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Let $X$ be a noetherian space. We say a subset $Z$ of $X$ is constructible in $X$, if it is a finite union of locally closed subsets of $X$.

There is the following theorem of Chevalley(we are not supposed to prove it in this thread).

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 $X$.

Hartshorne Exercise II. 3.19 (a) is as follows. Show that the above thorem can be reduced to the following proposition.

Let $X, Y$ be affine and integral noetherian schemes. Let $f\colon X \rightarrow Y$ be a dominant morphism of finite type. Then $f(X)$ is constructible in $Y$.

How do we show this?

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I think one can take scheme-theoretic image with reduced induced structure, this makes the irruducible component to be a integral scheme... –  Li Zhan Dec 16 '12 at 0:51

2 Answers 2

Here's a hint: you have to make a series of reductions, none of which are too difficult. Try to get as many of the necessary adjectives as you can and tell us which ones you're having trouble with. The two key points are that (1) constructibility is a topological property, so you really only care about the underlying spaces, and (2) if $X$ is a space, $\{X_i\}$ a finite cover by constructible subsets (in particular, open subsets or closed subsets), then $U \subseteq X$ is constructible iff $U \cap X_i$ is constructible in $X_i$ for each $i$. Since both schemes are noetherian, any open cover can be taken to be finite, and there are only finitely many irreducible components. (I'll post an answer if you want it, but these sorts of argument are standard enough to be worth taking a crack at yourself first.)

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How do you replace $Z$ with $X$? –  Makoto Kato Dec 16 '12 at 9:19

We reduce the theorem of Chevalley into the proposition stated in the exercise by several steps. To do so, we need some lemmas.

Notation Let $X$ be a scheme. We denote the ring of global sections of the structure sheaf $\mathcal{O}_X$ of $X$ by $\Gamma(X, \mathcal{O}_X)$. By abuse of notatation, we often write $\Gamma(X)$ instead of $\Gamma(X, \mathcal{O}_X)$ if there is no risk of ambiguity.

Lemma 1 Let $f\colon X \rightarrow Y$ be a morphism of schemes. Suppose $f$ is a morphism of finite type and $Y$ is a noetherian scheme. Then $X$ is a noetherian scheme.

Proof: Let $x$ be a point of $X$. Since $f$ is of finite type, there exist an affine open neighborhood $U$ of $x$ and an affine open neighborhood $V$ of $f(x)$ such that $f(U) \subset V$ and the induced ring map $\Gamma(V) \rightarrow \Gamma(U)$ is of finite type. Since $Y$ is a noetherian scheme, $\Gamma(V)$ is noetherian(Hartshorne II, Proposition 3.2). Since $\Gamma(V) \rightarrow \Gamma(U)$ is of finite type, $\Gamma(U)$ is noetherian. Hence $X$ is a locally noetherian scheme. Since $f$ is quasi-compact and $Y$ is quasi-compact, $X$ is quasi-compact. Hence $X$ is a noetherian scheme. QED

Lemma 2 Let $X$ be a noetherian scheme. Let $U$ be an open subset of $X$. Then the canonical injection $f\colon U \rightarrow X$ is a morphism of finite type.

Proof: It is clear that $f$ is locally of finite type. Since $X$ is noetherian, every subset of $X$ is quasi-compact. Hence $f$ is quasi-compact. Therefore $f$ is of finite type. QED

Lemma 3 Let $Z$ be a closed subscheme of a scheme $X$. Then the canonical injection $f\colon Z \rightarrow X$ is a morphism of finite type.

Proof: Let $V$ be an open affine subset $X$. Let $A = \Gamma(V, \mathcal{O}_X)$. Then the closed subscheme $Z \cap V$ of $V$ is isomorphic to Spec($A/I$) for some ideal $I$ of $A$. Hence $f$ is locally of finite type. Since Spec($A/I$) is quasi-compact, $f$ is quasi-compact. Hence $f$ is of finite type. QED

Lemma 4 Let $X$ be a noetherian scheme. Let $Z$ be a subscheme of $X$. Then the canonical injection $f\colon Z \rightarrow X$ is a morphism of finite type.

Proof: There exists an open subscheme $U$ of $X$ such that $Z$ is a closed subscheme of $U$. Let $h\colon Z \rightarrow U$ and $g\colon U \rightarrow X$ be the canonical injections. By Lemma 2, $g$ is of finite type. By Lemma 3, $h$ is of finite type. Hence $f = g\circ h$ is of finite type. QED

Lemma 5 Let $(X_i)$ be a finite family of schemes of finite type over a scheme $S$. Then the coproduct $\sqcup X_i$ is of finite type over $S$.

Proof: Clear.

Definition Let $X$ be a topological space. A finite union of locally closed subset of $X$ is called quasi-constructible. If $X$ is a noetherian topological space, a quasi-constructible subset of $X$ is called constructible.

Lemma 6 Let $X$ be a topological space. Let $W$ be a locally closed subset of $X$. Let $Z$ be a subset of $Z$. Then $Z$ is a locally closed subset of $X$ if and only if it is a locally closed subset of $W$.

Proof: Suppose $Z$ is a locally closed subset of $X$. There exist an open subset $U$ of $X$ and a closed subset $F$ of $X$ such that $Z = U \cap F$. Since $Z = Z \cap W = U \cap F \cap W$, $Z$ is a locally closed subset of $W$.

Conversely suppose $Z$ is a locally closed subset of $W$. There exist an open subset $U$ of $X$ and a closed subset $F$ of $X$ such that $Z = U \cap F \cap W$. Hence $Z$ is a locally closed subset of $X$.

Lemma 7 Let $X$ be a topological space. Let $W$ be a locally closed subset of $X$. Let $Z$ be a subset of $W$. Then $Z$ is a quasi-constructible subset of $X$ if and only if it is a quasi-constructible subset of $W$.

Proof: This follows immediately from Lemma 6.

Lemma 8 Let $X$ be a topological space. Let $(W_i)$ be a finite cover of $X$. Suppose each $W_i$ is a locally closed subset of $X$. Let $Z$ a subset of $X$. Then $Z$ is a quasi-constructible subset of $X$ if and only $Z \cap W_i$ is a quasi-constructible subset of $W_i$ for every $i$.

Proof: This follows immediately from $Z = \bigcup_i (Z \cap W_i)$ and Lemma 7.

Lemma 9 Let $f\colon X \rightarrow Y$ be a morphism of affine schemes. Then there exist a closed subscheme $Z$ of $Y$ and a morphism $g\colon X \rightarrow Z$ such that $f = j\circ g$ and $g(X)$ is dense in $Z$, where $j\colon Z \rightarrow X$ is the canonical morphism. Moreover, if $f$ is of finite type, $g$ is of finite type.

Proof: Suppose $X =$ Spec$(B)$, $Y =$ Spec$(A)$. Let $\psi\colon A \rightarrow B$ be a homomoprphism which induces $f$. Let $I$ be the kernel of $\psi$. Then $C = \psi(A)$ is canonically isomorphic to $A/I$. $\psi$ factors into $A \rightarrow C \rightarrow B$, where $C \rightarrow B$ is the canonical injection. If $B$ is of finite type over $A$, $B$ is of finitye over $C$. Letting $Z =$ Spec$(C)$ we are done. QED

We will reduce the theorem in several steps.

Step 1. We may assume $Z = X$.

Proof: Let $Z = (U_1 \cap F_1)\cup \cdots \cup (U_n \cap F_n)$, where $U_i$ is an open subset of $X$ and $F_i$ is a closed subset of $X$. Let $Z_i = U_i\cap F_i$ for each $i$. We consider each $Z_i$ as a reduced closed subscheme of $U_i$. By Lemma 4, the canonical morphism $Z_i \rightarrow X$ is of finite type. Let $Z' = \sqcup Z_i$ be a coproduct of schemes. Let $g\colon Z' \rightarrow X$ be the canonical morphism. By Lemma 5, $g$ is of finite type. Since $g(Z') = Z$, $fg(Z') = f(Z)$. Since $fg$ is of finite type, we may replace $f$ with $fg$.

Step 2. We may assume $Y$ is affine.

Proof: Let $(V_i)$ be a finite open affine cover of $Y$. By Lemma 8, $f(X)$ is constructible in $Y$ if $f(X) \cap V_i$ is constructible in $V_i$ for all $i$. Let $f_i\colon f^{-1}(V_i) \rightarrow V_i$ be the restriction of $f$. Then $f_i(f^{-1}(V_i)) = f(X) \cap V_i$. Clearly $f_i$ is of finite type. Hence it suffices to prove that $f_i(f^{-1}(V_i))$ is constructible in $V_i$ for each i.

Step 3. We may assume $X$ is affine.

Proof: Let $(U_i)$ be a finite open affine cover of $X$. Let $g_i\colon U_i \rightarrow X$ be the canonical ingection for each $i$. Let $f_i = f\circ g_i$. Since $X$ is a noetherian scheme by Lemma 1, $g_i$ is of finite type by Lemma 2. Since $f$ is of finite type, $f_i$ is of finite type. Since $f(X) = \bigcup f_i(U_i)$, it suffices to prove $f_i(U_i)$ is constructible in $Y$ for each $i$.

Step 4. We may assume $Y$ is irreducible.

Proof: Let $Y_1,\dots,Y_m$ be irreducible components of $Y$. By Lemma 8, it suffices to prove $f(X) \cap Y_i$ is constructible in $Y_i$ for each $i$. We regard $Y_i$ as a reduced closed subscheme of $Y$. Let $f_i\colon X\times_Y Y_i \rightarrow Y_i$ be the canonical morphism. Since $f$ is of finite type, $f_i$ is of finite type. Since $f(X) \cap Y_i = f_i(f^{-1}(Y_i))$ and $f^{-1}(Y_i) = X\times_Y Y_i$, we may replace $f$ with $f_i$.

Step 5. We may assume $X$ is irreducible.

Proof: Let $X_1,\dots,X_n$ be irreducible components of $X$. Since $f(X) = \bigcup f(X_i)$, it suffices to prove $f(X_i)$ is constructible in $Y$ for each $i$. We regard each $X_i$ as a reduced closed subscheme of $X$. Let $g_i\colon X_i \rightarrow X$ be the canonical morphism. Since $g_i$ is of finite type by Lemma 3, $fg_i$ is of finite type. Hence we may replace $f$ with $f_i = fg_i$.

Step 6. We may assume $f$ is dominant.

Proof: By Lemma 9, there exist a closed subscheme $Z$ of $Y$ and a morphism $g\colon X \rightarrow Z$ of finite type such that $f = j\circ g$ and $g(X)$ is dense in $Z$, where $j\colon Z \rightarrow X$ is the canonical morphism. By Lemma 7, if $f(X)$ is constructible in $Z$, $f(X)$ is constructible in $Y$. Hence we may replace $f$ with $g$.

Step 7. We may assume $X$ and $Y$ are integral.

Proof: We may replace $f$ with $f_{red}\colon X_{red} \rightarrow Y_{red}$.

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