introduction to algebraic geometry - dolgachev

De nition Two systems of algebraic equations S;S0 k[T] are called equivalent if SolS;K =SolS0;K for any k-algebra K.. The next result gives a simple criterion when two di erent systems

Lecture 1 SYSTEMS OF ALGEBRAIC EQUATIONSThe main objects of study in algebraic geometry are systems of algebraic equations and theirsets of solutions Let k be a eld and k[T1;::: ;Tn] = k[T] be the algebra of polynomials in nvariables overk A system of algebraic equations over kis an expression

fF = 0g F 2 S

where S is a subset ofk[T] We shall often identify it with the subsetS

LetK be a eld extension of k A solution ofS in K is a vector (x1;:::;xn)2Kn such thatfor allF 2S

F(x1;::: ;xn) = 0:

Let Sol(S;K) denote the set of solutions of S in K Letting K vary, we get dierent sets ofsolutions, each a subset ofKn For example, let

S=fF(T1;T2) = 0g:

be a system consisting of one equation in two variables Then

Sol(S;Q) is a subset of Q2 and its study belongs to number theory For example one ofthe most beautiful results of the theory is the Mordell Theorem (until very recently the MordellConjecture) which gives conditions for niteness of the set Sol(S;Q):

Sol(S;R) is a subset ofR2 studied in topology and analysis It is a union of a nite set and

an algebraic curve, or the wholeR2, or empty

Sol(S;C) is a Riemann surface or its degeneration studied in complex analysis and topology.All these sets are dierent incarnations of the same object, an ane algebraic variety over k

studied in algebraic geometry One can generalize the notion of a solution of a system of equations

by allowingK to be any commutativek-algebra Recall that this means thatK is a commutativeunitary ring equipped with a structure of vector space over k so that the multiplication law inK is

a bilinear mapKK !K The map k!K de ned by sendinga2k toa
1 is an isomorphismfrom kto a sub eld of K isomorphic tok so we can and we will identifyk with a sub eld ofK.The solution sets Sol(S;K) are related to each other in the following way Let  : K ! L

be a homomorphism of k-algebras, i.e a homomorphism of rings which is identical on k We canextend it to the homomorphism of the direct products n :Kn ! Ln Then we obtain for any

a= (a1;:::;an)2Sol(S;K),

 n(a) := ((a1);:::;(an))2Sol(S;L):

This immediately follows from the de nition of a homomorphism of k-algebras (check it!) Let

sol(S;) : Sol(S;K)! Sol(S;L)

be the corresponding map of the solution sets The following properties are immediate:

(i) sol(S;idK) = idSol( S ; K );where idA denotes the identity map of a setA;

(ii) sol(S; ) =sol(S; )sol(S;), where :L!M is another homomorphism ofk-algebras.RemarkOne can rephrase the previous properties by saying that the correspondences

K7! Sol(S;K); !sol(S;)

de ne a functor from the category of k-algebrasAlgk to the category of setsSets

De ned by x ! xp is regular andbijective (it is surjective becauseK is algebraically closed and it is injective becausexp=ypimplies

x=y) However, the inverse is obviously not regular

Sometimes, a regular map is called a polynomial map It is easy to see that it is a continuousmap of ane algebraick-sets equipped with the induced Zariski topology However, the converse

is false (Problem 7)

It follows from the de nition that a regular functionf :V !kis given by a polynomialF(T)which is de ned uniquely modulo the ideal I(V) ( of all polynomials vanishing identically on V).Thus the set of all regular functions on V is isomorphic to the factor-algebraO(V) = k[T]=I(V)

It is called the algebra of regular functions on V, or the coordinate algebra of V Clearly it isisomorphic to the coordinate algebra of the ane algebraic variety X de ned by the ideal I(V):

Any regular mapf :V !W de nes a homomorphism

f:O(W)! O(V); '7!'f;

and conversely any homomorphism  :O(W)! O(V) de nes a unique regular map f :V ! W

such that f= All of this follows from the discussion above

2 Prove that the variety de ned by the equationT1T2

?1 = 0 is not isomorphic to the ane line

k

5 Let X and Y be two ane algebraic varieties over a eld k, and let XY be its Cartesianproduct (see Problem 4 in Lecture 1) Prove thatO(XY)=O(X) k O(Y)

6 Prove that the correspondenceK ! O(n;K) ( = nn-matrices with entries in K satisfying

MT =M?1) is an abstract ane algebraick-variety

7 Give an example of a continuous map in the Zariski topology which is not a regular map

Lecture 4 IRREDUCIBLE ALGEBRAIC SETS AND RATIONAL FUNCTIONS

We know that two ane algebraick-setsV andV0are isomorphic if and only if their coordinatealgebrasO(V) andO(V0) are isomorphic Assume that both of these algebras are integral domains(i.e do not contain zero divisors) Then their elds of fractionsR(V) andR(V0) are de ned Weobtain a weaker equivalence of varieties if we require that the eldsR(V) andR(V0) are isomorphic

In this lecture we will give a geometric interpretation of this equivalence relation by means of thenotion of a rational function on an ane algebraic set

First let us explain the condition that O(V) is an integral domain We recall that V  Kn

is a topological space with respect to the induced Zariskik-topology of Kn Its closed subsets areane algebraic k-subsets of V From now on we denote byV(I) the ane algebraick-subset of

Kn de ned by the ideal I  k[T] If I = (F) is the principal ideal generated by a polynomial

F, we write V((F)) =V(F) An algebraic subsets of this form, where (F) 6=f0g;(1), is called ahypersurface

De nition. A topological space V is said to be reducible if it is a union of two proper non-emptyclosed subsets (equivalently, there are two open disjoint proper subsets ofV) OtherwiseV is said

to be irreducible By de nition the empty set is irreducible An ane algebraick-setV is said to bereducible (resp irreducible) if the corresponding topological space is reducible (resp irreducible).Remark 1. Note that a Hausdor topological space is always reducible unless it consists of atmost one point Thus the notion of irreducibility is relevant only for non-Hausdor spaces Alsoone should compare it with the notion of a connected space A topological spaces X is connected

if it is not equal to the union of two disjoint proper closed (equivalently open) subsets Thus anirreducible space is always connected but the converse is not true in general

For every ane algebraic set V we denote by I(V) the ideal of polynomials vanishing onV.Recall that, by Nullstellensatz,I(V(I)) = rad(I)

Proposition 1. An ane algebraic setV is irreducible if and only if its coordinate algebra O(V)has no zero divisors

Proof SupposeV is irreducible and a;b 2 O(V) are such that ab = 0 Let F;G 2 k[T] betheir representatives ink[T] Thenab=FG+I(V) = 0 implies that the polynomialFGvanishes

on V In particular,V V(F)[V(G) and henceV =V1

[V2 is the union of two closed subsets

V1 = V \V(F) and V2 = V \V(G): By assumption, one of them, say V1, is equal to V Thisimplies that V V(F), i.e., F vanishes on V, henceF 2I(V) anda= 0 This proves thatO(V)does not have zero divisors

Conversely, suppose thatO(V) does not have zero divisors LetV =V1

[V2 whereV1 andV2

are closed subsets LetF 2I(V1) andG2I(V2) ThenFG2I(V1

[V2) and (F+I(V))(G+I(V)) =

0 inO(V) Since O(V) has no zero divisors, one of the cosets is zero, sayF +I(V) This impliesthat F 2I(V) andI(V1)I(V), i.e.,V =V1 This proves the irreducibility ofV.

De nition. A topological spaceV is called Noetherian if every strictly decreasing sequenceZ1



Z2

:::Zk  of closed subsets is ... k-algebra K An equivalence class is called an ane algebraic variety over k

(or an ane algebraic< h3>k-variety) If X denotes an ane algebraic< h3>k-variety... ofk,

we may consider an ane algebraic< h3>k-set as an ane algebraic< h3>K-set This is often done when we

do not want to specify to which eld the coecients of the equations... class="page_container" data-page="5">

Lecture AFFINE ALGEBRAIC SETSLet X be an ane algebraic variety over k For dierentk-algebras K the sets of K-points

X(K)