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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.2012-12-19
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    @CarlMummert I waited for someone else to post his original proof, but noone else did. Anyway answering one's own question is perfectly legitimate in this site. Please don't downvote for such a post. It is against the site's policy.2012-12-19
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    @CarlMummert "Since Stack Overflow launched, we’ve been trying to explain that it’s not just a Q&A platform: it’s also a place where you can publish things that you’ve learned: recipes, FAQs, HOWTOs, walkthroughs, and even bits of product documentation, as long you format it as a question and answer. if you have a question that you already know the answer to if you’d like to document it in public so others (including yourself) can find it later it is OK to ask, and answer, your own question on a relevant Stack Exchange site." http://blog.stackoverflow.com/2012/05/encyclopedia-stack-exchange/2012-12-19
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    !Makoto Kato: it seems I disagree with their blog post. I do not view this site as a place for people to document new mathematics; there are more appropriate places to do so, such as personal websites and the arXiv.2012-12-19
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    @CarlMummert If you disagree with the site's policy, I recommend you not to use this site.2012-12-19
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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.2012-12-19
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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".2012-12-19
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    @CarlMummert So what kind of problem would it cause to this site if that was the case?2012-12-19
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    @CarlMummert "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." Please explain what's wrong with this question. It's OK that you don't like it. Everybody has his preference. However, it cannot be a proper reason for prohibiting such a question.2012-12-19
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    My original comment in this thread is the explanation for my downvote. The only reason I left the comment was to provide that explanation.2012-12-19
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    @CarlMummert "If many users did this sort of thing, the site would be much more difficult to use." Please explain why the site would be much more difficult to use.2012-12-20
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    @CarlMummert I'm waiting for your explanation.2012-12-20
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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.2012-12-20
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    @CarlMummert "I have explained everything I wish to explain, and there is nothing else that I wish to add at this time." So you don't wish to explain why the site would be much more difficult to use.2012-12-20
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    @CarlMummert One of my motivations for asking questions in this site is to provide answers for questions by other members by using my questions. http://math.stackexchange.com/questions/43929/why-are-projective-morphisms-closed http://math.stackexchange.com/questions/241711/how-can-i-prove-formally-that-the-projective-plane-is-a-hausdorff-space http://math.stackexchange.com/questions/238239/a-property-of-the-radical-closure-of-a-field2012-12-20
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    @CarlMummert I opened a meta thread on this issue. http://meta.math.stackexchange.com/questions/6925/whats-wrong-with-this-question2012-12-20
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    @CarlMummert Since the above meta thread was closed(they said it was not a real question), I opened a new one changing the question. http://meta.math.stackexchange.com/questions/6926/whats-wrong-with-answering-ones-own-question2012-12-21
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    @CarlMummert I deleted my answers.2012-12-21
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    @CarlMummert Again, my meta question was closed. This is a new one. http://meta.math.stackexchange.com/questions/6927/do-this-sort-of-questions-would-make-this-site-much-more-difficult-to-use2012-12-21
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    Insanity: Doing the same thing over and over again and expecting different results. -Albert Einstein2012-12-21
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    @robjohn I changed the questions to conform to their reasons for close. One reason that it was a duplicate is simply their misunderstanding.2012-12-21
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    @MakotoKato. **PLEASE** stop polluting the main site with what is *obviously* meta material.2012-12-21
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    @MarianoSuárez-Alvarez Don't you see that I was trying to bring the issue to meta?2012-12-22
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    I have voted to close as "too localized", due to: (1) the extremely narrow framing of the question (don't just prove it, but prove it in *this specific way*), (2) the presentation makes it unlikely that anyone actually interested in the question find it, unless they are searching by book/problem number, and (3) Makoto doesn't even need help with the problem, so not only is it unlikely to help future visitors, it is unlikely to help *current* visitors.2012-12-22
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    Actually, I do sometimes look for algebraic geometry results by searching for the corresponding exercise in Hartshorne...2012-12-23
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    I'm curious to know who voted to close, except of course Hurkyl.2012-12-23
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    @Hurkyl "I have voted to close as "too localized"" You seem to completely forget that this is an exercise of Hartshorne. You voted to close his question.2012-12-24
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    @Makoto: Hartshorne chose to publish his question in a medium other than MSE, and in a setting where the readers are expected to learning rather than teaching.2012-12-24
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    @Hurkyl "Hartshorne chose to publish his question in a medium other than MSE, and in a setting where the reader is one who is expected to learn rather than to teach." So? Please explain why this is the reason for voting to close his question.2012-12-24
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    Rest assured, I'm not holding your responsible for Hartshorne's decision to publish expository material in textbook format. However, I do take issue with your approach to MSE as if you're a assigning homework to the community, especially your insistence that answers be provided as if they were students handing in the answers to said homework rather than people offering help. That last point is likely the cause of the comments indicating that people believe you are taking a class, and trying to get people to give you answers that you can copy and submit as your own homework.2012-12-24
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    As for the reason to close, as I've indicated, I think I believe both the way you have asked the question and the way you are insisting people respond conspire to make the thread unlikely to ever be of use to anybody.2012-12-24
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    @Hurkyl "as if you're assigning homework to the community, especially your insistence that answers be provided as if they were students handing in the answers to said homework rather than people offering help." As I explained in my comment for Rankeya's answer, his answer does not correctly answer the question. That's why I don't accept it. Are you saying that if someone asks a question which is an exercise of a book, it should be closed?2012-12-24

2 Answers 2

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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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    "Hopefully, you are not expecting someone to post a complete proof of this question." May I ask the reason?2012-12-18
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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?".2012-12-18
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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?" I'm not insisting. I'm just asking. This is an exercise. We are supposed to prove it by ourselves. We are supposed to use (a) and (b) to prove (c).2012-12-18
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    Vakil and the Stacks Project's approaches to this theorem are different from Hartshorne's.2012-12-18
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    Try to show using Noetherian induction that if $X$ is a Noetherian topological space, then a subset $E \subset X$ is constructible if and only if for every closed irreducible subset $Y \subset X$, $E \cap Y$ contains a non-empty open subset of $Y$ or is nowhere dense in $Y$. You will need Noetherian induction for this.2012-12-18
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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?" Since we talked about a similar question before in Ex. II 3.19 (b), let me clear it. This is an exercise of Hartshorne. He expects us to prove it by ourselves, which means without looking up other sources. Understood?2012-12-19
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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.2012-12-19
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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.2012-12-19
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    @Makoto: So you're assigning homework to the community, under the pretense of actually needing help with a question, for nothing more than the pleasure of seeing people respond differently?2012-12-22
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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.2012-12-23
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    @Hurkyl To clutch at straws like your last comment you must really hate Makoto. I don't see why gaining "pleasure" from seeing different answers is a bad thing.2012-12-23
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    @soup: The use of MSE in such a manner is already controversial. It gets worse because he doesn't make his intentions clear: it wastes the time of anyone who answers questions on MSE because they want to helping people learn and solve problems, and IIRC there have been multiple occasions where there has been conflict because someone posts an answer that would have been reasonable (or even good) anywhere else, but didn't meet Makoto's intent.2012-12-23
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    @Hurkyl "it wastes the time of anyone who answers questions on MSE because they want to helping people learn and solve problems," By giving answers, they help people learn and solve problems. You seem to completely forget that there are users of the site other than an OP who may appreciate the answer.2012-12-24
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    @Hurkyl "and IIRC there have been multiple occasions where there has been conflict because someone posts an answer that would have been reasonable (or even good) anywhere else, but didn't meet Makoto's intent." Please point out the case if any.2012-12-24
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    Since there seem to be some misunderstanding, I will explain why I don't accept this answer. The title question(i.e. Hartshorne's question) asks us to prove (c) i.e. the theorem of Chevalley using (a) and (b). However, both Vakil and the stacks project do not use (a) and (b) to prove (c). Their approaches are different from Hartshorne's. While I appreciate Rankeya's answer, it does not answer the question correctly.2012-12-24
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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