Bryden Cais.  Algebraic Geometry, HW 2.  February 12, 2003

1. Let f : X → Y be a morphism of schemes and y ∈ Y a point. Recall that X_{y} is defined to be the fibre product X ×_{Y} Spec k(y), using the maps f and the natural may g : Spec k(y) → Y given by 𝔭 ↦ y for all 𝔭 ∈ Spec k(y). Without loss of generality, we may assume that the maps

X ×_{Y} Spec k(y) → Spec k(y),
X ×_{Y} Spec k(y) → X

are given by projection onto each component. We therefore see that X_{y} consists of all points (x, 𝔭) ∈ X × Spec k(y) such that f(x) = g(𝔭) = y, that is, the underlying topological space is f^{−1}(y) ⊂ X.

2. Let f : ℙ^{n} × ℙ^{m} → ℙ^{nm+n+m} be given by

((X_{0}, … X_{n}), (Y_{0}, …, Y_{m})) ↦ (X_{0}Y_{0}, X_{1}Y_{0}, …, X_{n}Y_{0}, X_{0}Y_{1}, …, X_{n}Y_{m}).

Observe first that f is well defined since some X_{i} ≠ 0 so that if X_{i}Y_{j} = 0 for all j then Y_{j} = 0 for all j. Moreover, it is not difficult to see that f is injective. Indeed, let X_{i}, Y_{j} ≠ 0. Then we set Z_{kl} = X_{k}Y_{l}. We thus have

(X_{0}, …, X_{n}) = (Z_{0j}/Z_{ij}, Z_{1j}/Z_{ij}, …, Z_{nj}/Z_{ij}).

Similarly, we can reconstruct the point (Y_{0}, …, Y_{m}) from the point (Z_{00}, … Z_{nm}) so that f is indeed injective. Next, observe that the image of f lies in the closed subset Z of ℙ^{nm+m+n} given by (Z_{uv}Z_{st} − Z_{ut}Z_{sv}) for u, s = 0, … n and t, v = 0, …, m. We show that f : ℙ^{n} × ℙ^{m} → Z is surjective. Without loss of generality, let (Z_{00}, …, Z_{nm}) ∈ Z and suppose that Z_{00} = 1. Then we have Z_{uv} = Z_{uv}Z_{00} = Z_{0v}Z_{u0} for all u, v. It follows that

(Z_{00}, …, Z_{nm}) = f((Z_{00}, Z_{10}, …, Z_{n0}), (Z_{00}, Z_{01}, …, Z_{0m}))

so that f : ℙ^{n} × ℙ^{m} → Z is surjective. It follows that f gives a closed embedding. If X, Y are projective varieties, then X, Y are closed subsets of ℙ^{n}, ℙ^{m} for some n, m. Thus, f(X × Y) is an isomorphism or X × Y with a closed subset of ℙ^{nm+n+m} so that the product of two projective varieties is again projective.

3. Let f : X → Y be a finite morphism and let U = Spec A be any open subset of Y such that f^{−1}(U) = Spec B with B an integral extension of ϕ(A), where ϕ : A → B is the ring homomorphism corresponding to f : Spec B → Spec A (that is, a finitely generated A algebra). We first show that f is a closed map. Indeed, let V (I) be a closed set in Spec B. Then we claim that f(V (I)) = V (ϕ^{−1}(I)). Indeed, let 𝔭 ∈ V (ϕ^{−1}(I)). Then ϕ^{−1}(I) ⊆ 𝔭 so that 𝔭̄ = 𝔭 mod ϕ^{−1}(I) is a prime ideal of A/ϕ^{−1}(I). Now since B is integral over ϕ(A), we see that B/I is integral over ϕ(A/ϕ^{−1}(A)). Then by the Going Up Theorem, there exists a prime ideal 𝔮̄ of B/I such that 𝔭̄ = ϕ^{−1}(𝔮̄). Using the correspondence between prime ideals of B/I and prime ideals of B containing I, we have a prime ideal 𝔮 of B with 𝔮 mod I = 𝔮̄ and ϕ^{−1}(𝔮) = 𝔭. Hence, 𝔮 ∈ V (I) and f(𝔮) = 𝔭, as desired.

We now claim that the fibres of f are finite sets. Without loss of generality, we may suppose that A ⊂ B (since we can just replace A by its image under ϕ). We claim that for every ideal I ⊂ B we have dim I ∩ A = dim I. Indeed, by the Going Up Theorem (since B is integral over A), any sequence

I ∩ A ⊆ 𝔭_{1} ⊆ 𝔭_{2} ⊆ … ⊆ 𝔭_{n}

of prime ideals in A gives rise to a sequence

I ⊆ 𝔮_{1} ⊆ 𝔮_{2} ⊆ … ⊆ 𝔮_{n}

of prime ideals of B with A ∩ 𝔮_{i} = 𝔭_{i} for each i. This gives dim I ≥ dim I ∩ A. The Going Down Theorem gives the reverse inequality, so we conclude that dim I = dim I ∩ A for any ideal I of B. In particular, it follows that dim Spec A = dim Spec B. Now let 𝔭 be any prime ideal of A and set I = 𝔭B. Then since I ∩ A = 𝔭, we have dim 𝔭B = dim 𝔭. Since A is noetherian, dim A < ∞ from which it follows that there are only finitely many prime ideals of B lying over 𝔭 ⊂ A. In other words, the fibres of f are finite.

4. Let Y = (xy + zt = 0) ⊂ 𝔸_{x,y,z,t}^{4} and let X be the union of U_{1}, U_{2} with

U_{1} = (y + z′t = 0) ⊂ 𝔸_{u,y,z′,t}^{4}
U_{2} = (x′y + t = 0) ⊂ 𝔸_{x′,y,v,t}^{4}

and let f : X → Y be given by

(x, y, z′, t) ↦ (u, y, uz′, t) = (x, y, z, t)
(x′, y, v, t) ↦ (vx′, y, v, t) = (x, y, z, t).

Let W = (y = z = 0) ⊂ Y. It is not difficult to see that codim(W, Y) = 1 (since W ⊂ Y, clearly, and codim(W, 𝔸^{4}) = 2 since it is given by two non-redundant equations, while codim(Y, 𝔸^{4}) = 1 since it is given by 1 equation). Moreover, we have

f^{−1}(W) = (uz′ = y = v = 0) ⊂ X.

We therefore let Z be the component of f^{−1}(W) given by (z′ = y = v = 0) ⊂ X. We claim that codim(Z, X) = 2. Indeed, Z ⊂ X is is given by the 2 non-redundant equations z′ = v = 0 (which force y = t = 0 when combined with the defining equations of X). Thus, 2 = codim(Z, X) > codim(W, Y) = 1. Observe, of course, that \overline{f(Z)} is just the line y = z = t = 0, which is a proper subset of W = (y = z = 0).

5. Let X, Y be varieties and f_{1}, f_{2} : X → Y two morphisms which agree on an open dense subset U ⊂ X. We define g : X → Y × Y by g(x) = (f_{1}, f_{2})(x). Let Δ : Y → Y × Y be the usual diagonal embedding; since Y is a variety, it is separated so that D = Δ(Y) ⊂ Y × Y is closed. Certainly Y = Δ^{−1}(D). Since f_{1}, f_{2} agree on U, let f = f_{1}|_{U} = f_{2}|_{U} and define g̃ : U → Y × Y by g̃ = (f, f) = g|_{U} = Δ ∘ f. We then have

g^{−1}(D) ⊇ g̃^{−1}(D) = f^{−1} ∘ Δ^{−1}(D) = f^{−1}(Y) = U.

Since D is closed and g is a morphism, we see that g^{−1}(D) ⊆ X is closed and contains the open dense subset U; it follows that g^{−1}(D) = X, and hence that g(X) ⊆ D, that is, f_{1}, f_{2} agree on all of X.

6. Let X be a variety and suppose that U, V ⊂ X are open affine subsets. Since X is a variety, it is separated, and hence the image of Δ(X) in X × X is closed. It follows that Δ(X) ∩ (U × V) is a closed subset of U × V. But U × V is affine; Since Δ(X) ∩ (U × V) ≃ U ∩ V is a closed subset of an affine set, it is affine.

7. Let X be a proper variety, and let f ∈ O_{X}(X). Consider the closed subset of Z = (x, y) of X × 𝔸^{1} given by f(x)y = 1. Since X is proper, the projection pr_{2} of Z to 𝔸^{1} is closed. Observe first that pr_{2}(Z) is a proper subset of 𝔸^{1} since it does not contain the point y = 0. Since the only proper closed subsets of 𝔸^{1} are finite sets, it follows that pr_{2}(Z) is finite, so that f takes only finitely many nonzero values on X. It follows that f must take on some value, say α ∈ K on an open dense subset of X. It then follows from Problem 5 that f = α identically on X, i.e. that O_{X}(X) = k.

8. Let X be a variety and U ⊂ X open and proper. Observe that the diagonal D of U × X is a closed subset of U × X (indeed, it consists of all points (u, x) ∈ U × X satisfying f(u, x) = u − x = 0). Since X is proper, the projection pr_{2} : U × X → X is a closed map; it follows that pr_{2}(D) = U is a closed subset of X. Since U is both open and closed, we must have U = X.