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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
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!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
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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
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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
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Insanity: Doing the same thing over and over again and expecting different results. -Albert Einstein –  robjohn Dec 21 '12 at 0:50

2 Answers 2

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
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Vakil and the Stacks Project's approaches to this theorem are different from Hartshorne's. –  Makoto Kato Dec 18 '12 at 18:24
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@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
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@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
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@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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