1 Some algebraic geometry 1.1 Basic definitions. 1.1.1 Zariski and standard topology. Let F be a field (almost always R or C). Then we say that a subset X F n is Zariski closed if there is a set of polynomials S F [x1, ..., xn] such that X = F n(S) = x F n s(x) = 0, s S . One checks that the set of Zariski closed sets, Z,f in 2F n satisfiesj the axioms2 g for the closed sets in a topology. That is F n and Z. • ; 2 If T Z then Y T Y Z. • \ 2 2 If Y1, ..., Ym Z then jYj Z. • 2 [ 2 The Hilbert basis theorem (e.g. A.1.2 [GW]) implies that we can take the sets S to be finite. If X is a subset of F n then the Z-topology on X is the subspace topology corresponding to the Zariski topology. If F = R or C then we can also endow F n with the standard metric 2 n 2 topology corresponding to d(x, y) = x y where u = ui for k k k k i=1 j j u = (u1, ..., un). We will call this topology the standard topology (in the literature it is also called the classical, Hausdorff or Euclidean topology).P Our first task will be to study relations be tween these very different topologies. Examples. n = 1. Then if Y Z then Y = F or Y is finite. n = 2.A Zariski closed subset2 is a finite union of plane curves or all of F 2. Thus the Zariski topology on F 2 is not the product topology. 1.1.2 Noetherian topologies. The Zariski topology on a closed subset, X, of F n is an example of a Noetherian topology. That is if Y1 Y2 ... Ym ... is a decreasing sequence of closed subsets of X then there exists N such that if i, j N ≥ then Yi = Yj (this is a direct interpretation of the Hilbert basis theorem). If X is a topological space that be cannot written X = Y Z with Y and Z closed and both are proper the X is said to be irreducible.[ Clearly an irreducible space is connected but the converse is not true. 1 Exercise. What are the irreducible Hausdorff topological spaces? Lemma 1 Let X be a Noetherian topological space. Then X is a finite union of irreducible closed subspaces. If X = X1 X2 Xm with Xi closed [ [···[ and irreducible and if Xi * Xj for i = j then the Xi are unique up to order. 6 For a proof see A.1.12 [GW].The decomposition X = X1 X2 Xm [ [···[ with Xi closed and irreducible and Xi * Xj for i = j is called the irredundant 6 decomposition of X into irreducible components. Each of the Xi is called an irreducible component. 1.1.3 Nullstellensatz. n If X is a subset of F then we set X = f F [x1, ..., xn] f X = 0 . If I n f 2 n j j g I F [x1, ..., xn] is an ideal then we set F ( ) = x F f(x) = 0, f . I f 2 j 2 Ig n Lemma 2 A closed subset, X, of F is irreducible if and only if X is a I prime ideal (i.e. F [x1, ..., xn] X is an integral domain). j We now recall the Hilbert Nullstellensatz. Theorem 3 Let F be algebraically closed and let be an ideal in F [x1, ..., xn]. Then I n 1. F ( ) = if and only if = F [x1, ..., xn]. I ; I 2. IX(I) = p . I We will give a proof of 1. in this theorem for the case of F = C since the method of proof will be consistent with the techniques of the later material to be presented. That 1. implies 2. is the trick of Rabinowitz which we defer to the references. We prove that if I is a proper ideal then Cn( ) = . Let be a maximal I 6 ; M proper ideal with . Then k = C[x1, ..., xn]/ is a field containing C. I M m M Since the image of the monomials x 1 xmn span k as a vector space over 1 ··· n C we see that dimC k is at most countable. Since C is algebraically closed, if k = C there must be t k that is transcendental over C. We assert that the 6 1 2 elements t a a C are independent over C. Indeed, if f j 2 g m c i = 0 t a i=1 i X 2 with a1, ..., am C and distinct and c1, ..., cm in C then multiplying by 2 m (t ai) i=1 Y we find that m cj (t ai) = 0. j=1 i=j X Y6 This defines a polynomial satisfied by t that is trivial if and only if all of the cj are 0 (evaluate at ai and get ci (ai aj)) Since C is not countable this j=i Y6 is a contradiction. We have proved that k is one dimensional over C with basis 1 + . Thus xi + = zi1 + with zi C. We have shown that M M M 2 xi zi for all i = 1, ..., n. Since the ideal z is maximal we see that 2 M n If g = z and since we have z C ( ). M If g I M 2 I 1.2 Affi ne varieties and dimension. 1.2.1 Affi ne varieties. Let X, Y be sets and let f : X Y be a mapping. If g is a function from Y ! to F then we set f g = g f which is a function on X. In this section we will assume that F is algebraically closed (you can assume it is C). An affi ne variety over F is a pair (X, R) with X a topological space and R and algebra over F of continuous functions from X to F (F is endowed with the Zariski topology) such that there exists a Z-closed subset n Z in F for some n and a homeomorphism, f, of X onto Z such that f : F [x1, ..., xn] Z R is an algebra isomorphism. A morphism affi ne varieties jj ! (X, R) and (Y, S) is a map f : X Y such that f S R. An isomorphism ! is a morphism that is one to one and onto and f is an isomorphism of algebras. We note that if (X, R) is an affi ne variety then R must be a finitely generated algebra over F and if a R is such that ak = 0 for some k > 0 then a = 0. That is, the only nilpotent2 element in R is 0. The converse is also true. Let R be a finitely generated algebra over F without nilpotents. Let r1, ..., rm generate R as an algebra over F . Then we have an algebra homomorphism : F [x1, ..., xm] R defined by (xi) = ri. Let = ker and set Z = n ! I F ( ). Then since R = F [x1, ..., xm]/ and R has no nilpotents we must I I 3 have p = . Thus = Z (by the nullstellensatz). Thus R is isomorphic I I I I with (Z) = F [x1, ..., xm] Z . Actually more is true. The topological space X isO also completely determinedj by R. Indeed, the nullstellensatz implies that the maximal ideals in (Z) are the deals z + for z Z. Thus the set Z can be identifiedO with the set of maximalIf g idealsI in 2(Z). The topology is determined as follows. If Y is a closed subset of Z andO is the ideal of elements in (Z) such that Y = z Z g(z) = 0, g J then O f 2 j 2 J g Y + z Z z for an ideal of R. Returning to R we may define f 2 jJ If gg I R to be the set of maximal proper ideals endowed with the topology that M has as its closed sets the sets R( ) = maximal with . We have just seen that if we use theM factI thatfMjR is isomorphic withI Mg(Z) and O the maximal ideals in (Z) are the images of the ideals z for z Z. This O If g 2 implies that if R then R/ = F 1 + . Thus if r R then we can define r( ) =Mc if 2r M= c1 + . WeM thereforeM have made R2into an algebra M M of F valued functions on R. The bottom line is that all of the information is in the algebra R. M Recall that if F = C and if Y Cn is Z-closed then we also have the standard (metric) topology on Y . If(X, R) is an affi ne variety and if (X, R) is isomorphic with (Y, C[x1, ..., xn] Y ) and (V, C[x1, ..., xm] V ) with Y Z-closed j j in Cn and V Z-closed in Cm. Then we have f : X Y and g : X V ! ! homeomorphisms in the Z-topology such that f : C[x1, ..., xn] Y R and 1 j ! g : C[x1, ..., xm] V R algebra isomorphisms. This g f : Y V is j ! 1 ! given by the map = (1, ..., m) with j = (g f )(xj V ) C[x1, ..., xn] Y . 1 j 2 j Thus g f is continuous in the standard metric topology. This implies that endowed with the standard topology Y and V are homeomorphic. We therefore see that this implies that (X, R) also has a natural metric topology which we will also call standard.