Chevalley Theorem

Proof. ($\Rightarrow$): For $Z$ Constructible and $X_{0}$ irreducible, $Z\cap X_{0}=\bigcup_{i=1}^{m}(U_{i}\cap F_{i})$ with $F_{i}$ irreducible, then $\overline{U_{i}\cap F_{i}}=F_{i}$. If $\overline{X_{0}\cap Z}=\bigcup_{i=1}^{m}F_{i}=X_{0}$, so $F_{i}=X_{0}$ for some $i$, Then $U_{i}\cap F_{i}=U_{i}\cap X_{0}$ is a nonempty open subset of $X_{0}\cap Z$.

($\Leftarrow$): We use the Noetherian induction of Noetherian space. For $Z=\varnothing$, the result is trivial. Assume all subsets of $Z$ satisfies the condition above is Constructible. Let $\bar{Z}=F_{1}\cup \cdots\cup F_{n}$ with $F_{i}$ irreducible. $\overline{F_{1}\cap Z}=F_{1}$, since $F_{1}$ is irreducible, there’s a closed subset $F_{1}'$ of $F_{1}$ such that $F_{1}-F_{1}'\subset Z$ is open in $Z$. Take $F^{*}=F_{1}'\cup F_{2}\cup \cdots\cup F_{n}$, then $Z=(F_{1}-F^{*})\cup (Z\cap F^{*})$. $F_{1}-F^{*}$ is obviously locally closed. For $Z\cap F^{*}$, if $X_{0}$ irreducible and $\overline{Z\cap F^{*}\cap X_{0}}=X_{0}$, then $X_{0}\subset F$ and $Z\cap F^{*}\cap X_{0}=Z\cap X_{0}$ open. Since $\overline{Z\cap F^{*}}\subset F^{*}\subset \bar{Z}$, by induction hypothesis, $Z\cap F^{*}$ is constructible, so $Z$ is constructible. ◻

Proof. Assume $\overline{f(Y)}=V(I)$ for some ideal $I$, then $I=\bigcap_{\mathfrak{p}\in Y}\varphi^{-1}(\mathfrak{p})=\varphi^{-1}(\mathfrak{N}_{B})$. For $V(I)=X$, $I=\mathfrak{N}_{A}$ and hence $\ker\varphi\subset\mathfrak{N}_{A}$.

Conversely, if $\ker\varphi\subset\mathfrak{N}_{A}$. Since $\varphi^{-1}(\mathfrak{N}_{B})\supset \mathfrak{N}_{A}\supset \varphi^{-1}(0)$. Take radicals then we get $\mathfrak{N}_{A}=\varphi^{-1}(\mathfrak{N}_{B})$. Thus $V(I)=\overline{f(Y)}$. ◻

Proof. We use the above proposition to prove the theorem. For any $X_{0}$ irreducible in $X$, let $X_{0}=V(\mathfrak{p})$ for some $\mathfrak{p}\in X$. Consider $f(Y)\cap X_{0}$, if $f(Y)\cap X_{0}$ is dense in $X_{0}$, then for $\varphi':A /\mathfrak{p}\to B /\mathfrak{p}B$, by lemma above, $\ker\varphi'\subset\mathfrak{N}_{A /\mathfrak{p}}$. We need to find an open subset of $f(Y)\cap X_{0}$. Let $A'=A /\mathfrak{p}$, $B'=B /\mathfrak{p}B$, $X'=\mathrm{Spec}(A /\mathfrak{p})$ and $Y'=\mathrm{Spec}(B /\mathfrak{p}B)$, $f^{*}:Y'\to X'$. Let $B'=A'[x_{1},\dots,x_{n}]$, and $A^{*}=A[x_{1},\dots,x_{r}]$ s.t. $x_{1},\dots,x_{r}$ algebraically independent and $x_{r+1},\dots,x_{n}$ algebraic over $A^{*}$. Then $\sum\limits_{i=1}^{n_{j}}g_{ji}x_{j}^{i}=0$ with $g_{ji}\in A^{*}$. Then $\prod\limits_{j=k+1}^{n}g_{jn_{j}}$ is a nonzero polynormial with coefficients in $A$. Let $a$ be a coefficient of $\sum\limits_{i=1}^{n_{j}}g_{ji}x_{j}^{i}=0$. Now we show $X_{a}$ is open in $f(Y)\cap X_{0}$. For $\mathfrak{p}\in X'$ with $a\notin \mathfrak{p}$, consider $\mathfrak{p}[x_{1},\dots,x_{r}]=\mathfrak{p}^{*}$, a prime ideal in $A^{*}$, then $a\notin \mathfrak{p}$. Thus $B'_{\mathfrak{p}^{*}}$ isintegral over $A^{*}_{\mathfrak{p}^{*}}$. By lying over theorem, $\exists \mathfrak{q}$ prime in $B'_{\mathfrak{p}^{*}}$ s.t. $\mathfrak{q}\cap A_{\mathfrak{p}^{*}}^{*}=\mathfrak{p}^{*}A_{\mathfrak{p}^{*}}^{*}$. So $\mathfrak{q}\cap A=(\mathfrak{q}\cap B)\cap A=\mathfrak{p}^{*}\cap A=\mathfrak{p}$. Thus $\mathfrak{p}=(\mathfrak{q}\cap B)\cap A\in f(Y)$. ◻

Proof. If $Y'=X_{f}\cap V(I)$ for some $f$ and $I$, just take $B'=S^{-1}(B /I)$, where $S=\{1,f,\dots\}$ and then $\mathrm{Spec}(B')\cong Y'\subset \mathrm{Spec}(B)$.

For any open set in Noetherian space, we can write it as the finite union of basis. So we can assume $Y'=\bigcup_{i=1}^{m}X_{f_{i}}\cap V(I_{i})$ for some $f_{i}$ and $I_{i}$. For each $X_{f_{i}}\cap V(I_{i})$, construct $B_{i}$ as above, then take $B'=\prod\limits_{i=1}^{m}B_{i}$. Then $\mathrm{Spec}(B')\cong Y'\subset \mathrm{Spec}(B)$ ◻