vector space and fix a vector v V . Thenv Vec(V) is defined by x v,x . It follows that

vf := f(x + tv) f(x)t

t=0

O(X)

which implies that this vector field is indeed algebraic. We claim that every alge-braic vector field on V is of this form. In fact, ifV = Cn then

Vec(Cn) =n

i=1

C[x1, . . . , xn]

xi

which means that every algebraic vector field onCn is of the form =

i hi

xiwhere hi C[x1, . . . , xn] =O(Cn). (This follows from the two facts that everyvector fieldon Cn is of this form with arbitrary functionshiand that(xi) =hi.)

Another observation is that for every vector field on Xthe correspondinglinear map:O(X) O(X) is a derivation, i.e. is alinear differential operator:

(f h) =f h + h f for all f , h O(X).Proposition4.14. The map sending a vector field to the corresponding linear

differential operator defines a bijectionVec(X) Der(O(X), O(X)) End(O(X)).

Proof. It remains to show that every derivation :O(X) O(X) is givenby an algebraic vector field. For this, define x := evx . Then the vector field(x)xXis algebraic and the corresponding linear map is .
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The Example4.9above shows that forX=Vwe have a canonical isomorphismT X X V, using the identifications TxX =V {x} V. Then p : T X Xis identified with the projection prX and algebraic vector fields correspond tomorphism: X X V of the form(x) = (x, x).

Proposition4.15. LetX Vbe a closed subset.(1) If Vec(V) then|Xdefines a vector field onX (i.e. x TxX for all

x X) if and only if f = 0 for all f I(X). Moreover, it suffices totest a system of generators of the idealI(X).

(2) There is a canonical bijectionT X {(x, )| TxX V} where the

latter is a closed subset ofXV. ThusT Xhas the structure of an affinevariety. Using coordinates, we get

T X {(x, a1, . . . , an) |

n

i=1ai

f

xi(x) = 0 for allf I(X)} XCn

(3) A vector fieldonXis algebraic if and only if: X T Xis a morphism.Proof. (1) We have x TxX for all xX if and only ifxf= 0 for all x

and all f I(X) which is equivalent to f|X= 0 for all f I(X).(2) We can assume that V = Cn andO(V) = C[x1, . . . , xn]. If I(X) =

(f1, . . . f m) then, by (1),

TX := {(x, x) X V| TxX}

= {(x, a1, . . . , an) |n

i=1

aifjxi

(x) = 0 for j = 1, . . . , m} X Cn

which shows that this is a closed subspace ofX Cn

. Now (2) follows easily.(3) Using the identification ofT X with the closed subvariety TXabove, an

arbitrary section : X T Xhas the form x =

hi(x) xi

with arbitrary func-tions hi on X. The vector field is algebraic if and only ifhi =xi is regular onXwhich is equivalent to the condition that :X T X is a morphism.

Example 4.10. Consider the curve H :=V(xy1) C2. Then I(H) =(xy 1). For a vector field = a(x, y)x+ b(x, y)y onC2 we get

(xy 1) =a(x, y)y+ b(x, y)x.Thus (xy 1)|H= 0 if and only ifay+bx = 0 on H. It follows that xx yydefines a vector field 0 on Hand that Vec(H) =O(C)0. (In fact, setting h :=ay|H= bx|Hwe get a|H=h x|H and b|H= h y|H.)The tangent bundleT H HC2 has the following description (see Proposi-tion4.15(1)):

T H= {(t, t1, , ) | t1 + t = 0} = {(t, t1, t2, | t C, C} HC.Example 4.11. Now consider Neils parabola C :=V(y2 x3) C2 (see

Example1.16). Then a vector field ax + by defines a vector field on Cif and onlyif

3ax2 + 2by= 0 on C.

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To find the solutions we use the isomorphismO(C) C[t2, t3], x t2, y t3(see Example2.7). Thus we have to solve the equation 3at= 2b inC[t2, t3]. This

is easy: Every solution is a linear combination (with coefficients in C[t2

, t3

]) of thetwo solutions (2t2, 3t3) and (2t3, 3t4). This shows that

0:= (2xx+ 3yy)|D and 1:= (2yx+ 3x2y)|Dare vector fields onCand that Vec(C) = O(C)0 +O(C)1. Moreover, x20 = y1.

Our calculation also shows that every vector field on Cvanishes in the singularpoint 0 of the curve. For the tangent bundle we get

T C= {(t2, t3, , ) | 3t4 + 2t3 = 0} C C2which has two irreducible components, namely

T C= {(t2, t3, 2, 3t) | t, C} {(0, 0)} C2

Exercise 4.12. Determine the vector fields on the curve D := V(y2 x2 x3) C2.Do they all vanish in the singular point ofD?

Exercise 4.13. Determine the vector fields on the curves D1 := {(t, t2, t3) C3 | t C}and D2 := {(t3, t4, t5) C3 | t C}.(Hint: For D2 one can use thatO(D2) C[t3, t4, t5] = C

i3 Ct

i.)

Proposition 4.16. The vector fields Vec(X) on X form a Lie algebra withLie bracket

[, ] := .Proof. By Proposition 4.14 it suffices to show that for any two derivations

, ofO(X) the commutator is again a derivation. But this is a generalfact and holds for any associative algebra, see the following Exercise 4.15.

Exercise 4.14. LetA be an arbitrary associativeC-algebra. ThenA is a Lie algebrawith Lie bracket [a, b] := ab ba, i.e., the bracket [ , ] satisfies the Jacobi identity

[a, [b, c]] = [[a, b], c] + [b, [a, c]] for all a,b, c A.Exercise 4.15. Let R be an associative C-algebra. If , : R R are both C-

derivations, then so is the commutator . This means that the derivationsDer(R) form a Lie subalgebra of EndC(R).

Exercise 4.16. LetX Cn be a closed and irreducible. Then dim T X 2dim X. IfXis smooth then T Xis irreducible and smooth of dimension dim T X= 2 dim X.(Hint: IfI(X) = (f1, . . . f m) thenT X Cn Cn is defined by the equations

fj = 0 and

ni=1

yifjxi

(x) = 0 for j = 1, . . . , m .

The Jacobian matrix of this system of 2mequations in 2n variables x1, . . . , xn, y1, . . . , ynhas the following block form

Jac(f1, . . . , f m) 0 Jac(f1, . . . , f m)

and thus has rank 2 rkJac(f1, . . . , f m) = 2(n dim X).)

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4.7. Differential of a morphism. Let : X Y be a morphism of affinevarieties and let x X.

Definition 4.12. The differential of inx is the linear map

dx : TxX T(x)Ydefined by dx() := .

If Z X is a closed subvariety and z Z, then we get for the inducedmorphism |Z: Z Y that d(|Z)z = dz|TzZ. Another obvious remark is thatthe differential of a constant morphism is the zero map.

Remark 4.13. Set y := (x). The comorphism :O(Y) O(X) defines ahomomorphism my mx and thus a linear map : my/m2y mx/m2x. It is easyto see that the differential dx corresponds to the dual map of

under the iso-

morphismsTxX Hom(mx/m2x,C) andTyY Hom(my/m2y,C) (see Lemma4.1).Example4.14. Using the identification T(x,y)XY =TxXTyY (see Propo-

sition4.3) we easily see that the differentiald(prX)x :T(x,y)XY TxXcoincideswith the linear projection prTxX.

Proposition4.17. Let = (f1, . . . , f m) :Cn Cm,fj O(Cn) = C[x1, . . . , xn].Then the differential

dx : TxCn = Cn T(x)Cm = Cm

of inx Cn is given by the Jacobi matrix

Jac(f1, . . . , f m)(x) = fjxi (x)i,j .Proof. The identification of the tangent space TxCn = Derx(O(Cn)) with Cn

is given by (x1, . . . , xn) (see Example4.2). This implies thatdx() = (( )(y1), . . . , ( )(ym)) = (f1, . . . , f m).

Now the claim follows since

fj =

ni=1

fjxi

(x) xi.

Exercise 4.17. Let : X Y and : Y Zbe morphisms of affine varieties andletx X. Then

d( )x = dy dxwherey := (x) Y.

In order to calculate explicitly differentials of morphisms we will again use theepsilonization (4.3). Recall that for TxXthe map := evx :O(X) C[]is a homomorphism of algebras and vice versa. If : X Y is a morphism and

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xX, y := (x) Y , then we obtain, by definition, the following commutativediagram:

O(X) evx C[]

O(Y)

evy dx()

IfX :=V and Y :=Ware vector spaces then a homomorphism :O(V) C[]corresponds to an elementx v V V where(f) =f(x + v), and so corresponds to the element (x+ v) W W . Thus we obtain the followingresult which is very useful for calculating differentials of morphisms.

Lemma 4.18. Let : V W be a morphism between vector spaces, and letx V andv TxV =V. Then we have

(x+ v) =(x) + dx(v) where both sides are considered as elements ofW W .

Example 4.15. The differential of the morphism ?m : Mn Mn,A Am, inE is m Id. In fact, (E+ X)m =E+ mX.

The differential of : M2 M2,(A) :=A2, in an arbitrary matrixA is givenbydA(X) =AX+ XA, because (A+ X)

2 =A2 + (AX+ XA).The differential of the matrix multiplication : Mn Mn Mn in (E, E) is

the addition: (E+ X)(E+ Y ) =E+ (X+ Y).

Exercise 4.18. Consider the multiplication : M2 M2 M2 and show:(1) d(A,B) is surjective, ifA or B is invertible.(2) If rk A= rk B = 1, then d(A,B) has rank 3.(3) We have rk d(A,0) = rk d(0,A)= 2rk A.

Exercise 4.19. Calculate the differential of the morphism : End(V)V V givenby (, v) (v), and determine the pairs (, v) where d(,v) is surjective.

4.8. Tangent spaces of fibers. Let : X Y be a morphism,x X andF :=1((x)) the fiber through x. Since|F is the constant map, its differentialin any point is zero and so TxF ker dx. This proves the first part of the followingresult.

Proposition4.19. Let : X Ybe a morphism, x XandF :=1((x))the fiber throughx.

(1) TxF ker dx.(2) If the fiberF is reduced inx, thenTxF= ker dx.

Proof. Puty := (x) Y. By definition the fiber is reduced in x if and onlyif the ideal in the local ring OX,x generated by(my ) is perfect which means thatOF,x = OX,x/(my)OX,x (see Definition2.6).

Now let TxXbe a derivation ofO(X) in x. If ker dx then =0. Hence , regarded as a derivation ofOX,x, vanishes on (my)OX,x and thusinduces a derivation ofOF,x in x, i.e., TxF.
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Example 4.16. Let X Cn be a closed subset and I(X) = (f1, . . . , f m).Consider the morphism = (f1, . . . , f m) :Cn Cm. Then X=1(0), and thisfiber is reduced in every point. Thus, for every x X,

TxX= ker dx = ker Jac(f1, . . . , f m)(x)

as we have already seen in Proposition4.4.

Exercise 4.20. For every point (x, y) X Y we have TxX= ker d(prY)(x,y) andTyX= ker d(prX)(x,y) where prX , prY are the canonical projections (see Proposition4.3).

Exercise 4.21. For the closed subset N M2 of nilpotent 2 2-matrices we haveI(N) = (tr, det).

4.9. Morphisms of maximal rank. The main result of this section is thefollowing theorem.

Theorem 4.20. Let : X Y be a dominant morphism between two irre-ducible varieties X and Y. Then there is a dense open set U X such thatdx :TxX T(x)Y is surjective for allx U.

We first work out an important example which will be used in the proof of theproposition above.

Example 4.17. LetYbe an irreducible affine variety and X YCan irre-ducible hypersurface. Assume that I(X) = (f) where f =

ni=0 fit

i O(Y)[t] =O(Y C) and fn= 1. Consider the following diagram:

X

p

Y C

prY

Y

Then the differentialdp(y,a) :T(y,a)X TyY is surjective if f

t(y, a) = 0, and this

holds on a dense open set ofX.

Proof. We haveT(y,a)X T(y,a)YC =TyYC, and this subspace is givenbyT(y,a)X= {(, ) | (, )f= 0}, becauseI(X) = (f). Now we have

(, )f =n

i=0

(fi ai + fi(y) i ai1 ) =n

i=0

fi ai + ft

(y, a)

Since dp(y,a)

(, ) =we see that dp(y,a)

is surjective if f

t(y, a)

= 0 which proves

the first claim. But ft cannot be a multiple offand thus does not vanish on X,proving the second claim.

The next lemma shows that the situation described in the example abovealways holds on an open set for every morphism of finite degree.

Lemma 4.21. LetX, Ybe irreducible affine varieties and : X Y a mor-phism of finite degree. Then there is a special open setU Yand a closed embed-ding: 1(U) UC with the following properties:

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(i) I((U)) = (f) wheref=n

i=0 fiti O(U)[t];

(ii) prU =|1(U).

1(U)

VUC(f)

p

UC

prU

U

Proof. We have to show that there is a non-zeros O(Y) such that O(X)sO(Y)s[t]/(f) with a polynomial f O(Y)s[t]. Then the claim follows by settingU :=Ys.

By assumption, the fieldC(X) is a finite extension ofC(Y) of degree n, say,

C(X) = C(Y)[h]C(Y)[t]/(f)

wheref=ni=0 fiti,fi C(Y) andfn= 1. There is a non-zero element s O(Y)such that

(a) fi O(Y)s for all i,(b) h O(X)s and(c)O(X)s = O(Y)s[h] =

n1i=0O(Y)shi.

In fact, (a) and (b) are clear. For (c) we first remark that O(Y)s[h] =n1

i=0O(Y)shi O(X)s, because of (a) and (b). Ifh1, . . . , hm is a set of generators ofO(X) we canfind a non-zero s O(Y) such that hi O(Y)s[h], proving (c).

Setting U := Ys we get 1(U) =Xs andO(Xs) =O(Ys)[t]/(f), by (c), and

the claim follows.

Proof of Theorem 4.20. By the Decomposition Theorem (Theorem 3.17)we can assume that is the composition of a finite surjective morphism and aprojection of the formYCr Y. Since the differential of the second morphismis surjective in any point we are reduced to the case of a finite morphism. Now theclaim follows from Lemma4.21above and the Example4.17.

Lemma 4.22. Let : X Y be a morphism, x X and y := (x) Y.Assume thatX is smooth inx anddx is surjective.

(1) Y is smooth iny.(2) The fiber 1(y) is reduced and smooth in x, and dimx F = dimx X

dimyY.

Proof. By assumption,

dim TxF dim ker dx= dim TxX dim TyY dim X dim Y dimx Fwhich implies that we have equality everywhere. In particular, X and F are bothsmooth in x.

If we denote by m O(X)/myO(X) the maximal ideal corresponding tox F one easily sees that m/m2 is the cokernel of the natural map my/m2ymx/m

2x induced by

. The duality between mx/m2x and TxX (see Lemma 4.1

and Remark 4.13) implies that dim ker dx = dimC m/m2. Since dimker dx =

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5. NORMAL VARIETIES AND DIVISORS 149

dimx F = dim O(X)x/myO(X)x it follows thatO(X)x/myO(X)x is a domain(Proposition4.11), and so F is reduced in x.

Corollary 4.23. For every morphism : X Y there is a dense specialopen setU Xsuch that all fibers of the morphism|U: U Y are reduced andsmooth.

Proof. One easily reduces to the case where X is irreducible. Then there isa special open set U Xwhich is smooth (Corollary4.12) and such that dx issurjective for all x U (Theorem4.20). Now the claim follows from the previousLemma4.22.

Corollary 4.24 (Lemma of Sard). Let :Cn Cm be a dominant mor-phism and setS :={xCn | dx is not surjective}. ThenS is closed and(S)is a proper closed subset ofCm. In particular, there is a dense open setU Cmsuch that all fibers1(y) fory

Uare reduced and smooth of dimensionn

m.

Proof. If = (f1, . . . , f m) then S ={x Cn | rkJac(f1, . . . , f m)(x) < m}and so S is closed in Cn. Moreover, the differential of|S:S Cm at any pointofSis not surjective. Therefore, by Theorem 4.20, the closure of the image (S)has dimension strictly less than m.

Exercise 4.22. Let f C[x1, . . . , xn] be a non-constant polynomial. ThenV(f )is a smooth hypersurface for almost all C.

Corollary 4.25. If : X Y is a morphism such that dx = 0 for allx X, then the image(X) is finite. In particular, ifX is connected then isconstant.

Proof. If X

X is an irreducible component and Y := (X), then the

induced morphism : X Y has the same property, namely dx = 0 for allxX. It follows now from Theorem4.20that dim Y = 0. Hence is constanton X.

Example 4.18. LetVbe a vector space and W Va subspace. IfX V isa closed irreducible subvariety such that TxX Wfor allx XthenX x + Wfor any x X.(This follows from the previous corollary applied to the morphism : X V /Winduced by the linear projection V V/W.)

5. Normal Varieties and Divisors

5.1. Normality.Definition 5.1. LetA B be rings. An element b B is integral overAifb

satisfies an equation of the form

bn =n1i=0

aibi whereai A.

Equivalently,b Bis integral over A if and only if the subring A[b] B is a finiteA-module.

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If every element fromB is integral over A we say that B is integral overA.

Exercise 5.1. LetA

B be rings. IfA is Noetherian and B finite overA, thenB is

integral overA.

Lemma 5.1. LetA B Cbe rings and assume thatA is Noetherian.(1) IfB is integral overA andCintegral overB, thenC is integral overA.(2) The set

B := {b B| b is integral overA}is a subring ofB.

Proof. (1) Let c C. Then we have an equationcm =m1j=0 bjcj with bj B.In particular, the coefficients bj are integral over A and so, by induction, A1 :=A[b0, b1, . . . , bm1] is a finitely generated A-module. Moreover, A1[c] is a finitely

generatedA1-module, hence a finitely generatedA-module. But then A[c] A1[c]is also finitely generated.

(2) Letb1, b2 B. ThenA[b1] is integral over Aandb2is integral overA, henceintegral over A[b1], and so A[b1, b2] is integral over A[b1]. Thus, by (1), A[b1, b2] isintegral overAwhich implies that b1 + b2 andb1b2 are both integral over A, hencebelong to B.

Exercise 5.2. Letf C[x] be a non-constant polynomial. ThenC[x] is integral overthe subalgebraC[f].

Definition 5.2. LetA be a domain with field of fraction K. We call A inte-grally closedif the following holds:

Ifx K is integral overA, thenx A.An affine varietyX isnormal ifX is irreducible and O(X) is integrally closed. Wesay that X is normal inx X if the local ringOX,x is integrally closed.

Example5.3. A unique factorization domain A is integrally closed. In partic-ular,Cn is a normal variety.(Let Kbe the field of fractions ofA and x K integral over A: xn = n1i=0 aixiwhere ai A. Write x = ab where a, b A have no common divisor. Thenan =b(

n1i=0 aib

ni1ai) which implies thatb is a unit in A and sox A.)Exercise 5.3. If the domain A is integrally closed, then so is every ring of fraction

ASwhere 1 S A is multiplicatively closed.Lemma 5.2. LetXbe an irreducible variety. ThenXis normal if and only if

all local ringsOX,x are integrally closed.Proof. If X is normal thenOX,x =O(X)mx is integrally closed (see the

Exercise above), and the reverse implication follows fromO(X) =xX OX,x(Exercise1.26).
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5.2. Integral closure and normalization.

Proposition5.3. LetA be a finitely generatedC-algebra with no zero-divisors= 0 and with field of fractionsK, and letL/Kbe a finite field extension. Then

A := {x L | x is integral overA} Ais a finitely generatedC-algebra which is finite overA.

Proof. We already know that A is aC-algebra (Lemma5.1(2)).(a) We first assume that A = C[z1, . . . , zm] is a polynomial ring and K =

C(z1, . . . , zm). Let L= K[x] where x is integral over A and [L: K] =:n. Denotebyx1:=x, x2, . . . , xnthe conjugates ofxin some Galois extensionL

ofK. Clearly,allxj are integral over A, because they satisfy the same equation as x.

Ify =

n1i=0 cix

i (ci K) is an arbitrary element ofL we obtain the conju-gates ofy in L in the form

yj =n1i=0

cixi

j for j = 1, . . . , n .

Then n-matrixX := (xij ) has determinantd =

j

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(1) X is normal;(2) is finite and birational.

ThenO(X) is the integral closure of(O(X)) inC(X) =(C(X)), and we havethe following universal property:

(P) IfY is a normal affine variety then every dominant morphism : Y Xfactors through : There is a uniquely determined : Y X such that= :

X

Y

X

Proof. Since is birational we have (O(X)) O(X) C(X) =(C(X)).By (2)

O(X) is finite, hence integral over (

O(X)), and by (1) it is the integral

closure of(O(X)).IfYis normal affine variety and : Y Xa dominant morphism then

O(X) (O(X)) O(Y) C(Y).Denote byO(X) the integral closure ofO(X) inC(X). SinceO(Y) is integrallyclosed it follows that (O(X))C(Y) is contained inO(Y). Since inducesan isomorphismO(X) O(X) there is a uniquely determined homomorphism :O(X) O(Y) which makes the following diagram commutative:

O(X)

O(X)

O(Y) O(X)

Clearly, the corresponding morphism :Y Xis the unique morphism such that= .

Definition 5.5. The morphism : X Xconstructed above is called nor-malization of X. It follows from Lemma 5.4 that it is unique up to a uniquelydetermined isomorphism.

Exercise

5.4.

If : X Y is a finite surjective morphism where X is irreducibleandYis normal, then #1(y) deg for ally Y. (See Proposition3.22and its proof.)Proposition5.5. LetXbe an irreducible variety. Then the set

Xnorm:= {x X| X is normal inx}is open and dense inX.

Proof. LetO(X) C(X) be the integral closure ofO(X) and definea:= {f O(X) | fO(X) O(X)}.

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Thenais a non-zero ideal ofO(X), because O(X) is finite over O(X), and Xnorm=X\ VX(a). In fact, for S:= O(X) \ mx we have

OX,x = O(X)S O(X)Sand the latter is the integral closure ofOX,x. On the other hand, O(X)S= O(X)Sif and only ifS a = which is equivalent to x / VX(a).

Exercise 5.5. Consider the morphism :C2 C4, (x, y) (x,xy,y2, y3).(1) is finite and :C2 Y :=(C2) is the normalization.(2) 0 Yis the only non-normal and the only singular point ofY.(3) Find defining equations forY C4 and generators of the ideal I(Y).

Exercise 5.6. IfXis a normal variety then so is X Cn.

5.3. Discrete valuation rings and smoothness. LetKbe a field.

Definition5.6. Adiscrete valuation of the fieldKis a surjective map : K

:=K\ {0} Z with the following properties:(a) (xy) =(x) + (y);(b) (x + y) min((x), (y)).

To simplify the notation one usually defines (0) := .Example 5.7. LetK=Q and p N a prime number. Define p(x) :=r Z

ifp occurs with exponent r in the rational number x= 0. Then p :Q Z is adiscrete valuation ofQ.

The following lemma collects some facts about discrete valuations. The easyproofs are left to the reader.

Lemma5.6

. LetKbe a field and: K

Z a discrete valuation.(1) A:= {x K| (x) 0} is a subring ofK.(2) m:= {x K| (x)> 0} A is a maximal ideal ofA.(3){x K| (x) = 0} are the units ofA.(4) For every non-zero x Kwe havex A orx1 A.(5) m= (x) for everyx K with(x) = 1.(6) mk = {x K| (x) k} and these are all non-zero ideals ofA.(7) Ifm= (x) then everyz Khas a unique expression of the formz = txk

wherek Z andt is a unit ofA.Definition 5.8. A domain A is called a discrete valuation ring if there is a

discrete valuation of its field of fractions K such that A={x K| (x) 0}.In particular, Ahas all the properties listed in Lemma 5.6 above.

Exercise 5.7. Let A be a discrete valuation ring with field of fraction K. IfB Kis a subring containingA then either B = A or B = K.

In the sequel we will use the following characterization of a discrete valuationrings (see [AM69, Proposition 9.2]).

Proposition5.7. LetA be a Noetherian local domain of dimension 1, i.e. themaximal idealm

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