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Morphisms 5Th May, 2020

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Morphisms 5Th May, 2020 MTH 517/617 : Algebraic Geometry Lecture 2 : Morphisms 5th May, 2020 Reference : Algebraic Geometry, Hartshorne, Chapter-I, Section-3. We now proceed to prove the result that for two affine varieties X; Y we have X ∼= Y if and only if A(X) ∼= A(Y ). We first prove a stronger result: Theorem 0.1. Let X be any variety and Y be an affine variety. Then there is a natural bijective mapping of sets α : Hom(X; Y ) ! Hom(A(Y ); O(X)) where the Hom set of the left is the set of morphism of varieties, whereas the second one denotes the homomorphism of k-algebras. Proof. First notice that for a morphism ' : X ! Y of varieties, the map 'b : O(Y ) !O(X) is defined as follows: take f : Y ! k regular. Then by definition of morphism, the map f ◦ ' : X ! k is regular. So we define 'b(f) = f ◦ '. It is routine verification that 'b is a k-algebra morphism. Now by Theorem 3.2 (Hartshorne, Pg.17) we have an isomorphism −1 T : O(Y ) ! A(Y ) of k-algebras. This defines the map α(') := 'b ◦ T . At this point it is worth noting that the map T −1 : A(Y ) !O(Y ) was defined in Theorem −1 n 3.2 (Hartshorne, Pg.17) by T (f + I(Y )) := fejY , where fe : A ! k is the global evaluation map. Now suppose we have a k-algebra homomorphism h : A(Y ) !O(X). Since Y is an affine variety, for a fixed n 2 N, let Y ⊆ An, which is a closed (and irreducible) subset. Then A(Y ) = k[x1; : : : ; xn]=I(Y ). Let xi be the natural images of xi in A(Y ), i.e., xi := xi + I(Y ) 2 A(Y ). Now consider ξi := h(xi) 2 O(X). This means that each of ξi : X ! k (1 ≤ i ≤ n) are regular functions. Using these we define a map : X ! An as (a1; : : : ; an) := ξ1(a1; : : : ; an); : : : ; ξn(a1; : : : ; an) (a1; : : : ; an) 2 X We claim that (X) ⊆ Y . First notice that Y = Z(I(Y )). So, for any (a1; : : : ; an) 2 X we need to prove that f( (a1; : : : ; an)) = 0 for any f 2 I(Y ). 1 P ν1 νn For this first write f = ν cνx1 : : : xn . Since f 2 I(Y ) we must have f(x1;:::; xn) = 0. Now, since h is a k-algebra morphism, we have f( (a1; : : : ; an)) = f h(x1)(a1; : : : ; an); : : : ; h(xn)(a1; : : : ; an) X ν1 νn = cνh(x1)(a1; : : : ; an) : : : h(xn)(a1; : : : ; an) ν X ν1 νn = h cνx1 ::: xn (a1; : : : ; an) ν = h f(x1;:::; xn) (a1; : : : ; an) = 0 So by restricting the image, we have : X ! Y is a map. We need to prove that is a morphism of varieties which follows from the following lemma. Lemma 0.2. Let X be a variety, and let Y ⊆ An be an affine variety. Then, a map : X ! Y is a morphism if and only if xi◦ is a regular function on X for each i = 1; : : : ; n, n where x1; : : : ; xn are regarded as the co-ordinate functions : A ! k. n Proof. The co-ordinate functions xi : A ! k are defined by xi(a1; : : : ; an) := ai. In other words, these are given by the global polynomial xi 2 k[x1; : : : ; xn] itself. Therefore, these are clearly regular functions. On the other hand, assume that xi ◦ : X ! k (1 ≤ i ≤ n) are regular functions on X. We need to show that : X ! Y is a morphism. P ν1 νn For any polynomial f 2 k[x1; : : : ; xn] writing f = ν cνx1 : : : xn we have X ν1 νn f ◦ = cν(x1 ◦ ) ::: (xn ◦ ) ν is a regular function, since O(X) is a k-algebra. Now we verify that is a continuous map. So let Z = Z(g1; : : : ; gr) \ Y ⊆ Y be a closed set. Then we have −1 n o (Z(g1; : : : ; gr) \ Y ) = Q 2 X : (Q) 2 Z(g1; : : : ; gr) \ Y n o = Q 2 X : gi( (Q)) = 0 for all i = 1; : : : ; r r \ −1 = (gi ◦ ) (0) i=1 which is a closed set from above discussion. This implies that is a continuous map. Next we need to show that for any open subset V ⊆ Y and for any regular function f : V ! k, −1 −1 the function f ◦ : (V ) ! k is regular. Let P0 2 (V ) an arbitrary point. Now using regularity of f at (P0) consider an open neighbourhood W ⊆ V of (P0) and polynomials p; q such that q 6= 0 on W and f = p=q on W . Then, p( (P )) (f ◦ )(P ) = for all P 2 −1(W ) q( (P )) 2 Notice that from above discussion, we have p ◦ ; q ◦ 2 O( −1(W )) with q ◦ 6= 0 on −1(W ). Therefore (q ◦ )−1 2 O( −1(W )) and hence f ◦ = (p◦ )(q ◦ )−1 2 O( −1(W )). This proves that f ◦ is locally regular at each point of −1(V ). Thus it must be regular. Back to proof of Theorem 0.1. So far we have proved that α is an well defined map and for any h : A(Y ) !O(X) we have constructed a . Let us prove that α( ) = b ◦ T −1 = h (this will prove that α is a surjective map). Since b ◦ T −1; h : A(Y ) !O(X) are k-algebra morphisms, it is enough to check the equality of these two functions at xi + I(Y ) 2 A(Y ) for −1 1 ≤ i ≤ n. Next for any such 1 ≤ i ≤ n, we have b ◦ T (xi + I(Y )); h(xi + I(Y )) : X ! k are regular maps. For any P 2 X we calculate, −1 b ◦ T (xi + I(Y ))(P ) = b(xeijY )(P ) = (xeijY ) ◦ (P ) = (xeijY )(ξ1(P ); : : : ; ξn(P )) = (xeijY )(h(x1 + I(Y ))(P ); : : : ; h(xn + I(Y ))(P )) = h(xi + I(Y ))(P ) −1 This shows that b ◦ T (xi + I(Y )) = h(xi + I(Y )) for every 1 ≤ i ≤ n and consequently α( ) = b ◦ T −1 = h. Final part is to show that α is a injective map. For 1; 2 : X ! Y assume that α( 1) = α( 2). By definition of α, this implies that b1 = b2 : O(Y ) !O(X). This means that for any f : Y ! k regular function we have c1(f) = f ◦ 1 = f ◦ 2 = c2(f) Now for any P 2 X, taking the global regular functions xeijY : Y ! k (see the notations in the previous lemma) we get xeijY ( 1(P )) = xeijY ( 2(P )). This means that 1(P ); 2(P ) are points in Y whose i-th co-ordinate are equal for each 1 ≤ i ≤ n. Therefore 1(P ) = 2(P ) for every P 2 X and hence 1 = 2. Corollary 0.3. Let X; Y are two affine varieties. Then X and Y are isomorphic if and only if A(X) and A(Y ) are isomorphic as k-algebras. Proof. Let ' : X ! Y be an isomorphism. Then notice that 'b : O(Y ) !O(X) is a k-algebra isomorphism: indeed if : Y ! X is its inverse, then IdO(X) = Idb X = [◦ ' = 'b ◦ b and similarly, IdO(Y ) = b◦'b. Now since both of X and Y are affine varieties, using Theorem 3.2 (Hartshorne, Pg.17) we have TX : O(X) ! A(X) and TY : O(Y ) ! A(Y ) are k-algebra −1 isomorphisms. This implies TX ◦ 'b ◦ TY : A(Y ) ! A(X) is a k-algebra isomorphism. On the other hand suppose h : A(Y ) ! A(X) be a k-algebra isomorphism. Then the map −1 TX ◦h : A(Y ) !O(X) is an isomorphism. Using Theorem 0.1, there exists a homomorphism ' : X ! Y of varieties such that −1 −1 'b ◦ TY = α(') = TX ◦ h We need to prove that ' : X ! Y is an isomorphism of varieties. 3 Switching the roles of X and Y , there exists : Y ! X so that −1 −1 −1 b ◦ TX = α( ) = TY ◦ h Now we have −1 −1 '[◦ = b ◦ 'b = b ◦ TX ◦ TX ◦ 'b ◦ TY ◦ TY −1 −1 −1 = TY ◦ h ◦ TX ◦ TX ◦ h ◦ TY = IdO(Y ) and similarly, [◦ ' = 'b ◦ b = IdO(X). Using the injectivity of α : Hom(Y; Y ) ! Hom(A(Y ); O(Y )) to −1 −1 α(' ◦ ) = '[◦ ◦ TY = IdO(Y ) ◦ TY = α(IdY ) we obtain ' ◦ = IdY . Similarly, we have ◦ ' = IdX . Thus ' is an isomorphism of varieties. The following result will be used in the exercises: Theorem 0.4. Let A be an integral domain which is a finitely generated k-algebra. Let K denote the quotient field of A and L=K be a finite extension. Then the integral closure B of A in L is a finitely generated A-module, and also a finitely generated k-algebra. Proof. For a proof see Basic Comm. Algebra, B. Singh, Theorem 14.1.3, Pg.225, except for the last statement "B is a finitely generated k-algebra", which is immediate from the facts: (i) A is a finitely generated k-algebra, and (ii) B is a finitely generated A-module. 4.
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