7-2 有限型・有限射・支配射

If $B$ is finitely generated, the single affine source is a finite affine cover, so $f$ is finite type.
Conversely suppose $f$ finite type. Then $\operatorname{Spec}B$ has a finite cover by principal affine opens $D(g_i)$ such that each $B_{g_i}$ is finitely generated over $A$。Choose finite algebra generators of every $B_{g_i}$ and write each as $b/g_i^{n}$。Adjoin to an $A$-subalgebra $B_0\subseteq B$ all numerators $b$, all $g_i$, and finitely many coefficients witnessing
$$ 1=\sum_i c_i g_i^{N} $$
for a common sufficiently large $N$; such a relation exists because the $D(g_i)$ cover.
After enlarging $B_0$ by finitely many elements, each localization map $(B_0)_{g_i}\to B_{g_i}$ is surjective. It is also injective because $B_0\subseteq B$。Thus it is an isomorphism. For any $b\in B$, its image belongs to $(B_0)_{g_i}$ for all $i$。Choose a common power $N$ with $g_i^Nb\in B_0$ for every $i$。Then
$$ b=\sum_i c_i g_i^N b\in B_0. $$
Hence $B=B_0$, which is finitely generated over $A$。□

Choose generators $b_1,\ldots,b_n$ and the surjection
$$ A[x_1,\ldots,x_n]\twoheadrightarrow B, \qquad x_i\mapsto b_i. $$
Hilbert basis theorem says $A[x_1,\ldots,x_n]$ is Noetherian. Therefore its kernel ideal is generated by finitely many elements $f_1,\ldots,f_r$。First isomorphism theorem gives the required presentation. □

For composition, write
$$ B=A[x_1,\ldots,x_m]/(f_1,\ldots,f_r), $$
$$ C=B[y_1,\ldots,y_n]/(g_1,\ldots,g_s). $$
Choose polynomial lifts $\widetilde g_j\in A[x_1,\ldots,x_m,y_1,\ldots,y_n]$。Then
$$ C\cong A[x_1,\ldots,x_m,y_1,\ldots,y_n]/ (f_1,\ldots,f_r,\widetilde g_1,\ldots,\widetilde g_s), $$
so the composite is locally of finite presentation. Finite affine covers compose, and intersections of source affines remain quasi-compact under the quasi-separated hypotheses, so quasi-compactness and quasi-separatedness are preserved.
For base change $A\to A'$,
$$ B\otimes_AA' \cong A'[x_1,\ldots,x_m]/(f_1',\ldots,f_r'), $$
where $f_i'$ is obtained by mapping coefficients to $A'$。Thus the same finite presentation works after affine base change. Base changes of finite affine covers and their quasi-compact intersections prove the global quasi-compact and quasi-separated conditions. □

For composition, affine-locally we have $A\to B\to C$ with
$$ B=A[b_1,\ldots,b_m], \qquad C=B[c_1,\ldots,c_n]. $$
Then $C=A[b_1,\ldots,b_m,c_1,\ldots,c_n]$。Thus local finite type is stable. Quasi-compact maps are stable under composition, so finite type is stable.
For base change $A\to A'$, if $B=A[b_1,\ldots,b_m]$, then
$$ B\otimes_AA'=A'[b_1\otimes1,\ldots,b_m\otimes1]. $$
Thus local finite type survives affine base change, and these calculations glue. Quasi-compactness is stable under base change: inverse images of affine quasi-compact opens are covered by base changes of a finite affine cover. Hence finite type survives. □

$X_y\to\operatorname{Spec}\kappa(y)$ is the base change of $f$ by $\operatorname{Spec}\kappa(y)\to Y$。Apply base-change stability. □

On affine charts $A\to B$, choose finite module generators $b_1,\ldots,b_n$。They also generate $B$ as an $A$-algebra, so the map is finite type. The determinant trick proved in Chapter 15 shows every $b\in B$ satisfies a monic polynomial over $A$, so the map is integral. □

1. If $B$ is finite over $A$ and $C$ finite over $B$, products of finite module generating sets generate $C$ over $A$。
2. If $B$ is generated over $A$ by $b_1,\ldots,b_n$, then $B\otimes_AA'$ is generated over $A'$ by $b_i\otimes1$。
3. Chapter 15 proves that an integral affine map sends $V(J)$ to
   $$ V(J\cap A) $$
   using lying over after quotient. Therefore it is closed on each affine target, hence globally closed.
4. Fiber algebra is
   $$ B\otimes_A\kappa(y), $$
   finite-dimensional over $\kappa(y)$ by (2). A finite-dimensional algebra over a field is Artinian: a descending chain of ideals is a descending chain of finite-dimensional vector subspaces and stabilizes. An Artinian ring has finitely many prime ideals, all maximal. Hence the fiber has finitely many points. □

finite implies finite type, and previous theorem says each fiber has finitely many points. □

First prove a lemma: a finite-type algebra $C$ over a field $K$ with finite spectrum is finite-dimensional over $K$。Let $\mathfrak p_1,\ldots,\mathfrak p_r$ be its finitely many minimal primes. For each $i$, Noether normalization gives a finite injective map
$$ K[t_1,\ldots,t_{d_i}]\hookrightarrow C/\mathfrak p_i. $$
Lying over makes $\operatorname{Spec}(C/\mathfrak p_i)\to\mathbb A_K^{d_i}$ surjective. If $d_i>0$, the target has infinitely many prime ideals: already $K[t_1]$ has infinitely many primes, since otherwise the product of representatives of all nonzero prime polynomials plus $1$ has a new irreducible factor. This contradicts finiteness of $\operatorname{Spec}C$。Thus all $d_i=0$, so every minimal prime is maximal; hence every prime of $C$ is maximal.
$C$ is Noetherian, so its nilradical $N$ is nilpotent. The finitely many maximal ideals are pairwise comaximal, and Chinese remainder theorem gives
$$ C/N\cong\prod_{\mathfrak m\in\operatorname{Spec}C}C/\mathfrak m. $$
Each residue field is a finite extension of $K$ by Zariski lemma, so $C/N$ is finite-dimensional. The filtration
$$ C\supset N\supset\cdots\supset N^r=0 $$
has quotients finite modules over $C/N$ because the ideals are finitely generated, so $C$ is finite-dimensional.
For base change, each affine fiber algebra $C$ is finite-dimensional over $\kappa(y)$ by the lemma. After a residue-field extension $K/\kappa(y)$ it becomes $C\otimes K$, still finite-dimensional and therefore has finite spectrum. Chapter 35's fiber base-change formula gives quasi-finiteness. Finite type is already base-change stable.
For composition $X\xrightarrow{f}Y\xrightarrow{g}Z$, a fiber over $z$ maps to the finite set $Y_z$。For every $y\in Y_z$, its inverse image is the finite set $X_y$。A finite union of finite sets is finite. Composition is finite type, hence quasi-finite. □

Every contraction of a prime of $B$ contains $\ker\varphi$, so the image lies in $V(\ker\varphi)$。
To prove density there, let $D(a)$ be a principal open meeting $V(\ker\varphi)$。Then $a\notin\sqrt{\ker\varphi}$, so no power of $\varphi(a)$ is zero. Thus $B_{\varphi(a)}$ is nonzero and has a prime ideal $\mathfrak q'$。Its contraction to $B$ is a prime $\mathfrak q$ not containing $\varphi(a)$。Then $f(\mathfrak q)\in D(a)$。Therefore every basic open of $V(\ker\varphi)$ meets the image, proving the closure formula.
$V(\ker\varphi)=\operatorname{Spec}A$ iff $\ker\varphi$ lies in every prime, i.e. in the nilradical. If $A$ reduced, nilradical is zero. □