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ALGEBRAIC GEOMETRY

Alexei Skorobogatov

December 16, 2003

Contents

1 Basics o commutative algebra 31.1 Integral closure. Noetherian rings . . . . . . . . . . . . . . . . 31.2 Noethers normalization and Nullstellensatz . . . . . . . . . . 7

2 Ane geometry 92.1 Zariski topology . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Category o ane varieties . . . . . . . . . . . . . . . . . . . . 102.3 Examples o rational varieties . . . . . . . . . . . . . . . . . . 14

2.4 Smooth and singular points . . . . . . . . . . . . . . . . . . . 172.5 Dimension. Application: Tsens theorem . . . . . . . . . . . . 19

3 Projective geometry 223.1 Projective varieties . . . . . . . . . . . . . . . . . . . . . . . . 223.2 Morphisms o projective varieties . . . . . . . . . . . . . . . . 24

4 Local geometry 264.1 Localization, local rings, DVR . . . . . . . . . . . . . . . . . . 264.2 Regular local rings . . . . . . . . . . . . . . . . . . . . . . . . 284.3 Geometric consequences o unique actorization in

OP . . . . . 29

5 Divisors 305.1 The Picard group . . . . . . . . . . . . . . . . . . . . . . . . . 305.2 Automorphisms oPnk and o A

nk . . . . . . . . . . . . . . . . 33

5.3 The degree o the divisor o a rational unction on a projectivecurve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5.4 Bezout theorem or curves . . . . . . . . . . . . . . . . . . . . 365.5 RiemannRoch theorem . . . . . . . . . . . . . . . . . . . . . 395.6 From algebraic curves to error correcting codes . . . . . . . . . 41

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A More algebra 43

A.1 Krulls intersection theorem . . . . . . . . . . . . . . . . . . . 43A.2 Completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43A.3 The topological space Spec(R) . . . . . . . . . . . . . . . . . . 44

B More geometry 46B.1 Finite morphisms . . . . . . . . . . . . . . . . . . . . . . . . . 46B.2 Functoriality o Pic . . . . . . . . . . . . . . . . . . . . . . . . 48

Basic reerences

Algebra:S. Lang. Algebra. Addison-Wesley, 1965.

M. Reid. Undergraduate commutative algebra. Cambridge University Press,1995.

B.L. Van der Waerden. Algebra. (2 volumes) Springer-Verlag.

Geometry:

M. Reid. Undergraduate algebraic geometry. Cambridge University Press,1988.

I.R. Shaarevich. Basic algebraic geometry. (2 volumes) Springer-Verlag.1994.

Further reerences

Advanced commutative algebra:

M.F. Atiyah and I.G. Macdonald. Introduction to commutative algebra.Addison-Wesley, 1969.

H. Matsumura. Commutative ring theory. Cambridge University Press, 1986

O. Zariski and P. Samuel. Commutative algebra. (2 volumes) Van Nostrand,1958.

Advanced algebraic geometry:

R. Hartshorne. Algebraic geometry. Springer-Verlag, 1977.

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1 Basics o commutative algebra

Let k be a eld. (Ane) algebraic geometry studies the solutions o systemso polynomial equations with coecients in k. Instead o a set o polyno-mials it is better to consider the ideal o the polynomial ring k[X1, . . . , X n]generated by them. The subset o kn consisting o common zeros o the poly-nomials o an ideal I k[X1, . . . , X n] is called the set o zeros o I. Suchsubsets o kn are called closed algebraic sets.

Hence our main object o study will be the polynomial ring k[X1, . . . , X n],its ideals, their sets o zeros in kn, and the quotient rings o k[X1, . . . , X n].Here is the list o principal acts that we prove in this chapter:

(1) Every ideal o k[X1, . . . , X n] is a nitely generated k[X1, . . . , X n]-module. (Hilberts basis theorem.)

(2) Every quotient ring o k[X1, . . . , X n] is o the ollowing orm: it con-tains a polynomial ring k[Y1, . . . , Y m] over which it is a nitely generatedmodule. (Emmy Noethers normalization lemma).

(3) I k is algebraically closed, then all maximal ideals o k[X1, . . . , X n]are o the orm (X1a1, . . . , X n an), ai k, that is, consist o polynomialsvanishing at a point (a1, . . . , an) kn.

(4) Let k be algebraically closed. I a polynomial f vanishes at all thezeros o an ideal ok[X1, . . . , X n], then some power f

m belongs to this ideal.

(Hilberts Nullstellensatz.)Fact (1) says that speaking about ideals and systems o (nitely many)

polynomial equations is the same thing. The meaning o (2) will be madeclear in the geometric part o the course (an important corollary o (2) isthe act that any algebraic variety is birationally equivalent to a hypersur-ace). Fact (3) speaks or itsel. Finally, (4) implies that certain ideals ok[X1, . . . , X n] bijectively correspond to closed algebraic sets.

1.1 Integral closure. Noetherian rings

Most o the time we assume that k is an algebraically closed eld. Whenk is not algebraically closed, k denotes a separable closure o k (unique upisomorphism, see [Lang]). Let R be a commutative ring with a 1. We shallalways consider ideals I R diferent rom R itsel. Then the quotient ringR/I is also a ring with 1. By denition, a ring is an integral domain i it hasno zero divisors.

IM is an R-module, then 1 R acts trivially on M. An R-module M isonite type i it is generated by nitely many elements, that is, i there exist

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a1, . . . , an

M such that any x

M can be written as x = r1a1 + . . . + rnan

or some ri R.I a module A over a ring R also has a ring structure (compatible with

that o R in the sense that the map R A given by r r.1A is a ringhomomorphism), then A is called an R-algebra. An R-algebra A is o nitetype (or nitely generated) i there exist a1, . . . , an A such that any x Acan be written as a polynomial in a1, . . . , an with coecients in R.

Prime and maximal ideals. An ideal I R is called prime i thequotient ring R/I has no zero divisors. An ideal I is maximal i it is notcontained in another ideal (diferent rom R). Then the ring R/I has no

non-zero ideal (otherwise its preimage in R would be an ideal containing I),hence every element x R/I, x = 0, is invertible (since the principal ideal(x) must concide with R/I, and thus contain 1), in other words, R/I is aeld. Since a eld has no non-trivial ideals, the converse is also true, so thatI R is maximal if R/I is a eld.

Integral closure. We shall consider various niteness conditions. Sup-pose we have an extension o integral domains A B. An element x B iscalled integral over A i it satises a polynomial equation with coecients inA and leading coecient 1.

Proposition 1.1 The ollowing conditions are equivalent:(i) x B is integral over A,(ii) A[x] is an A-module o nite type,(iii) there exists an A-module M o nite type such that A M B and

xM M.

The proo o (i) (ii) and (ii) (iii) is direct. Suppose we know (iii).Let m1, . . . , mn be a system o generators o M. Then xmi =

ni=1 bijmj,

where bij A.Recall that in any ring R we can do the ollowing determinant trick.

Let S be a matrix with entries in R. Let adj(M) be the matrix with entriesin R given by

adj(M)ij = (1)i+jdet(M(j,i)),where M(i, j) is M with i-row and j-th column removed. It is an exercise inlinear algebra that the product adj(M) M is the scalar matrix with det(M)on the diagonal.

We play this trick to the polynomial ring R = A[T]. For M we take then n-matrix Q(T) such that Q(T)ij = T ij bij. Let f(T) = det(Q(T))

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A[T] (this is the analogue o the characteristic polynomial o x). We have a

matrix identityadj(Q(T)) Q(T) = diag(f(T)).

We consider this as an identity between matrices over the bigger ring B[T].We are ree to assign T any value in B. Substitute T = x B, and applythese matrices to the column vector (m1, . . . , mn)

t. Then the let hand sideis zero. Hence f(x)mi = 0 or any i. Since the mi generate M the wholemodule M, is annihilated by f(x) B. In particular, f(x).1 = 0, that is,f(x) = 0. Now note that f(T) has coecients in A and leading coecient 1.QED

Remark. This proo does not use the act that A and B are integral. Iwe assume this we can remove the condition A M in (iii).Defnitions. Let A B be integral domains, then B is integral over A

i its every element is integral over A. The set o elements o B which areintegral over A is called the integral closure o A in B.

Important example. Let K be a number eld, that is, a nite extensiono Q. One denes the ring o integers OK K as the integral closure o Zin K.

Let us prove some basic properties o integral elements.

Proposition 1.2 (a) The integral closure is a ring.(b) Suppose thatB is integral overA, and is o nite type as an A-algebra.

Then B is o nite type as an A-module.(c) Suppose that C is integral over B, and B is integral over A, then C

is integral over A.

Proo. (a) Let x, y B be integral over A. Consider the A-modulegenerated by all the monomials xiyj, i, j 0. It is o nite type, and xy andx + y act on it.

(b) Suppose that B is generated by b1, ... , bn as an A-algebra, then B

is generated by monomials bi11 . . . binn as an A-module. All higher powers oeach o the bis can be reduced to nitely many o its powers using a monicpolynomial whose root is bi. There remain nitely many monomials whichgenerate B as an A-module.

(c) Let x C. Consider the A-subalgebra D C generated by x andthe coecients bi o a monic polynomial with coecients in B, whose rootis x. Then D is an A-module o nite type, as only nitely many monomialsgenerate it (the bi are integral, and the higher powers o x can be reduced tolower powers) Now use (iii) o the previous proposition. QED

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Defnition. A ring is integrally closed or normal i it is integrally closed

in its eld o ractions.

Examples. Z and Z[1+32

] are integrally closed, but Z[3] is not. I

k is a eld then k[x] and k[x, y] are integrally closed, but k[x, y]/(y2x2x3)is not.

Noetherian rings. Another important niteness property o rings isgiven in the ollowing denition.

Defnition-Proposition. A ringR satisying any o the ollowing equiv-alent properties is called Noetherian:

(i) any chain o ideals I1

I2

I3

. . . o R stabilizes (that is, there is

an integer m such that Im = Im+1 = Im+2 = . . . ),(ii) any set o ideals o R contains a maximal element,(iii) any ideal o R is generated by nitely many elements, that is, is an

R-module o nite type.

Proo. The equivalence o (i) and (ii) is completely ormal.(iii) (i): Let I = Ij, then I is an ideal which is generated, say, by

x1, . . . , xn as an R-module. Take m such that Im contains all the xi, thenthe chain stabilizes at Im.

(ii) (iii) is based on a trick called Noetherian induction. Supposethat I

R is an ideal which is not o nite type as an R-module. Consider

the set o subideals oI which are o nite type as R-modules. This set is notempty: it contains 0. Now it has a maximal element J = I. Take x I\ J,then the ideal J + (x) I is strictly bigger than J, but is o nite type asan R-module. Contradiction. QED

Examples o Noetherian rings: Fields, principal ideals domains, the ringo integers in a number eld.

An easy exercise: Quotients o a Noetherian ring are Noetherian.

Theorem 1.3 (Hilberts basis theorem) I R is Noetherian, then so arethe polynomial ring R[T] and the ormal power series ring R[[T]].

Sketch o proo. Let I R[T] be an ideal. We associate to it a series oideals in R:

A0 A1 A2 . . . ,where Ai is generated but the leading coecients o polynomials in I o degreei. Since R is Noetherian, this chain o ideals stabilizes, say, at Ar. Then wehave a nite collection o polynomials whose leading coecients generate A0,... , Ar. Then the ideal o R[T] generated by these polynomials is I. SeeLangs book or the proo o the other statement. QED

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1.2 Noethers normalization and Nullstellensatz

The elements r1, . . . , rn o a k-algebra R are algebraically independent (overk) i the only polynomial f(x1, . . . , xn) with coecients in k such thatf(r1, . . . , rn) = 0, is the zero polynomial.

Theorem 1.4 (E. Noethers normalization lemma.) Letk be any eld,and I k[T1, . . . , T n] be an ideal, R = k[T1, . . . , T n]/I. There exist alge-braically independent elements Y1, . . . , Y m R such that R is integral overk[Y1, . . . , Y m].

Proo. I I = 0 there is nothing to prove. Suppose we have a non-zero

polynomial f I. Let d be a positive integer greater than deg(f). Let uschoose new variables in the ollowing tricky way:

X2 = X2(X1)d, X3 = X3(X1)d2

, X4 = X4(X1)d3

, . . . , X n = Xn(X1)dn1

.

Substituting this into f we rewrite it as a linear combination o powers oX1 and a polynomial, say, g containing no pure powers o X1. We observethat the pure powers o X1 are o the orm i1 + di2 + d

2i3 + . . . + dn1in.

Since d > is all these integers are diferent, hence there is no cancellationamong the pure powers o X1. At least one such power enters with a non-zero coecient. On the other hand, any power o X1 in g is strictly lessthan the corresponding pure power. Thereore, we get a polynomial in X1with coecients in k[X2, . . . , X

n] and leading coecient in k. Normalizing

this polynomial we conclude that X1 is integral over R1 = k[X2, . . . , X

n]/I

k[X2, . . . , X n]. Hence R is integral over R1. We now play the same game

with R1 instead o R, and obtain a subring R2 over which R1 is integral.By Property (c) o integral ring extensions R is also integral over R2. Wecontinue like that until we get a zero ideal, which means that the variablesare algebraically independent. QED

Theorem 1.5 Let k be an algebraically closed eld. All maximal ideals o

k[X1, . . . , X n] are o the orm(X1 a1, . . . , X n an), ai k, that is, consisto polynomials vanishing at a point (a1, . . . , an) kn.

Proo. Any polynomial has a Taylor expansion at the point (a1, . . . , an).The canonical map

k[X1, . . . , X n] k[X1, . . . , X n]/(X1 a1, . . . , X n an)sends f to f(a1, . . . , an), hence is surjective onto k. It ollows that the ideal(X1 a1, . . . , X n an) is maximal.

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Let M a maximal ideal (recall that M

= k[X1, . . . , X n]), then K =

k[X1, . . . , X n]/M is a eld containing k. By Noetherian normalization K isintegral over its subring A = k[Y1, . . . , Y m]. But K is a eld, and we now showthat then A must also be a eld, in which case k[Y1, . . . , Y m] = k (no variablesat all), and hence K is integral over k. Indeed, let x A, then it is enough toshow that x1 K also belongs to A. Since x1 K is integral over A it issubject to a polynomial relation (x1)n + an1(x1)n1 + . . . + a1x1+ a0 = 0,or some ai A. Multiplying this by xn1 we express x1 as a polynomial inx with coecients in A, hence x1 A.

The k-algebra o nite type K is integral over k, hence by Proposition1.2 (b) K is a k-module (= vector space over k) o nite type (= o nite

dimension). Thus the eld K is an algebraic extension o k. Since k isalgebraically closed, we must have k = K. Now let ai k be the image oXiunder the map k[X1, . . . , X n] k = k[X1, . . . , X n]/M. Then M containsthe maximal ideal (X1 a1, . . . , X n an), hence coincides with it. QED

Remark. When k is not supposed to be algebraically closed, this prooshows that the quotient by a maximal ideal o k[X1, . . . , X n] is a nite ex-tension o k.

Corollary 1.6 Let k be an algebraically closed eld. I the polynomialso an ideal I k[X1, . . . , X n] have no common zeros in kn, then I =k[X

1, . . . , X n].

Proo. Assume I = k[X1, . . . , X n]. Hilberts basis theorem says thatk[X1, . . . , X n] is Noetherian. Then I is contained in a maximal ideal, sincethe set o ideals that contain I has a maximal element, by (ii) o Denition-Proposition above. Thereore I (X1 a1, . . . , X n an), or some ai k,since all the maximal ideals are o this orm by the previous result. Butthen all the polynomials o I vanish at the point (a1, . . . , an), which is acontradiction. QED

Theorem 1.7 (Nullstellensatz.) Let k be an algebraically closed eld. Ia polynomial f vanishes at all the zeros o an ideal I

k[X1, . . . , X n], then

fm I or some positive integer m.Proo. We know that I is generated by nitely many polynomials, say, I =

(g1, . . . , gr). Let T be a new variable. Consider the ideal J k[T, X1, . . . , X n]generated by g1, . . . , gr and T f1. We observe that these polynomials haveno common zero. The previous corollary implies that J = k[T, X1, . . . , X n],in particular, J contains 1. Then there exist polynomials p, p1, . . . , pr invariables T, X1, . . . , X n such that

1 = p(T f 1) + p1g1 + . . . +prgr.

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Note that this is an identity in variables T, X1, . . . , X n. Thus we can spe-

cialize the variables anyway we like. For example, we can set T = 1/f.Multiplying both sides by an appropriate power o f we get an identity be-tween polynomials in variables X1, . . . , X n, which gives that some power of belongs to I = (g1, . . . , gr). QED

2 Ane geometry

2.1 Zariski topology

Let k be any eld. Let us prove some easy acts about closed algebraic

sets. IX kn we denote by I(X) k[X1, . . . , X n] the ideal consisting opolynomials vanishing at all the points o X. We denote by Z(J) the set ozeros o an ideal J k[X1, . . . , X n]. It is a tautology that X Z(I(X)) andJ I(Z(J)). IX is a closed algebraic set, then X = Z(I(X)) (iX = Z(J),then I(Z(J)) J, hence Z(I(Z(J))) Z(J)).

Exercise. Show that i X Ank and Y Amk are closed subsets, thenX Y An+mk is a closed subset.

It is clear that the unction J Z(J) reverses inclusions; associates theempty set to the whole ring, and the whole ane space kn to the zero ideal;

sends the sum o (any number o) ideals to the intersection o correspondingclosed sets; and sends the intersection I1 I2 to Z(I1) Z(I2) (a part o thelast property is not completely obvious: i P / Z(I1) Z(I2), then f(P) = 0or some f I1 and g(P) = 0 or some g I2, but then (fg)(P) = 0, whereasf g I1 I2).

Because o these properties we can think o closed algebraic sets as theclosed sets or some topology on kn (any intersections and nite unions areagain closed, as are the empty set and the whole space). This topology iscalled Zariski topology. In the case when k = C or k = R we can compare itwith the usual topology on Cn where closed sets are the zeros o continuousunctions. Any Zariski closed set is also closed or the usual topology butnot vice versa. Hence the Zariski topology is weaker. Another eature is thatany open subset o kn is dense (its closure is the whole kn).

Defnition. A closed algebraic subset X kn is irreducible i there isno decomposition X = X1 X2, where X1 = X and X2 = X are closedalgebraic sets.

Proposition 2.1 A closed algebraic subset X kn is irreducible if I(X) isa prime ideal. Any closed set has a unique decomposition into a nite union

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o irreducible subsets X =

iXi such that Xi

Xj or i

= j (these Xis are

called the irreducible components o X).

Proo. Let us prove the rst statement. I we have X = X1 X2, thensince X = Z(I(X)) or any algebraic set, I(X) is a proper subset o I(Xi).I fi I(Xi) \ I(X), then f1f2 I(X), hence I(X) is not a prime ideal.Conversely, i I(X) is not prime, we can nd two polynomials f1 and f2not in I(X) such that f1f2 I(X), and dene Ii = (I(X), fi), Xi = Z(Ii).There exists a point P in X such that f1(P) = 0, hence P / X1 whichimplies X1 = X. Similarly we have X2 = X. Thereore X = X1 X2 is notirreducible.

Let us prove the second statement. I X is not irreducible, we have somedecomposition X = X1 X2, and then continue or X1 and X2. At somepoint we must stop because the chain o ideals I(X) I(X1) . . . stabilizessomewhere (the ring k[X1, . . . , X n] being Noetherian). I iXi = j Yj aretwo decompositions into irreducible subsets, then Xi = j(Xi Yj), andhence Xi = Xi Yj or some j (since Xi is irreducible). For the analogousreason we have Yj = Yj Xi or some i. Then Xi Xi, hence i = i. Itollows that Xi = Yj. Hence the two decompositions difer only in order.QED

Up till now we did not use Hilberts Nullstellensatz. Let k be an alge-

braically closed eld. Let us call an ideal I k[X1, . . . , X n] radicalifm Iimplies f I. A corollary o Hilberts Nullstellensatz is that radical idealsbijectively (via operations I and Z) correspond to closed algebraic sets. Themost important class o radical ideals are prime ideals. Again, by HilbertsNullstellensatz, these bijectively correspond to irreducible closed algebraicsets. A particular case o prime ideals are maximal ideals, they correspondto points o kn.

Zero sets o irreducible polynomials ok[X1, . . . , X n] are called irreduciblehypersuraces.

2.2 Category o ane varietiesAn ane varietyis a closed irreducible algebraic subset okn or some n. Thevariety kn will be also denoted Ank , and called the ane space o dimensionn.

Let X Ank be an ane variety. Let J = I(X) be the correspondingprime ideal. Let us denote k[X] := k[X1, . . . , X n]/J. Then k[X] is an integralk-algebra o nite type: k[X] contains no zero divisors. k[X] is called thecoordinate ring o X. The raction eld o k[X] is denoted by k(X), and is

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called the unction eld o X. Its elements are called rational unctions as

opposed to the elements o k[X] which are called regular unctions.The unction eld k(X) is an important object dened by X. Two ane

varieties X and Y are called birationally equivalent i k(X) = k(Y). Avariety X is called rational i k(X) is a purely transcendental extension ok, that is, k(X) = k(T1, . . . , T l). In other words, X is rational i and onlyi X is birationally equivalent to the ane space. It is a classical, and otena dicult problem o algebraic geometry to determine whether or not twogiven varieties are birationally equivalent.

Ane varieties orm a category, where a morphism X Y, X Ank , Y Amk , is given by a unction representable by m polynomials in n variables. The

varieties X and Y are called isomorphic i there are morphisms f : X Yand g : Y X such that fg and gf are identities.

Proposition 2.2 LetX Ank and Y Amk be ane algebraic varieties.(a) A morphism f : X Y denes a homomorphism o k-algebras f :

k[Y] k[X] via the composition o polynomials.(b) Any homomorphism o k-algebras : k[Y] k[X] is o the orm

= f or a unique morphism f : X Y.(c) f : X Y is an isomorphism o ane varieties i and only i f :

k[Y] k[X] is an isomorphism o k-algebras.

Proo. (a) ollows rom the act that the composition o polynomials is apolynomial.

(b) Let x1, . . . , xn be the coordinates on X, and t1, . . . , tm be the coor-dinates on Y. Let be the composition o the ollowing homomorphisms ok-algebras

k[t1, . . . , tm] k[Y] = k[t1, . . . , tm]/I(Y) k[X] = k[x1, . . . , xn]/I(X).

Let fi = (ti), i = 1, . . . , m. The polynomial map f = (f1, . . . , f m) mapsX to Amk . Let F(t1, . . . , tm) be a polynomial. Since we consider homomor-

phisms o k-algebras we have

F(f1, . . . , f m) = F((t1), . . . , (tm)) = (F(t1, . . . , tm)).

I F I(Y), then (F) = 0. Hence all the polynomials rom I(Y) vanish onf(X), that is, f(X) Z(I(Y)) = Y.

Finally, f = since these homomorphisms take the same values on thegenerators ti o the k-algebra k[Y].

(c) ollows rom (a) and (b). QED

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Examples. (1) Hypersuraces in Ank bijectively correspond to principal

ideals in k[x1, . . . , xn] (via the usual operations Z and I).(2) Square matrices o size n are parametrized by the points o An

2

k . Themultiplication o matrices is a morphism An

2

k An2k An2k . The determinantis a morphism An

2

k A1k.The closed subset given by det(M) = 1 is the special linear group SL(n).

Note that the inverse is an isomorphism SL(n) SL(n). The group op-erations are morphisms when that is the case then the group is called analgebraic group. SL(n) An2k is a hypersurace o degree n.

(3) The orthogonal group O(n) An2k given by the conditions MMt = Iis another example o a closed subset which is an algebraic group. It is dened

by n2 quadratic polynomials.Exercise. Show that O(n) is not irreducible. (Hint: what are the possi-

ble values o det(M)?)

Zariski topology on Ank induces a topology on a variety X Ank . Anopen subset U X is an intersection o X with an open set o Ank . Suchsets are called a quasi-ane varieties. An example o a quasi-ane variety isthe general linear group GL(n) (square matrices with non-zero determinant).Later well see that GL(n) is isomorphic to an ane variety (see the end othis subsection).

Defnition. A rational unction f k(X) is called regular at a point Po X i f = g/h, where g, h k[X] and h(P) = 0. A unction is regular onan open set U X i it is regular at every point o U.

The ring o regular unctions on an open subset U X is denoted byk[U]. Since k[X] k[U] k(X) the raction eld o k[U] i k(X).

To a rational unction f k(X) one associates the ideal o denomina-tors Df k[X] consisting o regular unctions h such that hf k[X] (it isclearly an ideal!). The set o all points P where f is regular is X\Z(Df). In-deed, we can write f = g/h, g, h k[X], h(P) = 0, i and only i P / Z(Df).

An immediate corollary o the Nullstellensatz says that i I

k[X] is an

ideal, and f k[X] vanishes at all the common zeros o I in X, then fs Ior some s > 0. (Apply Theorem 1.7 to the pre-image o I in k[x1, . . . , xn]under the natural surjective map.) Well oten use the Nullstellensatz in thisorm.

Lemma 2.3 Let X be an ane variety. The subset o k(X) consisting ounctions regular at all the points o X is k[X]. A unction is regular on theopen subset given by h = 0, or h k[X], i and only i f k[X][h1], inother words, i f = g/hs or some g k[X] and s > 0.

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Proo. Let f be such a unction. Then Z(Df) =

. By Corollary 1.6

Df must be the whole ring, hence contains 1, hence f k[X]. This provesthe rst statement. To prove the second statement we note that Z(Df) iscontained in the closed set given by h = 0. By Nullstellensatz i h vanisheson Z(Df), then a power o h is in Df. QED

One denes rational maps using rational unctions instead o regular ones.Rational maps are not everywhere dened, that is, are not unctions! Arational map is called dominant i its image (=the image o the set o pointswhere the map is actually dened, that is, is regular) is dense, that is, notcontained in a smaller subvariety. The ollowing proposition is proved alongthe same lines as Proposition 2.2.

Proposition 2.4 (a) A dominant rational map f : X > Y denes ahomomorphism o k-algebras f : k(Y) k(X).

(b) Any homomorphism o k-algebras : k(Y) k(X) is o the orm = f or a unique dominant rational map f : X > Y.

We need the condition that f is dominant, as otherwise the compositiono rational unctions is not always dened. (On substituting the fi or thecoordinates we may have to divide by zero! But this cant happen i the mapis dominant, as the fi then do not satisy a non-trivial polynomial condition.)

Finally, a morphism o quasi-ane varieties is an everywhere dened ra-tional map. An isomorphism is a morphism that has an inverse. For example,the map M M1 is a morphism GL(n) GL(n) (and in act, an iso-morphism). On the ane space o square matrices An

2

k this is just a rationalmap.

A rational map which has an inverse (at the same time let and right)is called a birational map or a birational equivalence. For example, the mapM M1 is a birational equivalence o An2k with itsel.

Proposition 2.5 The ollowing conditions are equivalent: a rational mapf : X > Y is(i) birational,(ii) f is dominant, and f : k(Y) k(X) is an isomorphism o k-

algebras,(iii) there exist open sets in X and Y such thatf denes an isomorphism

between them.

Proo. The only implication which is not immediate is (i) (iii). Butthis is also easy, see [Reid, UAG], (5.8).

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Theorem 2.6 Any ane variety is birationally equivalent to a hypersurace.

Proo. By Noethers normalization there exist algebraically independentelements y1, . . . , ym k[X] such that k[X] is integral over k[y1, . . . , ym].Then the eld extension k(y1, . . . , ym) k(X) is nite. In act, one canarrange that this extension is separable (this is automatic i the character-istic o k is 0, see [Reid, UAG], (3.16) or the case o nite characteristic).Any separable nite extension can be obtained by adding just one element(primitive element theorem), say, y. Let F(t) be an irreducible polynomialwith coecients in k(y1, . . . , ym) such that F(y) = 0. By ddling with thecoecients can assume that the coecients o F(t) are in k[y1, . . . , ym], and

that F is irreducible as a polynomial in y1, . . . , ym, t. Then k(X) is the rac-tion eld o k[y1, . . . , ym, t]/(F(t)). Let V Am+1k be the hypersuace givenby F = 0. Then k(V) = k(X). QED

The ollowing lemma allows us to consider only ane open neighbour-hoods.

Proposition 2.7 Every open neighbourhood o a point o an ane varietycontains a neighbourhood isomorphic to an ane variety (and not just quasi-ane).

Proo. It is enough to show how to remove zeros o polynomials. Let

X Ank be a closed set given by the polynomial equations f1 = . . . = fm = 0in variables T1, . . . , T n, and let g be another polynomial, g(P) = 0. LetX0 X be given by g = 0. Consider the closed subset X1 An+1k withcoordinates T0, T1, . . . , T n given by f1 = . . . = fm = T0g 1 = 0. Thenthe projection (T0, T1, . . . , T n) (T1, . . . , T n) denes an isomorphism o X1with X0. QED

Exercise. Show that GL(n) is isomorphic to a closed subset o an anespace.

2.3 Examples o rational varietiesLets do some concrete algebraic geometry.

Let X Ank be a hypersurace given by f(X1, . . . , X n) = 0. We shallalways assume that no linear change o coordinates X1, . . . , X n reduces f toa polynomial in n 1 variables, and that no linear change o coordinatesreduces f to a homogeneous polynomial. Then X is called non-conical. Weshall only consider non-conical hypersuraces. We call the degree o X thedegree o f.

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In this section we examine some examples o rational hypersuraces. By

denition X Ank is rational i there is a birational map Amk > X.Such a map is given by rational unctions in m variables. The meaning orationality is that the points o X can be parametrized by rational unctionssuch that this parametrization is an isomorphism on a non-empty open set.

Remark. We observe that a rational variety over any innite eld k hasinnitely many k-points. Indeed, it contains an dense open subset isomorphicto a dense open subset oAnk , and the latter contains innitely many k-points(easy exercise).

(1) Quadrics (degree 2). Ik is not algebraically closed there may be no

k-point on X, e.g. x2

+ y2

+ 1 = 0 over k = R. Then X is not rational.The ollowing proposition is a generalization o the classical parametriza-

tion o the conic x2 + y2 = 1 by rational unctions x = (t2 1)/(t2 + 1),y = 2t/(t2 + 1).

Proposition. A non-conical quadric with a k-point is rational.

Proo. The idea is to exploit the classical stereographic projection. Wecan assume that the k-point is N = (0, 0, . . . , 0) (N stands or the NorthPole ...). Then we can write

f = Q(x1, . . . , xn) + L(x1, . . . , xn),

where the homogeneous polynomials L and Q have degrees 1 and 2, respec-tively (there is no constant term). Chose a hyperplane H not passing throughN, say the one given by x1 = 1. The coordinates on H are x2, . . . , xn. Let Lbe the line passing through N and the point (1, x2, . . . , xn). Consider LX.The line L is the set (t,tx2, . . . , txn), t k, hence the k-points o L Xcorrespond to the roots o the ollowing equation in t:

t2Q(1, x2, . . . , xn) + tL(1, x2, . . . , xn) = 0.

The root t = 0 is the point N, but it is the other root t =

L/Q that is

interesting to us. Dene the rational map : An1k > X by sending(x2, . . . , xn) to the other (residual) intersection point:

(x2, . . . , xn) = L(1, x2, . . . , xn)Q(1, x2, . . . , xn)

(1, x2, . . . , xn).

The image is obviously in X. The inverse map sends (x1, x2, . . . , xn) to(x2/x1, . . . , xn/x1). (Check that this is indeed the inverse to , both rightand let.) QED

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(2) A cubic surace. Assume that the characteristic o k is not 3, and

that k contains a non-trivial cubic root o 1. Consider the hypersuraceX A3k given by the equation

x31 + x32 + x

23 = 1.

It contains two skew lines (non-parallel with empty intersection)

L0 = {(1, x, x)}, x k, L1 = {(y, y, 1)}, y k.Let us write Px = (1, x, x), Qy = (y, y, 1). Let Lx,y be the line passingthrough Px and Qy,

Lx,y = {tPx + (1 t)Qy}, t k.We now contruct a rational map : A2k > X by sending (x, y) to the third(residual) point o intersection oLx,y with X. More precisely, substitutingtPx + (1 t)Qy into the equation o X we get a cubic polynomial in t withcoecients in k(x, y). It has two obvious roots t = 0 (the point Qy) and t = 1(the point Px). Ater checking that its degree is exactly 3 (this is enough tocheck or a particular choice o x and y, say x = 1, y = 0) we can write it asc(x, y)t(t 1)(t (x, y)), where (x, y) k(x, y). Set

(x, y) = (x, y)Px + (1 (x, y))Qy.By construction this is rational map A2k > X.

Exercise. Find (x, y), then check that is a birational equivalence.There is a geometric construction o the inverse map. Let R be a point

o X\ (L0 L1). Let i be the plane spanned by R and Li. Then or R in anon-empty Zariski open subset o X the intersection 0 L1 is a single pointon L1, call it Q. Similarly, 1 L0 is a point on L0, call it P. We have awell-dened rational map

f : X

\(L0

L1)

> A2k, f(R) = (P, Q).

Check that f is inverse to .

(3) Conic bundles over the ane line. Let k be algebraically closed.Let

F(x0, x1, x2) =

ij0, i0+i1+i22fi0,i1,i2(t)x

i00 x

i11 x

i22

be a rank 3 quadratic orm over the eld K = k(t). By Tsens theorem (seeTheorem 2.15 below) the conic C A2K given by F(1, x1, x2) = 0 has a K-point. But we know rom (1) that a non-conical quadric with a rational point

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is rational. Hence C is rational over K, in other words, K(C) = K(y) =

k(t, y) is a purely transcendental extension o K, and hence o k as well.Multiplying the coecients o F by a common multiple, we can assume

that they are actually in k[t]. Hence the equation F(t; 1, x1, x2) = 0 denes asurace X A3k. I we assign to t a value in k we get a plain conic. Thus Xis a pencil (1-parameter amily) o conics, or a conic bundle. We thus provedthat any conic bundle over the ane line over an algebraically closed eld isrational.

Exercise. Let X be a smooth cubic surace, and L X be a straightline. Planes passing through L orm a 1-dimensional amily. Show thator almost all planes the intersection

X is the union o L and a conic.

Deduce that X is birationally equivalent to a pencil o conics. Now (3) givesanother proo o rationality o X.

2.4 Smooth and singular points

Let X Ank be an ane variety, and suppose that the ideal oX is generatedby polynomials f1, . . . , f m. Let P = (a1, . . . , an) be a point o X. We denethe partial derivatives fi/Tj as partial derivatives o a polynomial (in apurely algebraic way, this works over any eld). Then we get some constantsfi/Tj(P)

k. Recall that an ane subspace o Ank is a translation o a

vector subspace.Defnition. The ane subspace o kn given by the system o linear

equations in T1, . . . , T n

nj=1

fi/Tj(P)(Tj aj) = 0, i = 1, . . . , m,

is called the tangent space to X at P = (a1, . . . , an), as is denoted by TX,P.

Exercise. Let m = 1. Show that TX,P is the union o lines passing

through P such that the restriction o f to this line is a polynomial in onevariable with a multiple root at P.

All we need to know to compute TX,P are partial derivatives o the equa-tions deniting X at P. Thus it makes sense to dene TU,P, where U is anopen subset o X containing P, by the same ormula, so that TU,P = TX,P.

We have a unction rom X to non-negative integers given by dim(TX,P).

Lemma 2.8 (Upper semi-continuity) The subset o X given by the con-dition dim(TX,P) s is a closed subset o X.

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Proo. The condition dim(TX,P)

s is equivalent to the condition that

the rank o the matrix (fi/Tj(P)) is at most n s, which is equivalentto the vanishing o the determinants o all its square submatrices o sizen s + 1. But these determinants are polynomials in a1, . . . , an, whichimplies our statement. QED

It ollows that the subset o X consisting o points where dim(TX,P) takesits minimal value is open. It is non-empty, and X is irreducible, hence thissubset is dense. The points in this subset are called smooth or non-singular,and all the other points are called singular.

Exercises. When is the curve X A2k given by xa + yb, where a and bare positive integers, non-singular at (0, 0)?Show that any hypersurace dened by a homogeneous polynomial odegree at least 2 is singular at the origin.

Proposition 2.9 Let mP = (T1 a1, . . . , T n an) k[X] be the maximalideal o a point P in the coordinate ring o X. Then the tangent space TX,P(considered as a vector space with origin at P) is canonically dual to thequotient ring mP/m

2P.

Proo. We rst do an exercise in linear algebra. Let V be a vector space,V its dual space (that is, the space o linear orms V

k), and Sa subspace

o V. Consider the subspace

W = {v V|f(v) = 0, or any f S}.

Its dual space W can be identied with V/S (the restrictions o linearunctions on V). Since W is canonically isomorphic to (W) we concludethat W is canonically isomorphic to (V/S).

Ater a translation in kn we can assume without loss o generality thatai = 0, i = 1, . . . , n. Let V = k

n. The coordinates Ti orm a basis o thedual space V. Let S V be the subspace o linear terms o unctionsrom I(X), that is, linear combinations o

nj=1 fi/Tj(P)Tj . The k-vectorspace mP/m

2P consists o linear combinations o Tis modulo linear terms o

unctions rom I(X). This means that mP/m2P = V

/S. On the other hand,by denition TX,P V is the space o zeros o the unctions rom S. Nowthe exercise in linear algebra above gives the required isomorphism. QED

The proposition implies that the tangent space to X at P is an intrinsicinvariant o X, in the sense that it only depends on the isomorphism class oX, and not on the particular embedding.

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Exercise. Prove that computing mP/m2P or X and or an open subset

U X containing P gives the same thing. (The resulting vector spaces arecanonically isomorphic.)

Let p : Ank Amk be a collection o polynomials p1, . . . , pm in n variablesthat denes a morphism p : X Y. The mn-matrix o partial derivativesJ = (pi/xj) is called the Jacobian matrix. Using the chain rule or partialderivatives one easily checks that J denes a linear map p : TX,P TY,Q,where Q = p(P). The construction o p rom p is unctorial in the sensethat i q : Y Z is a morphism, then (qp) = qp. In particular, i p isan isomorphism, then so is p. This is a practical way to check that a givenmap is not an isomorphism at a given point.

2.5 Dimension. Application: Tsens theorem

Defnition. Let X be an ane variety over a eld k. The transcendencedegree o k(X) over k is called the dimension o X. The dimension o aclosed ane set is dened as the maximum o the dimensions o its irreduciblecomponents.

For example, dim(Ank ) = n since k(Ank ) is the purely transcendental eld

extension k(x1, . . . , xn).

Proposition 2.10 LetX Ank be a variety, Y X a subvariety, Y = X.Then dim(Y) < dim(X).

Proo. Let tr.deg.kk(Y) = m, and choose u1, . . . , um k[X] such thattheir images in k[Y] are algebraically independent. Then u1, . . . , um arealgebraically independent in k(X). In particular, tr.deg.kk(X) m. Forcontradiction assume that tr.deg.kk(X) = m. Since Y = X the idealI(Y) k[X] is non-zero. Any u I(Y), u = 0, must be algebraicallydependent on u1, . . . , um. Thus there exists a polynomial F(t, t1, . . . , tm) =

iai(t1, . . . , tm)t

i such that F(u, u1, . . . , um) is zero in k[X]. We can as-

sume that the ai are polynomials, and that F is irreducible in all the vari-ables t, t1, . . . , tm. In particular, the constant term a0(t1, . . . , tm) is notthe zero polynomial. Since the image o u in k[Y] is zero, the image oa0(u1, . . . , um) is zero in k[Y]. This gives an algebraic relation between theimages o u1, . . . , um, contrary to our choice o u1, . . . , um. This contradic-tion proves that tr.deg.kk(X) < m. QED

Corollary 2.11 A subvariety X Ank has dimension n 1 i and only iX is a hypersurace. Then I(X) is a principal ideal o the polynomial ringk[T1, . . . , T n].

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Proo. It remains to show that dim(X) = n

1 implies that X is a

hypersurace. Since the dimensions are diferent we have X = Ank . Let F bea non-zero element in I(X). We write F as a product o irreducible actorsF = F1 . . . F m. Then X = (X {Fi = 0}). Since X is irreducible we haveX = X {Fj = 0} or some j, hence Fi I(X). Replace F by Fj. ThenX is contained in the irreducible hypersurace given by F = 0. Since thedimensions are equal, these varieties coincide by Proposition 2.10. QED

Theorem 2.12 The dimension o the tangent space at a smooth point o Xequals dim(X).

Proo. We know that X is birationally equivalent to a hypersurace V km+1 given by some (non-constant) polynomial F = 0. Since k(X) = k(V)we conclude that dim(X) = dim(V).

The dimension o the tangent space to X at any smooth point is thesame, and can be computed in any non-empty open subset o X. Wee canarrange that the birational map X > V is an isomorphism on this openset. This reduces the whole computation to V.

By Proposition 2.10 dim(V) m. On the other hand, the transcendencedegree ok(V) is at least m. Indeed, F depends on at least one variable, sayx1. Then the images o x2, . . . , xm+1 in k[V] are algebraically independent(as otherwise we would have a polynomial G(x2, . . . , xm+1) in the principalideal (F) which is impossible since F depends on x1 whereas G does not).Thereore, dim(V) = m.

On the other hand, the dimension o TV,P or P in a certain dense opensubset o V is m: i all the partial derivatives o F vanish everywhere on V,they must belong to the principal ideal (F) by Hilberts Nullstellensatz. Sincethey have degrees less than the degree o F, they must be zero polynomials.I the characteristic o k is zero, this implies that F is a constant, whichis a contradiction. I the characteristic is p, then F is a p-th power, whichcontradicts the irreducibility o F. QED

We quote the ollowing result without proo (see Ch. 1 o [Shaarevich]or Ch. 11 o [Atiyah-McDonald]).

Theorem 2.13 Let k be an algebraically closed eld. Let X Ank be avariety, F be a polynomial taking both zero and non-zero values on X. Thendim(X {F = 0}) = n 1.

Corollary 2.14 Let F1, . . . , F m, m n, be homogeneous polynomials in nvariables. Then the closed ane set X given by F1 = . . . = Fm = 0 hasdimension at least n m. In particular, X is not empty.

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Proo. Note that the all-zero point is in X. Thus at each successive

intersection the dimension drops at most by one. QED

The natural place o this statement is in intersection theory o subvarietieso the projective space. To demonstrate its orce we now deduce just onecorollary.

Recall that k is algebraically closed. Let

F(x0, x1, x2) =

ij0, i0+i1+i22fi0,i1,i2(t)x

i00 x

i11 x

i22

be a homogeneous polynomial o degree 2 with coecients in k[t].

Theorem 2.15 (Tsen) There exist non-zero polynomials p0(t), p1(t), p2(t)with coecients ink such thatx0 = p0(t), x1 = p1(t), x2 = p2(t) is a solutiono the equation F(x0, x1, x2) = 0.

Proo. Let m be a positive integer. Polynomials o degree m orm anm + 1-dimensional vector space over k. Hence the dimension o the vectorspace o coecients o p0(t), p1(t), p2(t) is 3m + 3. We observe that theconditions on these coecients that must be satised in order or x0 = p0(t),x1 = p1(t), x2 = p2(t) to be a solution, are given by homogeneous (quadratic)polynomials. Let us compute the number o these conditions.

Let be the maximum o the degrees o the fi0,i1,i2(t). Suppose that thedegree opj (t), j = 1, 2, 3, is at most m. Then the degree oF(p0(t), p1(t), p2(t))is at most + 2m. This means that the coecients o the pj(t) must satisy + 2m + 1 homogeneous polynomial conditions (1 must be added to provideor the zero constant term). For large m we have 3m + 3 > + 2m +1, hencethe number o variables exceeds the number o equations. By Corollary 2.14the dimension o the closed ane set o polynomials that are solutions, ispositive. We conclude that there exist such non-zero polynomials. QED

A more general approach to dimension. Let R be a ring. The Krulldimension o R is dened as the supremum o all integers n such that there

exists a chain I0 I1 . . . In o distinct prime ideals o R. For example,the Krull dimension o a eld is 0. The dimension o Z and, more generaly,o the ring o integers in a number eld is 1. The dimension o k[T] is also1. The dimension o k[X, Y] is 2, etc.

In act all given denitions o dimension are equivalent.

Theorem 2.16 LetX be an ane variety, then the Krull dimension o k[X]equals the transcendence degree o k(X).

This can be deduced rom Theorem 2.13.

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3 Projective geometry

3.1 Projective varieties

Below is the list o main notions, and the outline o diferences with the anecase.

The projective space Pnk is the set o equivalence classes o points o An+1k \

{(0, . . . , 0)}, where two points are equivalent i they difer by a commonnon-zero multiple. The equivalence class o (x0, x1, . . . , xn) is denoted by(x0 : x1 : . . . : xn).

The zeros o homogeneous polynomials (also called orms) are closed pro-

jective sets. One denes open sets as their complements. This gives rise toZariski topology on Pnk . The condition Ti = 0 denes an open subset o Pnkisomorphic to the ane space Ank with coordinates T0/Ti, . . . , T n/Ti. We getn + 1 ane spaces which provide an open covering o Pnk .

Let f(T1, . . . , T n) be a polynomial o degree d. It can be written as thesum f = f0 + . . . + fd, where fi is a orm o degree i. The homogenizationof is the orm o degree d in n + 1 variables given by

F(T0, T1, . . . , T n) = F = Td0 f0 + T

d10 f1 + . . . + fd.

I X

Ank is a closed ane set, then associating to polynomials in the ideal

o X their homogenizations denes the projective closure o X.

There are serious reasons or working with projective varieties:(1) The set o complex-valued points o Pn is compact in the usual com-

plex topology (e.g. P1 over C is just the Riemann sphere),(2) classications are simpler (e.g. quadratic orms are classied only by

their rank),(3) intersection theory is simpler (any two curves in the plane, or more

generally, any n hypersuraces in Pnk have a common point).

An ideal J k[T0, . . . , T n] is called homogeneous i whenever f Jwe also have fi J, where fi is the homogeneous part o f o degree i.Homogeneous ideals are generated by homogeneous polynomials.

The ane cone o a closed projective set X is the set o points in An+1kgiven by the same (homogeneous) equations as X. To a closed projective setX one associates the ideal I(X) o all polynomials in k[T0, . . . , T n] vanishingon the ane cone o X. The ideal I(X) is clearly homogeneous. To anyhomogeneous ideal J one associates its set o zeros Z(J) Pnk . This alwaysgives a non-empty set unless J = 1 (empty ane cone) or J = (T0, . . . , T n)(the ane cone consists o the zero point). Hence the projective variant

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o Nullstellensatz, which immediately ollows rom the ane Nullstellensatz,

reads as ollows.

Theorem 3.1 (Projective Nullstellensatz) I J is a homogeneous ideal,then

(1) Z(J) = if the radical o J contains the ideal (T0, . . . , T n) (themaximal ideal o the zero point in An+1k ),

(2) i Z(J) = , then I(Z(J)) is the radical o J.

Projective variety is an irreducible closed projective set. (The denitiono irreducible is the same as in the ane case.) Quasi-projective varieties are

dense open subsets o projective varieties. Rational unctions on a projectivevariety X are ractions o orms o equal degree FG

, where G I(X), modulonatural equivalence: F

G= F1

G1i F G1 F1G I(X). The rational unction

FG

is called regular at a point P X i G(P) = 0. The eld o rationalunctions on X is again denoted by k(X), but it is not the eld o ractionso the ring o regular unctions on X! Indeed, the only regular unctions onP1k are constants

1: k[P1k] = k (easy exercise, in act every regular unctionon A1k P1k is a polynomial, and every non-constant polynomial has a poleat innity).

A rational map f : X

> Pnk is (a not necessarily everywhere de-

ned unction) given by (F0, . . . , F n), where Fi k(X), dened up to anoverall multiple rom k(X). A rational map f is regular at P X i thereexists a representative (F0, . . . , F n), such that all the Fis are regular at P,and (F0(P), . . . , F n(P)) = (0, . . . , 0). A morphism is an everywhere regularrational map.

Examples o projective varieties, rational maps and morphisms(a) Rational normal curves. This is a map f : P1k C Pn given by

f : (X : Y) (Xn : Xn1Y : . . . : XYn1 : Yn).

One checks that f is a morphism, whose image is given by equations T0T2 =T21 , T1T3 = T22 , and so on. The inverse map is given by

g : (T0 : . . . : Tn) (T0 : T1) = (T1 : T2) = . . . = (Tn1 : Tn)

which is everywhere dened (check!). Hence f is an isomorphism o P1k witha closed subvariety C Pn, called the rational normal curve o degree n.For n = 2 one recovers the rational parametrization o the conic.

1The same is true for any projective variety, see page 20.

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(b) The Veronese embedding o Pnk . This is a natural generalization o the

previous example, where one considers all monomials o degree d. This denesan isomorphism oPnk with a closed subvariety o P

Nk , where N = C

n+dn . For

d = 2 and n = 2 one gets the Veronese surace in P5k.(c) Any quadric in Q P3k is isomorphic to P1k P1k. (Think o two

amilies o P1ks on a quadric). The stereographic projection rom a pointP Q denes a birational map Q > P2k which is not a morphism (notregular at P). Neither is the inverse map a morphism (two lines o Q passingthrough P are contracted to points).

(d) The Segre embedding Pnk Pmk Pnm+n+mk . This map associates totwo vectors their tensor product. The map is well dened everywhere, and

is a bijection with the image. The image can be interpreted as the set onon-zero matrices o rank one. In particular, when n = m = 1 we get aquadric in P3k.

The Segre embedding can be used to dene the structure o a projectivevariety on PnkPmk . As a consequence we realize the product o two projectivevarieties as a closed subset o some projective space.

(e) Elliptic curve. These are smooth plane cubic curves with a k-point.It can be proved that such a curve is isomorphic to a projective curve y2z =x3 + axz2 + bz3. The map (x : y : z) (x : y : z) is a non-trivialautomorphism with exactly our xed points (there are three obvious xed

points with y = 0, z = 1, and also (0 : 1 : 0)). This shows that the curveis not P1k, as any element o P GL(2) that xes three diferent point is anidentity. This ollows rom the act that Aut(P1k) =PGL(2), see Proposition5.4 below.

3.2 Morphisms o projective varieties

Examples show that i f : X Y is a morphism o ane varieties, thenf(X) Y need not be a closed subset. A standard example is X A2kgiven by xy = 1, mapped to A1k by the morphism (x, y) x. It is anotherpleasant eature o projective varieties that the image o a projective variety

under a morphism is always closed! We start with a lemma.

Lemma 3.2 LetX andY be quasi-projective varieties. The graph f o anymorphism f : X Y is closed in X Y.

Proo. It is enough to consider the case Y = Pnk . Consider the morphism(f, Id) : XPnk Pnk Pnk , and let Pnk Pnk be the diagonal (the grapho the identity map). Then f = (f, Id)

1(). It is clear that the preimageo a closed subset is closed. Thus it is enough to prove that Pnk Pnk is

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closed. But is given by TiXj = TjXi or all i and j, and hence is closed.

QED

Theorem 3.3 Let X be a projective variety, and Y be a quasi-projectivevariety. Then the projection to the second actor p : XY Y maps closedsubsets to closed subsets.

Corollary 3.4 Let f : X Y be a morphism, where X is a projectivevariety. Then f(X) is closed in Y.

Proo. Apply the theorem to p(f) = f(X). QED

Proo o the theorem. We can assume that X = Pnk . Next, the closedness

can be checked locally in a small ane neighbourhood o every point. Thuswe can assume that Y is a closed subset oAmk . But then Y can be replaced byAmk . All in all, we see that the general statement ollows rom the statementor the projection Pnk Amk Amk . Let us prove it.

Let a closed subset in Pnk Amk be given by equationsgi(T0, . . . , T n; Y1, . . . , Y m) = 0, i = 1, . . . , s,

where the gis are homogeneous polynomials in variables T0, . . . , T n whose co-ecients are polynomials in variables Y1, . . . , Y m. We must prove that the setU oy = (y1, . . . , ym)

km such that the ideal Jy = (gi(T0, . . . , T n; y1, . . . , ym))

has no zeros in Pnk , is open. By the projective Nullstellensatz, Z(Jy) = ifall Tis are contained in the radical o Jy, that is, T

lii Jy or some li. Let

l = l0+. . .+ln, then in any monomial o degree l at least one variable Ti entersin the power greater or equal to li. Hence Jy contains the ideal Il generatedby all monomials o degree l. Let Ul be the set o y = (y1, . . . , ym) km suchthat Jy Il. Then U is the union o all Uls, l = 1, 2, . . . , hence it is enoughto prove that each Ul Amk is open.

I one can represent a monomial as a linear combination o homogeneouspolynomials, then the coecients can be chosen to be homogeneous polyno-mials. Let di be the degree o gi(T0, . . . , T n; y1, . . . , ym). Then y Ul if theproducts o the gi(T0, . . . , T n; y1, . . . , ym) with all monomials o degree l dispan the vector space o orms o degree l. Equivalently, the correspondingmatrix has maximal rank, which is the condition on the non-vanishing o thedeterminants o its square submatrices o maximal size. This cleary describesan open subset. (Which may well be empty, but it does not matter!) QED

This result hints at the ollowing denition.

Defnition. A morphism f : X Y o quasi-projective varieties isproper i it is a composition o a closed embedding X Pnk Y and theprojection Pnk Y Y.

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It ollows rom the previous theorem that f(X)

Y is closed provided

f is proper. It is clear that i f is proper, and P Y, then f1(P) is aprojective variety.

Proposition 3.5 Any regular unction on a projective variety is constant.Any morphism o a projective variety X to an ane variety sends X to apoint.

Proo. The second statement clearly ollows rom the rst one. A regularunction gives rise to a morphism f : X P1k whose image does not containthe point at innity. Hence f(X) = P1k is a union o nitely many points.Since X is irreducible, f(X) is just one point. QED

4 Local geometry

4.1 Localization, local rings, DVR

Let R be a ring. A subset S R is called multiplicative i it is closed undermultiplication and contains 1. The localization S1R o R with respect to Sis dened as the set o ormal ractions a

b, with a R and b S, up to the

equivalence relation: ab

= a1b1

if (ab1a1b)s = 0 or some s S. When R hasno zero divisors, the natural map a

a

1 is an injective homomorphism orings, so that we can think o R as a subset o S1R. Then S1R is simplythe ractions with restricted denominators. Note that i I R is an ideal,then S1I is an ideal in S1R.

Exercise. A localization o a Noetherian ring is Noetherian.

Examples. (1) I S = R \ {0}, then S1R is just the eld o ractions.(2) Let P R be a prime ideal, then S = R \ P is a multiplicative

system. Then S1R is denoted RP and is called the localization o R at P.The ring RP has a very important property: S

1P is its only maximal ideal(every element not in S1P is by denition invertible, hence S1P containsall other ideals). Such rings have a name.

Defnition. Rings with just one maximal ideal are called local rings.

Examples: (a) rational numbers such that p does not divide the denomi-nator,

(b) p-adic integers Zp,(c) rational unctions in one variable over a eld k such that the denom-

inator does not vanish at 0,(d) ormal power series k[[T]],

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(e) rational unctions in two variables such that the denominator does

not vanish at (0, 0).

In all these examples, except the last one, the maximal ideal is principal.Such ring orms the simplest class o local rings.

Defnition. A local ring whose maximal ideal is principal is called adiscrete valuation ring (DVR).

Defnition. Any generator o the maximal ideal o a DVR is called auniormizer or a local parameter.

Lemma 4.1 Let R be a Noetherian integral domain, t

R\

R. Theni=1(ti) = 0.

Proo. ([Reid, UCA], 8.3) For contradiction let x = 0 be contained in(ti), or any i 1. We write x = tixi, then (x) (x1) (x2) . . . is anascending chain o ideals. Then (xi+1) = (xi) = (txi+1) or some i. Hencexi+1 = taxi+1, but this implies t R since xi+1 = 0 and R is an integraldomain. Contradiction. QED

Proposition 4.2 LetR be a DVR with maximal ideal m = () and the eldo ractions K. Then

(1) there exists a discrete valuation on K dened by R, that is, a ho-momorphism v : K Z such that v(x + y) min{v(x), v(y)}, andR \ {0} = {x K|v(x) 0}, m \ {0} = {x K|v(x) 1}.

(2) All non-zero ideals o R are principal ideals mi = (i), i 1.

Proo. (1) By the previous lemma every x R, x = 0, is in mi \ mi+1or some i 0. Then x = iu, where u R must be a unit. We also writey = ju with u R. Then xy = i+j uu, hence v(xy) = v(x) + v(y).Suppose that i j, then x + y = i(u + jiu), hence v(x + y) v(x) =min{v(x), v(y)}. We now can extend v to K by the ormula v(x/y) =v(x)

v(y). The remaining properties are clear.

(2) Let s be the innum o v on the ideal I R, then there existsx I such that v(x) = s. Then ms = (x) I. On the other hand,v(I \ {0}) {s, s + 1, . . . }, and ms \ {0} = {x K|v(x) s}, henceI ms. All in all we have I = ms. QED

Observe that (1) implies that R = {x K|v(x) = 0}.A DVR, like any other PID, is a UFD (note my excellent style).

Exercise. Prove that any DVR is normal (=integrally closed in its eldo ractions).

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The local ring o a subvariety. Let X

Pnk be a variety, and Y

X

a subvariety. We dene the local ring OY o X at Y as the subring o k(X)consisting o the rational unctions that are regular on an open set with anon-trivial intersection with Y. It is easy to check the ring axioms. An idealmY OY consisting o unctions vanishing on Y is maximal since OY/mYis the eld k(Y). Any unction in OY \ mY is invertible in OY, so that mYis the unique maximal ideal. Thus OY is a local ring.

Let U X be an ane open subset o X. Then OY is the localizationo k[U] at the prime ideal I(Y U) (i.e. we divide by the unctions that donot vanish on Y).

4.2 Regular local rings

Defnition 4.3 A Noetherian local ring R with maximal idealm and residueeld k is regular i the Krull dimension o R is dimk(m/m

2).

Key example. It ollows rom Proposition 2.9, Theorem 2.12 and Theorem2.16 that i P is a smooth point o X, then OP is a regular local ring.

A theorem o Auslander and Buchsbaum says that a regular local ring isa UFD ([Matsumura], 20.3). A very important corollary is

Theorem 4.4 The local ring o a smooth point o an algebraic variety is aUFD.

See Appendix A.2 or a sketch o proo o this theorem.

Exercise. Let X be a curve in P2k, and P X a smooth point. Here isa low level proo that OP is a DVR (and hence also a UFD). The questionbeing local we can assume that X A2k, P = (0, 0). To x ideas supposethat TX,P is the line y = 0. Then X is given by

y + i+j2

aijxiyj = 0.

The maximal ideal mP OP is mP = (x, y). All we need to do is to showthat it is principal. I claim that mP = (x). Indeed, on X we have

y = x

i+j2,i1 aijxi1yj

1 +

j2 a0jyj1 ,

and the raction is regular at P, that is, belongs to OP. Hence y (x).

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4.3 Geometric consequences o unique actorization in

OPThe ollowing statement is a generalization o Corollary 2.11.

Theorem 4.5 In a suciently small neighbourhood o a smooth point anysubvariety o codimension 1 can be given by one equation.

Proo. Let P be a smooth point o X contained in a subvariety Y X ocodimension 1. Replacing X by a neighbourhood oP we can assume that Xis ane with coordinate ring k[X]. Choose a non-zero element f I(Y) k[X]. Now

OP

k(X) is a UFD by Theorem 4.4, so that f is a product o

prime actors f = f1 . . . f r, fi OP. By urther shrinking X we can assumethat fi k[X], i = 1, . . . , r. We have

Y = (Y {f1 = 0}) . . . (Y {fr = 0}).Y is irreducible hence Y = Y {fi = 0} or some i. Replacing f by this fiwe can assume without loss o generality that f is a prime element o OP.

Consider the closed subset Z = Z(f) X. By Theorem 2.13 the codi-mension o Z is 1. Now Proposition 2.10 implies that Y is an irreduciblecomponent o Z, so that we can write Z = Y Y, where Y is the uniono irreducible components o Z other than Y. I P /

Y we replace X by

a small ane neighbourhood o P that does not intersect with Y. ThenY = Z(f). Let us show that f is an equation o Y, that is, the ideal I(Y) isgenerated by f. (Recall that we shrinked X a several times by now, so thatf does not have to be a global equation o Y in the original X). Indeed, letg I(Y) = I(Z(f)). By the Nullstellensatz there exists a positive integer msuch that gm is divisible by f in k[X]. Then the same thing is also true inOP. Since OP is a UFD we conclude that g is divisible by f.

It remains to exclude the possibility that P Y. Choose h I(Y) \I(Y), h I(Y) \ I(Y), then hh I(Z) whereas h, h / I(Z). By theNullstellensatz there exists a positive integer n such that (hh)n is divisible

by f in k[X]. Then the same thing is also true in OP. Since OP is a UFD weconclude that h or h is divisible by f, hence vanishes on Z. Contradiction.QED

Corollary 4.6 LetX be a smooth variety. Then the local ring o a subvarietyY X o codimension 1 is a DVR.

Proo. The maximal ideal mY OY is generated by any local equation oY in X, which exists by the previous theorem. Hence mY is principal whichmeans that OY is a DVR. QED

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Recall that a rational map is regular on a non-empty open set. For smooth

projective varieties there is a better result:

Corollary 4.7 A rational map o smooth projective varieties is regular awayrom a closed subset o codimension at least 2.

Proo. Without loss o generality we assume that the target space is Pnk .So let f : X > Pnk be a rational map. There exist a dense open setU X such that f restricted to U is regular, and U is maximal with thisproperty. Suppose that Z X is a subvariety o codimension 1 contained inX\ U. Write f = (f0, . . . , f n), where fi k(X). Multiplying all the fi by acommon multiple does not change f. We can choose this common multipleso that all the fi are in the DVR OZ and have no common actor. Thenat least one fi is not divisible by the generator o the maximal ideal o OZ,hence is non-zero at some point o Z. Thereore, f = (f0, . . . , f n) is regularon an open set which has a non-trivial intersection with Z. This contradictsthe act that U is the largest open set on which f is regular. QED

Corollary 4.8 Any rational map rom a smooth and projective curve to aprojective variety is a morphism.

Corollary 4.9 A birational map between smooth and projective curves is an

isomorphism.

Hence birationally equivalent smooth projective curves are isomorphic.The same is very ar rom being true in higher dimensions (there are manyexamples, e.g. the projection P2k > P1k rom a point (0 : 0 : 1) given by(x0 : x1 : x2) (x0 : x1), or the stereographic projection o a quadric). Thesmoothness assumption is also very important. Indeed, let C be the imageo the morphism f : P1k P2k dened by (x0 : x1) (x30 : x0x21 : x31). Thenf : P1k C is a birational map, but C is not isomorphic to P1k (C containsa singular point (1 : 0 : 0), whereas P1k is smooth).

5 Divisors

5.1 The Picard group

Let X be a smooth variety. Let Div(X) = {Y nYY} be the ree abeliangroup generated by all subvarieties o codimension 1 Y X. The elementso Div(X) are called divisors. Divisors o the orm Y, where Y X isa subvariety o codimension 1, are called irreducible. A divisor is called

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efective i all the coecients nY

0, and at least one coecient is positive.

Then one writes

Y nYY > 0. The union o the subvarieties Y such thatnY = 0 is called the support o

Y nY.

Corollary 4.6 allows us to dene the divisor o a rational unction f k(X) as

div(f) =

codim(Y)=1

valY(f).Y,

where valY : k(X) Z is the valuation attached to the irreducible divisor

Y X. The ollowing lemma shows that this sum is nite.

Lemma 5.1 Letf

k(X). For almost all subvarieties Y

X o codimen-

sion 1 we have valY(f) = 0.

Proo. Let U be a dense open set where f is regular, and U be a denseopen set where f1 is regular. Then f OY or any subvariety Y Xo codimension 1 that has a non-empty intersection with U U. Hence ivalY(f) = 0, then Y is one o the nitely many irreducible components oX\ (U U) o codimension 1. QED

The divisors o rational unctions are called principal.

Proposition 5.2 Letf

k(X) be such that div(f) = 0. Then f is regular

on X.

Proo. Let P X be a point where f is not regular. Since OP is aUFD we can write f = ufi11 . . . f

inn , where u OP, and the fi are irreducible

elements. Since f is not regular at P we must have is < 0 or some s. Butfs comes rom a regular unction on some afne open neighbourhood U oP. Let Y U be given by fs = 0. Its Zariski closure is a subvariety o X ocodimension 1 (c. Theorem 2.13). But then valY(f) = ns < 0 contrary toour assumption. QED

Corollary 5.3 Let X be a smooth and projective variety. Then a rationalunction is determined by its divisor up to a constant.

Proo. The ratio o two unctions with the same divisor is regular byProposition 5.2. But Proposition 3.5 says that any regular unction on aprojective variety is a constant. QED

Defnition. The group o classes o divisors modulo principal divisors iscalled the Picard group o the variety X, and is denoted by Pic(X).

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Remark. For a smooth and projective variety Xthere is an exact sequence

o abelian groups

1 k k(X) Div(X) Pic(X) 0.

The words exact sequence mean that the kernel o each homomorphism isthe image o the previous one.

Remark. Note the analogy with the construction o the class group Cl(K)o a number eld K. In that case we have an exact sequence

1 OK K Frac(K) Cl(K) 0,

where Frac(K) is the group o ractional ideals o K with respect to multi-plication.

The Picard group contains a lot o inormation about X.

Examples. Pic(Ank ) = 0.Pic(Pnk ) = Z is generated by the class o a hyperplane.Pic(Pn1k . . . Pnmk ) = Zm.Pic(P2k \ C) = Z/d where C is a curve o degree d.The Picard group o a smooth cubic surace is isomorphic to Z7 (and is

generated by 27 lines on it, c. [Reid, UAG], Ch. 3).

Functoriality o the Picard group. Associating the abelian groupPic(X) to a smooth variety X is a natural construction in the sense that toany morphism f : X Y there corresponds a homomorphism f : Pic(Y) Pic(X). This f is unctorial: i g : Y Z is another morphism, then(gf) = fg. We only outline the construction o f or smooth and projec-tive curves. See Appendix B.2 or a sketch o a more complicated denitiono f in the case o varieties o arbitrary dimension.

Consider a surjective morphism f : X Y o smooth and projectivecurves. We know by Corollary 4.6 that the local ring at any point o X or Yis a DVR. This makes it possible to dene the inverse image homomorphism

f : Div(Y) Div(X) as ollows. Let P Y, Q X such that f(Q) = P.Then we have an injective homomorphism o local rings f : OP OQ whichallows us to think about OP as a subring oOQ. Let uP be the local parameterat P, and let valQ be the valuation dened by the local ring OQ k(Y). Wedene

f(P) =

QX,f(Q)=PvalQ(uP).Q,

and then extend this to Div(Y) by linearity. It can be checked directly thati g k(Y) is a rational unction, then div(g f) = f(div(g)). Thus f

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gives rise to a homomorphism o Picard groups:

f : Pic(Y) Pic(X).

Example. Let X be the conic in P2k given by x0x1 = x22, Y = P

1k, and

f : X Y be given by (x0 : x1 : x2) (x0 : x1).Let P = (1 : 0) Y. We compute f(P). It is clear that f1(P) = Q =

(1 : 0 : 0). We rst check that x2/x0 is a local parameter at Q. Indeed, inthe ane plane given by x0 = 0 where the coordinates are t1 = x1/x0 andt2 = x2/x0, the equation o X is t1 = t

22. Hence the maximal ideal o Q is

(t1, t2) = (t2). Next, t1 is a local parameter at P. In k(X) we have t1 = t22,

thus valQ(uP) = 2, and nally f(P) = 2Q.Now we want to compute f(R), where R = (1 : 1) Y. It is clear thatf1(R) = {Q+, Q}, where Q = (1 : 1 : 1). Obviously t1 1 is a localparameter at R. Let us nd a local parameter at Q+. The maximal ideal atthis point is (t1 1, t2 1). But in k(X) we have t1 1 = (t2 1)(t2 + 1) sothat (t1 1, t2 1) = (t2 1) = (t1 1) since t2 1 is regular and invertibleat Q+. A similar computation shows that t1 1 is also a local parameter atQ. Thus valQ(uP) = 1, hence f

(P) = Q+ + Q.

5.2 Automorphisms o Pnk

and o Ank

As an application o the act that Pic(Pnk ) is isomorphic to Z we prove theollowing useul statement.

Proposition 5.4 Any automorphism o Pnk is given by a linear transorma-tion o the corresponding n + 1-dimensional vector space, so that Aut P1k =PGL(n + 1).

Proo. The group o non-degenerate matrices GL(n + 1) acts on kn+1.Lines through the origin are mapped to lines through the origin, henceGL(n + 1) also acts on Pnk . The only matrices that act trivially on the lines

through the origin are scalar matrices. Hence PGL(n + 1) = GL(n + 1)/kacts aithully on Pnk (the only element acting trivially is the identity). Wemust show that any automorphism o Pnk is o such a orm.

We rst consider P1k. Any rational map P1k > P1k is given by a

bijective rational unction t f(t). Then f(t) = (at + b)/(ct + d) or somea,b,c,d k. (It is easy to check that other rational unctions are neverbijective.) This morphism clearly comes rom the linear transormation withmatrix

a bc d

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Since Pic(Pnk ) = Z the induced morphism f sends a generator to a

generator. Let H Pnk be the projective subspace o codimension 1, alsocalled a hyperplane. We denote by [H] its class in Pic(P1k). The classes [H]and [H] are the only classes that generate Pic(Pnk ). But the hyperplaneclass [H] Pic(Pnk ) = Z contains efective divisors, whereas [H] does not.Hence f([H] ) = [H]. Thereore the inverse image o a hyperplane is ahyperplane. Now any rational map Pnk > Pnk is given by a collectiono n rational unctions f1, . . . , f n in the ane coordinates t1, . . . , tn. Thecondition f(H) = H says that

aifi = a denes a hyperplane or any

a, a1, . . . , an k, where not all o the ai are equal to 0. Write each fi as araction in lowest terms Fi/Gi. Then, or a, a1, . . . , an general enough, we

get a hyperplane only i all the Gi are o degree 1 and proportional, and theFi are o degree 1 (the details o this argument are let to the reader). Thisis precisely what we needed to prove. QED

Exercise. Compute Aut(A1k) by proving that every automorphism oA1k

extends uniquely to an automorphism o P1k which xes the point at innity.

Remark. In contrast, Aut(Ank ) or n 2 is very big. Note that (x, y) (x, y +g(x)), where g(x) is any polynomial, is an automorphism oA2k. Thereare also automorphisms o A2k dened by the elements o PGL(2) that pre-serve the line at innity. It can be proved that these automorphisms generatethe whole group. See [Shaarevich] or more details.

The Jacobian Conjecture. Let F(x, y), G(x, y) be polynomials in xand y. When is the map f : A2k A2k given by (x, y) (F(x, y), G(x, y))an automorphism o the ane plane?

Consider the Jacobian matrix

J(f) =

Fx

Fy

Gx

Gy

I g : A2k

A2k is another polynomial map, then one checks by using the

chain rule or partial diferentiation that J(fg) = J(f)J(g). I f is an au-tomorphism, then we can nd g such that fg is the identity map, so thatJ(f)J(g) is the identity matrix. This implies that det(J(f)) det(J(g)) = 1.This says that the polynomial det(J(f)) is invertible in the polynomial al-gebra k[x, y], and so must be a constant. The amous Jacobian Conjectureasserts that i det(J(f)) k, then f Aut(A2k). Many cases o polynomi-als o small degree are known, but the general case o this hard conjectureremains open.

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5.3 The degree o the divisor o a rational unction on

a projective curve

From this point on all our varieties are smooth and projective curves. Forcurves one denes the degree o the divisor

nPP as the integer

nP. It

is clear that this produces a surjective homomorphism deg : Div(X) Zwhich sends any point to 1.

Lemma 5.5 Letf : X Y be a surjective morphism o curves. Then k(X)is a nite extension o k(Y).

The proo is omitted. The number [k(X) : k(Y)] = dimk(

Y)k(X) is called

the degree o f, and is denoted by deg(f). A sketch o proo o the ollowingtheorem can be ound in Appendix B.1.

Theorem 5.6 Let f : X Y be a surjective morphism o smooth andprojective curves. Then the degree o the divisor f(P) equals deg(f), orany P Y.

In the notation o the example in the end o Subsection 5.1 we havedeg(f(P)) = deg(2Q) = 2, deg(f(R)) = deg(Q+ + Q) = 2. This agreeswith the act that the degree o the map f in that example is 2: indeed,

k(X) = k(t1) is a quadratic extension o k(Y) = k(t1).Corollary 5.7 The degree o the divisor o a non-zero rational unction ona smooth projective curve is zero.

Proo. I f is constant there is nothing to prove. By Corollary 4.8 anyrational non-constant unction f k(X) gives rise to a dominant morphismf : X P1k. Since X is projective, f(X) is closed, hence f(X) = P1k. Acomparison o denitions shows that the divisor div(f) equals f(0)f().By Theorem 5.6 deg(f(0)) and deg(f()) both equal to deg(f), hencedeg(div(f)) = 0. QED

Due to Corollary 5.7 deg descends to a surjective homomorphism deg :Pic(X) Z. Dene Pic0(X) as the kernel o deg.

Proposition 5.8 LetX be a smooth and projective curve.(i) The divisorPQ is principal or some P = Q i and only iX = P1k.(ii) Pic0(X) = 0 i and only i X = P1k.

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Proo. We already know that Pic(P1k) = Z is generated by the class o a

point. Hence deg is an isomorphism in this case, so that Pic0(P1k) = 0.To prove (i) let P = Q be points on X such that P Q = div(f). As

in the proo o Corollary 5.7 f denes a morphism f : X P1k. I we writediv(f) = f(0) f(), then the divisors f() and f(0) are efective anddisjoint. This implies that f(0) = P. By Theorem 5.6 the degree o themorphism f is 1, hence f is birational, hence is an isomorphism (Corollary4.9).

The degree o P Q is 0. Thus (ii) ollows rom (i). QEDAmong the plane curves this proposition applies to lines and conics.

5.4 Bezout theorem or curves

Let X Pnk be a smooth curve, and let F be a homogeneous orm in n + 1variables not vanishing identically on X. Cover X by open subsets Ui =X {Gi = 0}, where deg(Gi) = deg(F). Then F/Gi is a regular unction onUi. We dene the divisor o F by the ormula

div(F) =PX

valP(F/Gi).P,

where in the term corresponding to P we take any Gi such that P

Ui,or equivalently, Gi(P) = 0. The denition does not depend on what Ui wechoose or a given point P, because Gi/Gj is an invertible rational unctionon Ui Uj, so that or P Ui Uj we have valP(F/Gi) = valP(F/Gj ).Similarly, one shows that another amily o Gis produces the same divisordiv(F).

Defnition. The intersection index (X.F) is the degree o div(F).

Since the unction F/Gi is regular on Ui we see rom the denition thatdiv(F) 0, thereore (X.F) 0. The intersection index counts the num-ber o intersection points o X with the hypersurace F = 0 with correct

multiplicities.

Defnition. The degree o X in Pnk , denoted by deg(X), is the intersec-tion index o X with the hyperplane.

O course, we need to show that this denition makes sense, that is, itdoes not matter which hyperplane we take.

Theorem 5.9 (Bezout) The intersection index (X.F) depends only on Xand the degree o F. We have (X.F) = deg(X)deg(F).

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Proo. I F is another orm o the same degree, then div(F)

div(F)

is clearly the divisor o the rational unction F/F. Thus (X.F) = (X.F)since by Corollary 5.7 deg(div(F/F)) = 0. This shows that the intersectionindex depends only on deg(F) and not F itsel. To compute (X.F) we cannow replace F by any other orm o the same degree, or example by Hdeg(F),where H is linear. This proves the second statement. QED

Exercise. Let X P2k be a plane curve, P X be a smooth point, andL P2k be a line passing through P. Let div(L) be the divisor on X givenby a linear orm dening L.

1. Show that the multiplicity o P in div(L) is 1 i and only i L is notthe tangent line to X at P. Solution: This question is local. Assume thatP = (0, 0) X A2k, then X is given by 0 = ax + by+ terms o higherdegree in x and y. Since P is smooth, a and b are not both 0. To x ideasassume that b = 0. Then x is a local parameter at P. Indeed, the maximalideal o OP is (x, y), but modulo the equation o X we can write y as theproduct ox and a rational unction that is regular at P. Hence mP = (x). IL = x + y, then L = ( a

b)x+terms o degree at least 2 in x. Thereore,

i L is a linear orm such that L(P) = 0, then valP(L/L) = 1 i and only ib a= 0, that is, when the vectors (a, b) and (, ) are not proportional.On the other hand, TX,P is given by ax + by = 0.

2. The set o lines in P2k has a natural structure o the projective variety

P2k. Show that the map P TX,P denes a rational map f : X > P2k(called the Gauss map). Conclude that dim(f(X)) = 1.

3. By comparing the dimensions we see that P2k \ f(X) = . Thus thereare innitely many lines L such that div(L) is the sum o distinct points takenwith multiplicity 1. (Recall that k is algebraically closed hence innite.)

Thereore, deg(X) equals the maximal number o points in the intersectionX L. This is also true or the curves in Pnk , where we need to replace linesby hyperplanes. See [Shaarevich] or details. For plane curves, however, wecan also say that deg(X) equals the degree o the orm by which X is dened.Since a curve in Pnk is dened by two or more orms, this property has no

higher dimensional analogue. (Warning: two orms may not be enough todene a curve in P3k! An example is provided by the rational normal cubiccurve, that is, the image o P1k under the map (x : y) (x3 : x2y : xy2 : y3).)

Exercise. Let X P2k be a plane curve, P X be a smooth point. LetL P2k be the tangent line to X at P. Show that the multiplicity o P indiv(L) is 2 i and only i the Hessian matrix oX at P is non-degenerate. The

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Hessian matrix o the curve given by f(x0, x1, x2) = 0 is dened as ollows:

H(f) =

2f

x20

2f

x0x1

2f

x0x22f

x1x0

2f

x21

2f

x1x22f

x2x0

2f

x2x1

2f

x22

Solution (char(k) = 2): The question is local. Assume that P = (0, 0) X A2k and that L is given by y = 0. Then X is given by 0 = y +ax

2+bxy +cy2+terms o higher degree in x and y. As in the previous example one shows thatx is a local parameter at P. A direct computation shows that the Hessiano f at P is non-degenerate i and only i a

= 0. On the other hand, y can

be written as the product o the polynomial in one variable ax2+terms ohigher degree in x, and a rational unction that is regular at P.)

We dene deg(X), where X is a reducible curve, as the sum o the degreeso components. We then extend the Bezout theorem to reducible curves X,and get the same ormula (X.F) = deg(X)deg(F). Note that (X.F) is onlydened when F does not vanish identically on some component o X.

In the case o plane curves the Bezout theorem says that the number ocommon points o two (possibly reducible) curves without common compo-nents is the product o their degrees, provided we count points with correct

multiplicities. As an illustration o this we now prove Pascals mysterioustheorem.

Theorem 5.10 (Pascal) LetC P2k be a conic. For any six distinct pointso C consider the hexagon with vertices in these points. Then the commonpoints o the three pairs o opposite lines o the hexagon are collinear.

Proo. (Plucker) Let the sides o the hexagon be l1, m2, l3, m1, l2, m3 inthis order. We denote their equations by the same letters. Then Qx =l1l2l3 + xm1m2m3 is a cubic orm that depends on the parameter x k.We observe that or any x this cubic passes through the six vertices o the

hexagon. Let P C be a point diferent rom these six points. Thenli(P) = 0, mi(P) = 0 or i = 1, 2, 3, as no line contains more than two pointso a conic. Thus we can nd x k such that Qx(P) = 0. Write Q = Qx.Either C is a component o Q, or it isnt. In the second case by the Bezouttheorem C Q consists o at most six points, but there are visibly sevenpoints in this intersection. Hence the cubic orm dening Q is a product othe quadratic orm dening C and some linear orm L. Let Ai = li mi,i = 1, 2, 3. Since a line intersects a conic in at most two points, we haveAi / C. On the other hand, Ai Q, hence these points must be in L. QED

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Let X

P2k be a smooth curve o degree d given by f = 0. A point

P X is a called a ex i the tangent line TX,P intersects X in P withmultiplicity at least 3. We have seen that P is a ex o X i and only Pbelongs to the closed set given by det(H(f)) = 0, called the Hessian HX oX. It can be shown that i d 3, then the Hessian HX is a curve o degree3(n 2). By applying the Bezout theorem we see that every curve o degreeat least 3 has a ex, and at most 3n(n 2) o them. (Check that X is not acomponent o HX .)

We now study smooth plane cubics X P2k in some more detail.Exercise. (char(k) = 2, 3) Choose a ex P on smooth plane cubic X.

Suppose that P = (0 : 1 : 0), and TX,P

has the equation z = 0. Showthat X is then given by the ane equation y2 + axy = f(x), where f(x)is a polynomial in x o degree 3. Now show how to reduce to the equationy2z = x3 + axz2 + bz3.

Let us consider Pic0(X) where X is a smooth plane cubic. In the ollowingproposition we do not assume that k is algebraically closed.

Proposition 5.11 LetX be a smooth cubic curve in P2k with a k-point P0.Then the map P P P0 is a bijection rom the set o k-points X(k) toPic0(X).

Proo. By virtue o example (e) in Subsection 3.1 X is not isomorphicto P1k. (Note that the previous exercise shows how to reduce the equationo X to y2z = x3 + axz2 + bz3, at least when k is algebraically closed.) ByProposition 5.8 (i) the divisor P P0 is never equivalent to 0 or P = P0.This proves injectivity. To prove surjectivity use secants and tangents todecrease the number o points in the support o the divisor. QED

For any smooth and projective curve X it can be proved that Pic0(X) isisomorphic to the group o k-points on the so called Jacobian variety o X.

Further exercises: Directly prove the associativity o the group law on acubic curve.

5.5 RiemannRoch theorem

Defnition. Let X be a smooth projective curve, D Div(X). The space ounctions associated with D is the subset L(D) k(X) consisting o 0 andrational unctions f such that div(f) + D 0 (that is, is 0 or an efectivedivisor).

Note that any unction f L(D) is regular away rom the support o D.

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Proposition 5.12 LetD be a divisor on a smooth and projective curve X.

(i) Let g k(X). Then the map f f/g identies L(D) with L(D +div(g)).

(ii) L(D) = 0 i deg(D) < 0.(iii) L(0) = k.(iv) L(D) is a vector space o dimension at most deg(D) + 1.

Proo. (i) is obvious.(ii) ollows rom Corollary 5.7 and the trivial observation that an efective

divisor has positive degree.(iii) reects the act that div(f) = 0 implies that f is a constant unction.

(iv) Let D =

P nPP. Then f L(D) i and only i valP(f) nP.The ultrametric inequality (Proposition 4.2) shows that L(D) is closed underaddition; and L(D) is obviously closed under multiplication by constants.

Now we prove dim(L(D)) deg(D) + 1 or all D o non-negative degreeby induction on deg(D). Because o (i) we can assume that D is efectiveor zero. In the last case we conclude by (ii). Now D is efective, henceo positive degree. Suppose that the inequality is proved or all divisors odegree at most deg(D) 1. We can write D = nQQ +

P=Q nPP where

nQ > 0 and nP 0. Then D = D Q 0, and L(D) L(D). Itsuces to show that the codimension o L(D) in L(D) is 0 or 1. Choose alocal parameter uQ at Q. Then it is immediate rom the denition o L(D)that funQQ is regular at Q. Consider the linear orm on L(D) given by the

value o funQQ at Q. The set o zeros o this orm is precisely L(D

), hencethe codimension o this subspace is at most 1. Now we us