Finite type and finite fibers implies quasi-finite

Question

I am trying to understand different finiteness conditions, in particular I am looking at the following exercise from Algebraic Geometry and Arithmetic Curves by Qing Liu:

Let $f:X\rightarrow Y$ be a morphism of schemes. We say that $f$ has finite fibers if $f^{-1}(y)=X_y$ is a finite set for every $y\in Y$. We say that $f$ is quasi-finite if moreover $\mathcal{O}_{X_y,x}$ is finite over $k(y)$ for every $x\in X_y$. Show that a morphism of finite type with finite fibers is quasi-finite. Give an example of a morphism with finite fibers that is not quasi-finite.

I think one is supposed to use that finite type morphisms are closed under base change. Also, any intuition about these things are very welcome!

Answer 1

I recall the following definitions from Vakil's FOAG:

A morphism $f:X\to Y$ is locally of finite type if for every affine open subset $\operatorname{Spec}B=V$ of $Y$, and every affine open subset $\operatorname{Spec}A=U$ of $f^{-1}(V)$, the induced morphism $f^{*}_{|U}:B\to A$ expresses $A$ as a finitely generated $B$-algebra.

A morphism $f$ is of finite type if it is locally of finite type and quasicompact.

Without change the notations, let $f:X\to Y$ be a morphism of finite type; considering the diagram: \begin{equation} \require{AMScd} \begin{CD} U @>f_{|U}>> V\\ @VVV & @VVV\\ X @>>f> Y\\ \end{CD}; \end{equation} one has (obviously) the morphism of rings $f^{*}_{|U}:B=\mathcal{O}_Y(V)\to A=\mathcal{O}_X(U)$, and $A$ is a finitely generated $B$-algebra (via $f^{*}_U$).

By Qing Liu Algebraic Geometry and Arithemtic Curves:

Let $f:X\rightarrow Y$ be a morphism of schemes. $f$ has finite fibers if $f^{-1}(y)=X_y$ is a finite set for every $y\in Y$. $f$ is quasi-finite if moreover $\mathcal{O}_{X_y,x}$ is finite over $k(y)$ for every $x\in X_y$.

Let $y\in V$ and let $f$ be also a morphism of finite fibres by construction, $f^{-1}(y)=X_y=X\times_Y\operatorname{Spec}\kappa(y)$ and by hypothesis $X_y=\{x_1,\dots,x_n\}$. Considering the following diagram of schemes: \begin{equation} \begin{CD} X_y @>\overline{f}>> \operatorname{Spec}\kappa(y)\\ @VVV & @VVV\\ f^{-1}(V)@>>> V\\ @VVV & @VVV\\ X @>>f> Y \end{CD}; \end{equation} one has the $\operatorname{Spec}\kappa(y)$-scheme $X_y$, with structural morphism $\overline{f}=f\times i$, where $i$ is the inclusion of $\operatorname{Spec}\kappa(y)$ in $Y$.

For any $x\in X_y\subseteq U$, let $\mathfrak{p}_x$ be the corresponding prime ideal of $A$; one has that $\mathcal{O}_{X,x}$ is a finitely generated $\mathcal{O}_{Y,y}$-algebra (via $f^{*}_x$, the localization of $f^{*}_{|U}$ on $x$).

From all this, because $\mathcal{O}_{X_y,x}=\mathcal{O}_{X,x}\otimes_{\mathcal{O}_{Y,y}}\kappa(y)$: $\mathcal{O}_{X_y,y}$ is finite over $\kappa(y)$; because this statement holds for any $x\in X_y$, by definition, $f$ is a quasi-finite morphism of schemes!

Remark: more in general, a morphim of schemes locally of finite type with finite fibres is quasi-finite!

Example: considering the ring morphism $i:\mathbb{Q}\hookrightarrow\overline{\mathbb{Q}}$, the inclusion of $\mathbb{Q}$ in its algebraic closure; then $i^{\sharp}:\operatorname{Spec}\overline{\mathbb{Q}}\to\operatorname{Spec}\mathbb{Q}$ is a morphism of schemes with finite fibre but it is not of finite type.

Answer 2

Lemma 1: Let $k$ be a field and $A$ a finite-type $k$-algebra. If $A$ has only finitely many maximal ideals then $A$ is finite-dimensional over $k$.

Lemma 2: Let $k$ be a field and $A$ a finite-dimensional $k$-algebra. For any ${\frak m}\in\operatorname{Spec}A$ the localization $A_{\frak m}$ is also finite-dimensional.

As finite type morphisms are stable under base change we have that $X_y=X\times_Y \operatorname{Spec}k(y)\to \operatorname{Spec}k(y)$ is of finite type. Let $x\in X_y$ and $U$ an affine subset of $X_y$ containing $x$ (in fact $X_y$ is affine which follows e.g. from lemma 2.4.3 and exercise 2.5.11 (b)). Then by the first lemma $U\to k(y)$ is finite and from the second lemma it follows that $\mathcal{O}_{U,x}=\mathcal{O}_{X_y,x}$ is finite over $k(y)$ which is what we wanted to show.

Proof of lemma 1: By Noether normalization there is some $n\geq0$ and a finite injective map $k[T_1,\dots,T_n]\hookrightarrow A$. It is easy to see that $k[T_1,\dots,T_n]$ has infinitely many maximal ideals if $n\geq1$. In that case by lying-over the same would hold for $A$ which contradicts our assumption, i.e. $n=0$ and we win.

Alternative argument: If ${\frak m}_1,\dots{\frak m}_n$ are the maximal ideals of $A$, then $N=\bigcap_{i=1}^n {\frak m}_i$ is the nilradical of $A$ and the fields $A/{\frak m}_i$ are finite-dimensional over $k$ (both statements follow from the Nullstellensatz). Hence $A/N$ is finite-dimensional over $k$. Using $N^m=0$ for some $m$ it follows by a filtration-argument that $A$ is finite over $k$.

Proof of lemma 2: $A$ is Artinian, so $A\cong\prod_{i=1}^m A/{\frak m}_i^n$ for some $n$ where the ${\frak m}_i$ are all the prime (=maximal) ideals of $A$. We see that for ${\frak m}\in \operatorname{Spec}A$ the localization $A_{\frak m}$ is just one of the factors in the product above and thus clearly finite-dimensional over $k$.