# Hartshorne Exercise II. 3.19 (a)

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

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