# Hartshorne Exercise II. 3.19 (c)

Hartshorne Exercise II. 3.19 (c) is as follows. Prove the following theorem of Chevalley by using Exercise II. 3.19 (a) and (b) and noetherian induction on $Y$. How do we prove this?

Theorem of Chevalley Let $X$ be a scheme. Let $Y$ be a noetherian scheme. Let $f\colon X \rightarrow Y$ be a morphism of finite type. Then $f(Z)$ is constructible in $Y$ for every constructible subset $Z$ of $X$.

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This is the third part of this question you have asked on this site only to write an answer to it yourself. I would prefer to see all users make such posts on their personal blogs, rather than making this site a sort of public notebook. If many users did this sort of thing, the site would be much more difficult to use. For this reason I have downvoted this question. – Carl Mummert Dec 19 '12 at 2:01
!Makoto Kato: It is not this site's policy; the policies of the math SE are determined by its users. We are free to decide as a community that we do not welcome a certain type of post even if it is welcome on other SE sites. – Carl Mummert Dec 19 '12 at 2:33
To be clear, it is not answering your own question that is the issue. It is using the site as a kind of notebook to record your work. If a user occasionally answers their own question, especially after getting a hint, that is wonderful. But the purpose of the site is not for asking questions that the asker is already able to solve; I feel that distorts the meaning of "question". – Carl Mummert Dec 19 '12 at 2:38
I have explained everything I wish to explain, and there is nothing else that I wish to add at this time. – Carl Mummert Dec 20 '12 at 17:20
Insanity: Doing the same thing over and over again and expecting different results. -Albert Einstein – robjohn Dec 21 '12 at 0:50

## 2 Answers

Hopefully, you are not expecting someone to post a complete proof of this question (later edit: because this is a well known theorem whose proof you can look up). Check section $8.4$ of Foundations of Algebraic Geometry by Ravi Vakil (notes available here: http://math.stanford.edu/~vakil/216blog/). He develops the machinery to prove this theorem in this section and has many exercises (in particular, I think he uses Noetherian induction). If you are looking for a hint, he gives you one and asks you to make the argument precise.

Also, you can look here in the Stacks Project: http://stacks.math.columbia.edu/tag/054K.

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With all due respect Makoto, why do you insist on people providing you with complete proofs of well known facts that you can look up in a variety of sources? The other thing is that your question does not adequately reflect what you desire. You ask "How do we prove this?". I gave you a source that goes through the basic strategy of proving Chevalley's theorem in detail, and another source that actually gives you a proof. In my opinion this addresses your question "How do we prove this?". – Rankeya Dec 18 '12 at 5:25
Vakil and the Stacks Project's approaches to this theorem are different from Hartshorne's. – Makoto Kato Dec 18 '12 at 18:24
@Makoto Kato: if your goal is to prove it by yourself, it makes no sense to ask other people for the proof by posting the question on a public question and answer site. – Carl Mummert Dec 19 '12 at 2:03
@CarlMummert My goal is not to prove it by myself. I'm expecting the members of this site prove it by themselves. I would like to see their proofs which are possibly different from each other. – Makoto Kato Dec 19 '12 at 2:11
@Hurkyl There are several reasons for asking questions in this site. As I wrote, one of which is to use the results in my answers for other members questions. Of course, anybody can use them. – Makoto Kato Dec 23 '12 at 13:11

Lemma 1 Let $X, Y$ be integral noetherian affine schemes. Let $f\colon X \rightarrow Y$ be a dominant morphism of finite type. Then $f(X)$ contains a non-empty open subset of $Y$.

Proof: Let $X = Spec(B), Y = Spec(A)$. Since $X, Y$ are integral noetherian schemes, $A$ and $B$ are noetherian integral domains. Since $f$ is a dominant morphism, we may assume that $A$ is a subring of $B$. Since $f$ is a morphism of finite type, $B$ is a finiely generated $A$-algebra. Taking $b = 1$ in Exercise II. 3.19 (b), there exists a non-zero element $a$ of $A$ with the following property. If $\psi\colon A \rightarrow \Omega$ is any homomorphism of $A$ to an algebraically closed field $\Omega$ such that $\psi(a) \neq 0$, then $\psi$ extends to a homomorphism $\phi\colon B \rightarrow \Omega$.

Since $a \neq 0$, $D(a)$ is not empty. We claim that $D(a) \subset f(X)$. Let $P \in D(a)$. Let $K$ be the field of fractions of $A/P$. Let $\Omega$ be an algebraic closure of $K$. Let $\psi\colon A \rightarrow \Omega$ be the composition $A \rightarrow A/P \rightarrow K \rightarrow \Omega$. Since $\psi(a) \neq 0$, $\psi$ extends to a homomorphism $\phi\colon B \rightarrow \Omega$. Let $Q$ be the kernel of $\phi$. Then $Q$ is a prime ideal of $B$ lying over $P$. Hence $P \in f(X)$. Hence $D(a) \subset f(X)$ as desired. QED

Lemma 2 Let $X, Y$ be affine noetherian schemes. Suppose $Y$ is irreducible. Let $f\colon X \rightarrow Y$ be a dominant morphism of finite type. Then $f(X)$ contains a non-empty open subset of $Y$.

Proof: Suppose $X = X_1\cup\cdots\cup X_n$, where each $X_i$ is an irreducible closed subset of $X$. Then $f(X) = f(X_1)\cup\cdots\cup f(X_n)$. Hence $Y = \overline {f(X)}$ $= \overline{f(X_1)} \cup\cdots\cup \overline{f(X_n)}$. Since $Y$ is irreducible, $Y = \overline{f(X_i)}$ for some $i$. We regard $X_i$ as a reduced closed subscheme of $X$. Let $f_i\colon X_i \rightarrow Y$ be the composition $X_i \rightarrow X \rightarrow Y$. Applying Lemma 1 to $(f_i)_{red}\colon (X_i)_{red} \rightarrow Y_{red}$, we are done. QED

Lemma 3 Let $f\colon X \rightarrow Y$ be a morphism of affine schemes. Let $Z$ be a closed subscheme of $Y$. Then $p\colon X\times_Y Z \rightarrow X$ is a closed immersion and $p(X\times_Y Z) = f^{-1}(Z)$.

Proof: Suppose $X =$ Spec$(B), Y =$ Spec$(A), Z =$ Spec$(A/I)$. Then $X\times_Y Z$ = Spec$(B/IB)$ and we are done.

Proof of the theorem of Chevalley By Exercise II. 3.19 (a), we may assume that $X$ and $Y$ are integral noetherian affine schemes and $Z = X$. By noetherian induction, it suffices to prove the following assertion. Let $F$ be a closed subset of $Y$. If for every closed subset $G$ of $Y$ such that $G$ is a proper subset of $F$, $f(X) \cap G$ is constructible in $Y$, then $f(X) \cap F$ is constructible in $Y$.

Clearly we may assume $F$ is irreducible. Suppose $f(X) \cap F$ is not dence in $F$. Let $G$ be the closure of $f(X) \cap F$ in $F$. Since $G \neq F$, $f(X) \cap G$ is constructible in $Y$ by the induction assumption. Since $f(X) \cap F \subset f(X) \cap G \subset f(X) \cap F, f(X) \cap F = f(X) \cap G$. Hence $f(X) \cap F$ is constructible in $Y$.

Suppose $f(X) \cap F$ is dence in $F$. By Lemma 3, we regard $f^{-1}(F)$ as a closed subscheme of $X$. Then $f$ induces a morphism $g\colon f^{-1}(F) \rightarrow F$. Since $f(X) \cap F = f(f^{-1}(F))$, $g$ is dominant. Hence by lemma 2, $f(X) \cap F$ contains a non-empty open subset $U$ of $F$. Then $f(X) \cap F = U \cup (f(X) \cap (F - U))$. By the induction assumption, $f(X) \cap (F - U)$ is constructible in $Y$. Hence $f(X) \cap F$ is constructible in $Y$ as desired. QED

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