Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 101 / 2

A ring extension $A\subseteq B$ is an integral extension when every $b\in B$ is an integral element over $A$: there is a monic polynomialwith all $a_i\in A$.

A ring extension $A\subseteq B$ is finite when $B$ is a finitely generated module over $A$. Every finite extension is integral by the determinant trick.

The prime ideal correspondence for localization identifies primes of $B_{\mathfrak{p}}$ with primes $\mathfrak{q}$ of $B$ disjoint from $A\setminus \mathfrak{p}$. Passing to the quotient by $\mathfrak{p}{B}_{\mathfrak{p}}$ retains exactly those containing $\mathfrak{p}{B}_{\mathfrak{p}}$. Thus the image is the fiber of the map on spectra:

The claim fails for a general extension. If $k$ is an infinite field, all the infinitely many maximal ideals $(x-a)$ of $k[x]$ contract to $(0)$ in $k$.

It still fails for an integral extension. Take a finite field $k={\mathbb{F}}_q$ andEvery $b\in B$ satisfies the monic equation $b^q-b=0$, so $B$ is integral over the diagonal copy of $k$. The coordinate kernels are infinitely many distinct maximal ideals, all lying over $(0)$.

The claim is true for a module-finite ring extension. The primes above $\mathfrak{p}$ correspond to the primes of the fiber ringThis is a finite-dimensional algebra over the residue field $\kappa (\mathfrak{p})$, hence an Artinian ring, and an Artinian ring has only finitely many prime ideals.

Form the finite-dimensional $k$-algebraBecause a finite extension is integral, the Lying-over theorem supplies a prime of $B$ above $\ker f$; after localization and extension of the residue field to $k$, this shows $C\ne 0$. Therefore $C$ has at least one maximal ideal.

As an Artinian ring, $C$ has only finitely many maximal ideals. For each such ideal $\mathfrak{n}$, the quotient $C/\mathfrak{n}$ is a finite field extension of the algebraically closed field $k$, so it equals $k$. Consequently the quotient maps $C\to k$ are in bijection with the required extensions $g:B\to k$. The set of extensions is therefore finite and nonempty.