Solutions to Hartshorne's Algebraic Geometry/Separated and Proper Morphisms

The reference for this section is EGA II.5, EGA II.6, EGA II.7. For the discrete vaulation ring questions at the end see Samula and Zariski's Commutative Algebra II.

Exercise II.4.1
Let $$f: X \to Y$$ be the finite morphism. Finite implies finite type so we only need to show that $$f$$ is universally closed and separated.

$$f$$ is separated. We want to show that $$X \to X \times_Y X$$ is a closed immersion. To check that a morphism is a closed immersion it is enough to check for each element of an open cover of the target. Let $$\{ Spec\ B_i\}$$ be an open affine cover of $$Y$$. The pull-back of $$X \to X \times_Y X$$ along each $$Spec\ B_i \to Y$$ is $$Spec\ A_i \to Spec\ A_i \otimes_{B_i} A_i$$ where $$Spec\ A_i = f^{-1}Spec\ B_i$$. The ring homomorphism corresponding to these morphisms of affine schemes is surjective, and so they are all closed immersions according to Exercise II.2.18(c).

$$f$$ is universally closed. The proof of Exercise II.3.13(d) goes through to show that finite morphisms are stable under base change (in fact, the proof becomes easier). Secondly, we know that finite morphisms are closed (Exercise II.3.5) and therefore finite morphisms are universally closed.

Exercise II.4.2
Let $$U$$ be the dense open subset of $$X$$ on which $$f$$ and $$g$$ agree. Consider the pullback square(s):



Since $$Y$$ is separated, the lower horizontal morphism is a closed immersion. Closed immersions are stable under base extension (Exercise II.3.11) and so $$Z \to X$$ is also a closed immersion. Now since $$f$$ and $$g$$ agree on $$U$$, the image of $$U$$ in $$Y \times_S Y$$ is contained in the diagonal and so the pullback is, again $$U$$ (at least topologically. But this means that $$U \to X$$ factors through $$Z$$, whose image is a closed subset of $$X$$. Since $$U$$ is dense, this means that $$sp\ Z = sp\ X$$. Since $$Z \to X$$ is a closed immersion, the morphism of sheaves $$\mathcal{O}_X \to \mathcal{O}_Z$$ is surjective. Consider an open affine $$V = Spec\ A$$ of $$X$$. Restricted to $$V$$, the morphism $$Z \cap V \to V$$ continues to be a closed immersion and so $$Z \cap V$$ is an affine scheme, homeomorphic to $$V$$, determined by an ideal $$I \subseteq A$$. Since $$Spec\ A / I \to Spec\ A$$ is a homeomorphism, $$I$$ is contained in the nilradical. But $$A$$ is reduced and so $$I = 0$$. Hence, $$Z \cap V = V$$ and therefore $$Z = X$$.


 * 1) Consider the case where $$X = Y = Spec\ k[x,y] / (x^2, xy)$$, the affine line with nilpotents at the origin, and consider the two morphisms $$f,g: X \to Y$$, one the identity and the other defined by $$x \mapsto 0$$, i.e. killing the nilpotents at the origin. These agree on the complement of the origin which is a dense open subset but the sheaf morphism disagrees at the origin.
 * 2) Consider the affine line with two origins, and let $$f$$ and $$g$$ be the two open inclusions of the regular affine line. They agree on the complement of the origin but send the origin two different places.

Exercise II.4.3
Consider the pullback square



Since $$X$$ is separated over $$S$$ the diagonal is a closed immersion. Closed immersions are stable under change of base (Exercise II.3.11(a)) and so $$U \cap V \to U \times_S V$$ is a closed immersion. But $$U \times_S V$$ is affine since all of $$U, V, S$$ are. So $$U \cap V \to U \times_S V$$ is a closed immersion into an affine scheme and so $$U \cap V$$ itself is affine (Exercise II.3.11(b)).

For an example when $$X$$ is not separated consider the affine plane with two origins $$X$$ and the two copies $$U, V$$ of the usually affine plane inside it as open affines. The intersection of $$U$$ and $$V$$ is $$\mathbb{A}^2 - \{0\}$$ which is not affine.

Exercise II.4.4
Since $$Z \to S$$ is proper and $$Y \to S$$ separated it follows from Corollary II.4.8e that $$Z \to Y$$ is proper. Proper morphisms are closed and so $$f(Z)$$ is closed.

$$f(Z) \to S$$ is finite type. This follows from it being a closed subscheme of a scheme $$Y$$ of finite type over $$S$$ (Exercise II.3.13(a) and (c)).

$$f(Z) \to S$$ is separated. This follows from the change of base square and the fact that closed immersions are preserved under change of base.

$$f(Z) \to S$$ is universally closed. Let $$T \to S$$ be some other morphism and consider the following diagram

Our first task will be to show that $$T \times_S Z \to T \times_S f(Z)$$ is surjective. Suppose $$x \in T \times_S f(Z)$$ is a point with residue field $$k(x)$$. Following it horizontally we obtain a point $$x' \in f(Z)$$ with residue field $$k(x') \subset k(x)$$ and this lifts to a point $$x \in Z$$ with residue field $$k(x) \supset k(x')$$. Let $$k$$ be a field containing both $$k(x)$$ and $$k(x'')$$. The inclusions $$k(x''), k(x) \subset k$$ give morphisms $$Spec\ k \to T \times_S f(Z)$$ and $$Spec\ k \to Z$$ which agree on $$f(Z)$$ and therefore lift to a morphism $$Spec\ k \to T \times_S, Z$$ giving a point in the preimage of $$x$$. So $$T \times_S Z \to T \times_S f(Z)$$ is sujective.

Now suppose that $$W \subseteq T \times_S f(Z)$$ is a closed subset of $$T \times_S f(Z)$$. Its vertical preimage $$(f')^{-1}W$$ is a closed subset of $$T \times_S Z$$ and since $$Z \to S$$ is universally closed the image $$s' \circ f'((f')^{-1}(W))$$ in $$T$$ is closed. As $$f'$$ is surjective, $$f'((f')^{-1}(W)) = W$$ and so $$s' \circ f'((f')^{-1}(W)) = s'(W)$$. Hence, $$T \times_S f(Z)$$ is closed in $$T$$.

Exercise II.4.5
But by the valuative criterion for separability this diagonal morphism (if it exists) is unique. Therefore, the center, if it exists, is unique.
 * 1) Let $$R$$ be the valuation ring of a valuation on $$K$$. Having center on some point $$x \in X$$ is equivalent to an inclusion $$\mathcal{O}_{x,X} \subseteq R \subseteq K$$ (such that $$\mathfrak{m}_R \cap \mathcal{O}_{x,X} = \mathfrak{m}_x$$) which is equivalent to a diagonal morphism in the diagram
 * 1) Same argument as the previous part.

with $$S$$ a valuation ring of function field $$L$$. If the image of the unique point of $$Spec\ L$$ is not the generic point of $$X$$ then let $$Z$$ be the closure of its image with the reduced structure. We have a diagram The scheme $$Z$$ is an integral $$k$$-scheme of dimension less than $$n$$ and so the square on the lest admits a lifting, which gives a lifting for the outside rectangle. Moreover, as closed immersions are proper, any lifting of the outside rectangle factors uniquely through $$Z$$ by the valuative criteria and so the lifting is unique.
 * 1) The argument for the two cases is the same so we will prove: Suppose that every valuation ring $$R$$ of $$K$$ has a unique center in $$X$$, then $$X$$ is proper. This is clearly true for integral $$k$$-schemes of finite type of dimension zero. Suppose that it is true for integral $$k$$-schemes of dimension less than $$n$$ and that $$X$$ is an integral $$k$$-scheme of dimension $$n$$. We will use the valuative criteria. Suppose that we have a diagram

Now suppose that the image of the point of $$Spec\ L$$ is the generic point of $$X$$. Then we have a tower of field extensions $$L / K / k$$ and the valuation on $$L$$ induces a valuation on $$K$$. We then have the following diagram. By assumption the valuation ring $$R$$ has a unique center $$x$$ on $$X$$ and so there is a unique extension of the diagram above Hence, there is a unique lifting of our original square. By the valuative criteria, the scheme $$X$$ is then proper.


 * 1) Suppose that there is some $$a \in \Gamma(X, \mathcal{O}_X)$$ such that $$a \not\in k$$. Consider the image $$a \in K$$. Since $$k$$ is algebraically closed, $$a$$ is transcendental over $$k$$ and so $$k[a^{-1}]$$ is a polynomial ring. Consider the localization $$k[a^{-1}]_{(a^{-1})}$$. This is a local ring contained in $$K$$ and therefore there is a valuation ring $$R \subset K$$ that dominates it. Since $$\mathfrak{m}_R \cap k[a^{-1}]_{(a^{-1})} = (a^{-1})$$ we see that $$a^{-1} \in \mathfrak{m}_R$$.

Now since $$X$$ is proper, there exists a unique dashed morphism in the diagram on the left. Taking global sections gives the diagram on the right which implies that $$a \in R$$ and so $$v_R(a) \geq 0$$. But $$a^{-1} \in \mathfrak{m}_R$$ and so $$v_R(a^{-1}) > 0$$ This gives a contradiction since $$0 = v_R(1) = v_R(\frac{a}{a}) = v_R(a) + v_R(\frac{1}{a}) > 0$$.

Exercise II.4.6
Since $$X$$ and $$Y$$ are affine varieties, by definition they are integral and so $$f$$ comes from a ring homomorphism $$B \to A$$ where $$A$$ and $$B$$ are integral. Let $$K = k(A)$$. Then for valuation ring $$R$$ of $$K$$ that contains $$\phi(B)$$ we have a commutative diagram Since $$f$$ is proper, the dashed arrow exists (uniquely, but we don't need this). From Theorem II.4.11A the integral closure of $$\Phi(B)$$ in $$K$$ is the intersection of all valuation rings of $$K$$ which contain $$\phi(B)$$. As the dashed morphism exists for any valuation ring $$K$$ containing $$\phi(B)$$ so it follows that $$A$$ is contained in the integral closure of $$\phi(B)$$ in $$K$$. Hence every element of $$A$$ is integral over $$B$$, and this together with the hypothesis that $$f$$ is of finite type implies that $$f$$ is finite.

Exercise II.4.8
\tbd{\mathfrak{m}arginpar{Should really check that the all the claims made about pullbacks in here are true.}} Therefore $$f \times f'$$ has property $$P$$.
 * Let $$X \stackrel{f}{\to} Y$$ and $$X' \stackrel{f'}{\to} Y'$$ be the morphisms. The morphism $$f \times f'$$ is a composition of base changes of $$f$$ and $$f'$$ as follows:


 * Same argument as above but we should also note that since $$g$$ is separated the diagonal morphism $$Y \to Y \times_Z Y$$ is a closed embedding and therefore satisfies $$P$$.

The morphism $$X_{red} \to X \to Y$$ is a composition of a closed immersion and a morphism with property $$scP$$ and therefore it has property $$P$$. Therefore the vertical morphism out of the fibre product is a base change of a morphism with property $$P$$ and therefore, itself has property $$P$$. To se that $$f_{red}$$ has property $$P$$ it therefore remains only to see that the graph $$\Gamma_{f_{red}}$$ has property $$P$$ for then $$f_{red}$$ will be a composition of morphisms with property $$P$$. To see this, recall that the graph is following base change But $$Y_{red} \times_Y Y_{red} = Y_{red}$$ and $$\Delta = id_{Y_{red}}$$ and so $$\Delta$$ is a closed immersion. Hence, $$\Gamma$$ is a base change of a morphism with property $$P$$.
 * Consider the factorization

Exercise II.4.9
Let $$X \stackrel{f}{\to} Y \stackrel{g}{\to} Z$$ be two projective morphisms. This gives rise to a commutative diagram where $$f'$$ and $$g'$$ (and therefore $$id \times g'$$) are closed immersions. Now using the Segre embedding the projection $$\mathbb{P}^r \times \mathbb{P}^s \times Z \to Z$$ factors as

$$ \mathbb{P}^r \times \mathbb{P}^s \times Z \to \mathbb{P}^{rs + r + s} \times Z \to Z $$

So since the Segre embedding is a closed immersion then we are done since we have found a closed immersion $$X \to \mathbb{P}^{rs + r + s}_Z$$ which factors $$g \circ f$$.

Exercise II.4.10
Chow's Lemma is in EGA II.5.6.

Exercise II.4.11
See Samula and Zariski's Commutative Algebra II.

Suppose that $$L = K(t)$$. Then define:

$$ \mathfrak{m}_R = \{ a_0 + a_1 t + \dots + a_n t^n \in \mathcal{O}[t] : a_0 \in \mathfrak{m} \} $$

The ring $$\mathcal{O}[t]$$ is a discrete noetherian local domain with maximal ideal $$\mathfrak{m}_R$$ and quotient field $$L$$. By induction then, we can reduce to the case when $$L$$ is a finite field extension of $$K$$. Now consider a set of generators $$\{x_1, \dots, x_n\}$$ of $$\mathfrak{m}$$ such that $$x_1 \not\in \sqrt{(x_2, \dots, x_n)}$$\mathfrak{m}arginpar{does such a set always exist?} (if $$\mathfrak{m}$$ is principal wait for the next step). We claim that the ideal $$(x_1)$$ is not the unit ideal in $$\mathcal{O}' = \mathcal{O}[\frac{x_2}{x_1}, \dots, \frac{x_n}{x_1}]$$. If it were then there would be some polynomial $$f$$ of degree, say $$d$$, in the $$\frac{x_i}{x_1}$$ such that $$1 = x_1 f$$. Let $$f_0$$ be the degree 0 part of $$f$$ and $$f_1$$ be the higher degree part. Since $$x_1 \in \mathfrak{m}$$ the element $$1 - x_1 f_0$$ has an inverse, say $$a$$. Now with this in mind, our equality $$1 = x_1 f_0 + x_1 f_1$$ implies that $$1 = a x_1 f_1$$ which then implies that $$x_1^d = a x_1^{d + 1} f_1$$. Since $$f_1$$ is made up of terms of degree higher than zero, the element $$a x_1^{d + 1} f_1 \in (x_2, \dots, x_n)$$ which implies that $$x_1 \in \sqrt{(x_2, \dots, x_n)}$$ contradicting our assumption. So $$(x_1)$$ is not the unit ideal in $$\mathcal{O}'$$. Now let $$\mathfrak{p}$$ be a minimal prime ideal of $$(x_1)$$, and consider the localization $$(\mathcal{O}')_\mathfrak{p}$$.

Exercise II.4.12
See Samuel and Zariski's Commutative Algebra II.