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 ϕ : X → Y of varieties, the map ϕb : O(Y ) → O(X) is defined as follows: take f : Y → k regular. Then by definition of morphism, the map f ◦ ϕ : X → k is regular. So we define ϕb(f) = f ◦ ϕ. It is routine verification that ϕb 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(ϕ) := ϕb ◦ 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 )) := fe|Y , 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 ∈ 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 ) ∈ A(Y ). Now consider ξi := h(xi) ∈ O(X). This means that each of ξi : X → k (1 ≤ i ≤ n) are regular functions. Using these we deﬁne a map ψ : X → An as ψ (a1, . . . , an) := ξ1(a1, . . . , an), . . . , ξn(a1, . . . , an) (a1, . . . , an) ∈ X

We claim that ψ(X) ⊆ Y . First notice that Y = Z(I(Y )). So, for any (a1, . . . , an) ∈ X we need to prove that f(ψ(a1, . . . , an)) = 0 for any f ∈ I(Y ). 1 P ν1 νn For this ﬁrst write f = ν cνx1 . . . xn . Since f ∈ I(Y ) we must have f(x1,..., xn) = 0. Now, since h is a k-algebra morphism, we have f(ψ(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 ψ : X → Y is a map. We need to prove that ψ 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 ψ : X → Y is a morphism if and only if xi◦ψ 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 ∈ k[x1, . . . , xn] itself. Therefore, these are clearly regular functions.

On the other hand, assume that xi ◦ ψ : X → k (1 ≤ i ≤ n) are regular functions on X. We need to show that ψ : X → Y is a morphism.

P ν1 νn For any polynomial f ∈ k[x1, . . . , xn] writing f = ν cνx1 . . . xn we have

X ν1 νn f ◦ ψ = cν(x1 ◦ ψ) ... (xn ◦ ψ) ν is a regular function, since O(X) is a k-algebra.

Now we verify that ψ is a continuous map. So let Z = Z(g1, . . . , gr) ∩ Y ⊆ Y be a closed set. Then we have −1 n o ψ (Z(g1, . . . , gr) ∩ Y ) = Q ∈ X : ψ(Q) ∈ Z(g1, . . . , gr) ∩ Y n o = Q ∈ X : gi(ψ(Q)) = 0 for all i = 1, . . . , r r \ −1 = (gi ◦ ψ) (0) i=1 which is a closed set from above discussion. This implies that ψ 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 ◦ ψ : ψ (V ) → k is regular. Let P0 ∈ ψ (V ) an arbitrary point. Now using regularity of f at ψ(P0) consider an open neighbourhood W ⊆ V of ψ(P0) and polynomials p, q such that q 6= 0 on W and f = p/q on W . Then, p(ψ(P )) (f ◦ ψ)(P ) = for all P ∈ ψ−1(W ) q(ψ(P )) 2 Notice that from above discussion, we have p ◦ ψ, q ◦ ψ ∈ O(ψ−1(W )) with q ◦ ψ 6= 0 on ψ−1(W ). Therefore (q ◦ψ)−1 ∈ O(ψ−1(W )) and hence f ◦ψ = (p◦ψ)(q ◦ψ)−1 ∈ O(ψ−1(W )). This proves that f ◦ ψ is locally regular at each point of ψ−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 ψ. Let us prove that α(ψ) = ψb ◦ T −1 = h (this will prove that α is a surjective map). Since ψb ◦ 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 ) ∈ A(Y ) for −1 1 ≤ i ≤ n. Next for any such 1 ≤ i ≤ n, we have ψb ◦ T (xi + I(Y )), h(xi + I(Y )) : X → k are regular maps. For any P ∈ X we calculate, −1 ψb ◦ T (xi + I(Y ))(P ) = ψb(xei|Y )(P ) = (xei|Y ) ◦ ψ(P ) = (xei|Y )(ξ1(P ), . . . , ξn(P )) = (xei|Y )(h(x1 + I(Y ))(P ), . . . , h(xn + I(Y ))(P )) = h(xi + I(Y ))(P ) −1 This shows that ψb ◦ T (xi + I(Y )) = h(xi + I(Y )) for every 1 ≤ i ≤ n and consequently α(ψ) = ψb ◦ T −1 = h.

Final part is to show that α is a injective map. For ψ1, ψ2 : X → Y assume that α(ψ1) = α(ψ2). By deﬁnition of α, this implies that ψb1 = ψb2 : O(Y ) → O(X). This means that for any f : Y → k regular function we have

ψc1(f) = f ◦ ψ1 = f ◦ ψ2 = ψc2(f) Now for any P ∈ X, taking the global regular functions xei|Y : Y → k (see the notations in the previous lemma) we get xei|Y (ψ1(P )) = xei|Y (ψ2(P )). This means that ψ1(P ), ψ2(P ) are points in Y whose i-th co-ordinate are equal for each 1 ≤ i ≤ n. Therefore ψ1(P ) = ψ2(P ) for every P ∈ X and hence ψ1 = ψ2.

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 ϕ : X → Y be an isomorphism. Then notice that ϕb : O(Y ) → O(X) is a k-algebra isomorphism: indeed if ψ : Y → X is its inverse, then

IdO(X) = Idb X = ψ[◦ ϕ = ϕb ◦ ψb and similarly, IdO(Y ) = ψb◦ϕb. 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 ◦ ϕb ◦ 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 ϕ : X → Y of varieties such that −1 −1 ϕb ◦ TY = α(ϕ) = TX ◦ h We need to prove that ϕ : X → Y is an isomorphism of varieties. 3 Switching the roles of X and Y , there exists ψ : Y → X so that −1 −1 −1 ψb ◦ TX = α(ψ) = TY ◦ h Now we have −1 −1 ϕ[◦ ψ = ψb ◦ ϕb = ψb ◦ TX ◦ TX ◦ ϕb ◦ TY ◦ TY −1 −1 −1 = TY ◦ h ◦ TX ◦ TX ◦ h ◦ TY = IdO(Y ) and similarly, ψ[◦ ϕ = ϕb ◦ ψb = IdO(X). Using the injectivity of α : Hom(Y,Y ) → Hom(A(Y ), O(Y )) to −1 −1 α(ϕ ◦ ψ) = ϕ[◦ ψ ◦ TY = IdO(Y ) ◦ TY = α(IdY ) we obtain ϕ ◦ ψ = IdY . Similarly, we have ψ ◦ ϕ = IdX . Thus ϕ 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.