Hartshorne Exercise II. 3.19 (c)

Question

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 $X$.

Answer 1

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.

Answer 2

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 finitely 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 dense 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 dense 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