Chevalley's Theorem: a step in the proof from Vakil's FOAG

Question

I'm trying to solve the following exercises (7.4.M and 7.4.N) from Vakil's FOAG, on the proof of Chevalley's theorem, but I'm not sure if what I've done is correct. I would like to someone to provide feedback on the correctness of my solution, and if it's wrong, how I can solve it with the material developed by Vakil so far in the book.

Here's my attempt:

$X$ is Noetherian and thus, compact. Pick a finite affine cover of $X$, which I'll denote by $\{U_i\}_{i\in I}$. Any subset $Z$, of $X$ is constructible iff each $Z\cap U_i$ is constructible. Since $\pi (Z) = \cup_{i\in I}\pi (Z\cap U_i)$, it suffices to prove that each $\pi (Z\cap U_i)$ is constructible. This lets us assume that $X$ is affine. This proves the latter part of 7.4.N.

Suppose $X = \text{Spec }A$. Since $X$ is Noetherian, we may write $Z$ as the disjoint union of some locally closed subsets, say $Z_j\cap V_j$, where $Z_j$ are closed and $V_j$ are open. Since each $V_j$ is compact, being a subset of a Noetherian space, we may cover each $V_i$ by finitely many affines, and by an argument similar to the one in the previous paragraph, we may assume that $V_j$ itself is affine. Now, $Z_j = \text{Spec } A/I$, for some ideal $I$ of $A$. Thus, $Z_j\cap V_j$ is actually an open subscheme of $Z_j$. Since it suffices to show that $\pi (Z_j\cap V_j)$ is constructible, and we're taking finite union of these individual "pieces" we've built along the way, it in fact suffices to show that the image of a scheme under the given hypotheses is constructible. This finishes the proof of 7.4.M.

To finish the first part of 7.4.N, we break Y up into finitely many affine "pieces" and use a similar argument as before.

One of the reasons I'm a bit unsure whether this proof is true is the existence of solutions at chevalleys theorem reduces to showing image of whole scheme is constructible and Hartshorne Exercise II. 3.19 (a), both of which seem to use stuff which Vakil hasn't covered yet. For instance, he's not done closed subschemes yet, but I guess we may work in a closed subset of an affine scheme, which is something that would like $\text{Spec }A/I$ above. Further, I seem to require the latter part of 7.4.N for 7.4.M.

I would be very grateful is someone could point out where I've gone wrong, or assure me that what I've done is right.

Thank you.

Answer 1

7.4.M. : Let $Z$ be a constructible subset of $X$ and $\pi: X\to Y$.

Assume that image of any finite type morphism of noetheian schemes in constructibe. Now $Z$ is a disjoint union of $Z_i$ where each $Z_i$ is a locally closed subset of $X$. So to show $\pi(Z)$ is constructible, it suffices to show each $\pi(Z_i)$ is constructible. Now $Z_i=U_i\cap F_i$ where $U_i$ is an open subset and $F_i$ is a closed subset of $X$. Put the reduced induced structure on $F_i$, that induces a scheme structure on the open subset $Z_i=U_i\cap F_i\subset F_i$. Now consider the map $$\phi|_{Z_i}:Z_i\to Y $$ given by the restriciton of the map $$\phi: F_i\to X\to Y.$$

Now by our assumption image of $\phi|_{Z_i}=\pi(Z_i)$ is constructible. So this proves. 7.4.M.

7.4.N. : (First reduction) Now we want to reduce to the case $Y$ is affine. We have $\pi:X\to Y$. Now cover $Y$ by affine open subsets $Spec ~A$'s. Now $$\pi(X)=\bigcup_i \pi( \pi^{-1}(Spec~ A_i)).$$

So it suffices to show image of $\pi: \pi^{-1}(Spec~ A_i)\to Spec ~ A_i$ is constructible.

So we have reduced to the case where $Y$ is affine.

(second reduction):
Now we want to reduce to the case $X$ is affine. We have $\pi:X\to Y$ with $Y$ affine.

Again cover $X$ be affine open $Spec ~B$'s. Now $$\pi(X)=\bigcup_i\pi( Spec~B_i).$$ So it suffices to consider $X$ affine.

I hope this clears up the sequence of reductions.