Finite Morphisms have Finite Fibers

A morphim $\pi: X \to Y$ is finite if for every affine open $\text{Spec} B$ of $Y$, $\pi^{-1}(\text{Spec}B)$ is the spectrum of a $B$-algebra that is a finitely generated $B$-module.

Notice that finite morphisms are automatically affine.

We will show that all finite morphisms have finite fibers. However, the converse is not true.

Example. Finite fibers does NOT imply finite.
Let $\pi: \mathbb{A}^2 - \{(0,0)\} \to \mathbb{A}^2$ be the open embedding. It is not affine as  $\pi^{-1}(\mathbb{A}^2) =  \mathbb{A}^2 - \{(0,0)\}$ is not affine. So it cannot be finite. However it does have finite fibers.

Example. Finite fibers and affine does NOT imply finite.
Let $\pi: \mathbb{A}_{\mathbb{C}}^1 - \{0\} \to \mathbb{A}_{\mathbb{C}}^1$ be the open embedding. It clearly has finite fiber, and it is affine as $\pi^{-1}  \mathbb{A}^1 = \text{Spec} \mathbb{C}[x]_{(x)}.$ However, it is not finite as $\mathbb{C}[x]_{(x)}$ is not a finite $\mathbb{C}[x]$-module.

Proposition. Finite morphisms have finite fibers.

Proof. Let $\pi: X \to Y$ be a finite morphism. Suppose $Y = \text{Spec} B$. Then as $\pi$ is affine, $X$ must equal $\text{Spec} A$ for some $A$ that is finitely generated as a $B$-module. Let $[q]$ be a point of $Y$. We want to show that $\pi^{-1}(q)$ is finite.


Claim. We can assume $B$ is an integral domain and that $q = (0)$ is the generic point of $Y$.

Indeed, notice that $\pi^{\sharp}: B \to A$ induces $B/q \to A/ \pi^{\sharp}(q) A$ and thus a morphism
$\pi_1: \text{Spec} (A/\pi^{\sharp}(q) A) \to \text{Spec}(B/q).$  The fiber over $0$ in $\pi_1$ consists of exactly the prime ideals of $A$ containing $\pi^{\sharp} q$ that contracts to $q$. Thus $\pi_1^{-1}(0) \leftrightarrow \pi^{-1}(q)$.

Now suppose $B$ is an integral domain and $q = (0).$

Notice that $[q] $ lies in $U = \text{Spec} B_{q} \subset \text{Spec} B$ (here we think of $\text{Spec} B_{q} $ as the  set of prime ideals of $B$ contained inside $q$). So $\pi^{-1}(q)$ is the same as fiber over $q$ of the map $\pi: \pi^{-1}(U) \to U$.

Actually, as $B_{q} $ is a field, $U = \{[q]\}.$

Now $\pi^{-1}(U)$ are prime ideals of $A$ whose contractions under $\pi^{\sharp}$ is $q = (0)$. These are prime ideals $p$ such that $p \cap \pi^{\sharp} B = \{0\}$, i.e. prime ideals of $(\pi^{\sharp} B)^{-1} A = A'$.

As $A$ is a finite $B$-module, we must have $A'$ is also a finite $B_{(0)}$-module. Indeed suppose $a_1, \ldots, a_n$ generate $A$ over $B$. Let $a/\pi^{\sharp}b$ be in $A'$. Then $a$ is generated by $a_i$'s over $B$ and $1/\pi^{\sharp}(b) = 1/b \cdot 1$ is generated by $1$ over $B_{(0)}.$

As the map $\text{Spec} A' \to \text{Spec} B_{(0)}$ is finite and $ B_{(0)}$ is a field, we must have $\text{Spec} A'$ is a finite discrete space.