Induced map on spectra of rings

Question

Let $B$ be a ring containing $A$, and the ring extension is integral. Furthermore, $B$ is a finitely generated $A$-algebra. Then how to show that the induced map on spectra of the rings is a finite map, i.e., inverse image of finite sets is finite.

Answer 1

The fiber under $\text{Spec} (B)\to \text{Spec} (A)$ of a point $P=[\mathfrak p]\in \text{Spec} (A)$ is the spectrum of the ring $R=B\otimes_A \kappa (P)$, where $\kappa (P)=\text {Frac}\:(A/\mathfrak p)$.
Since finiteness of modules is preserved under base change, the algebra $R$ is finite as a module over the field $\kappa (P)$ (in other words is finite-dimensional!) and thus has finite spectrum by the following result which I will display for future reference :

Lemma
An algebra $R$ of finite dimension over a field $k$ has only finitely many prime ideals and these prime ideals are all maximal.
Proof
Given a prime $\mathfrak m\subset R$ the quotient $R/\mathfrak m$ is necessarily a field, so that $\mathfrak m$ is maximal.
The Chinese remainder theorem applied to the surjective morphism $R\twoheadrightarrow \prod_{\mathfrak m} R/\mathfrak m$ (where we take finitely many maximal ideals $\mathfrak m$) immediately implies that the number of those maximal ideals is finite and $\leq [R:k]$.

Answer 2

A prime ideal $\mathfrak q$ in Spec $B$ lies over a prime ideal $\mathfrak p$ in Spec $A$ precisely if $\mathfrak q\cap A = \mathfrak p$. This implies that $\mathfrak q$ contains $\mathfrak p B$. It is not equivalent to the latter; if $\mathfrak q$ contains $\mathfrak p B$, then $\mathfrak q \cap A$ might contain a larger prime ideal than $\mathfrak p$.

What you can check is that $\mathfrak q \cap A = \mathfrak p$ precisely if $\mathfrak q$ induces a non-unit ideal in $\kappa(\mathfrak p) \otimes_A B$ (where $\kappa(\mathfrak p)$ is the fraction field of $A/\mathfrak p$).

Indeed, this is a general description of the points in the fibre over $\mathfrak p$ (for any morphism $A \to B$); they are in canonical bijection with the points of $\kappa(\mathfrak p)\otimes_A B$.

Now suppose that $A \to B$ is a finite morphism. Then $\kappa(\mathfrak p) \to \kappa(\mathfrak p)\otimes_A B$ is a finite morphism. From this, can you prove that the fibre over $\mathfrak p$ is finite? (What can you say about the Spec of any finite-dimensional $k$-algebra, for a field $k$?)