Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 4 / 5

The characteristic of $K$ is either a prime $p$ or zero. In characteristic $p$, it is a finitely generated algebra over ${\mathbb{F}}_p$. The course's Zariski lemma says that a field finitely generated as an algebra over a field is a finite algebraic extension of that field. ThereforeIt remains to exclude characteristic zero. Write $K=\mathbb{Z}[a_1,\dots ,a_m]$. Because $K$ is then a field containing $\mathbb{Q}$, also $K=\mathbb{Q}[a_1,\dots ,a_m]$. Zariski lemma makes $K/\mathbb{Q}$ finite algebraic. Each $a_i$ has a monic equation with rational coefficients. Choose a positive integer $N$ clearing all their coefficient denominators. Each $a_i$ is then integral over $A=\mathbb{Z}[1/N]$, so $A[a_1,\dots ,a_m]$ is integral over $A$. This algebra equals $K$: it contains the given integer algebra $K$, and $1/N$ already lies in the field $K$.

An integral field extension forces the base domain to be a field. To see it here, for a nonzero $a\in A$ its inverse in $K$ satisfiesMultiplying by $a^{r-1}$ givesThus $A$ would be a field. But choose a prime $\ell \nmid N$. Reduction modulo $\ell$ is a well-defined map $\mathbb{Z}[1/N]\to {\mathbb{F}}_{\ell }$, so the nonzero element $\ell$ is not invertible in $A$. This contradiction rules out characteristic zero. Therefore the finite-field theorem for finitely generated integer algebras yieldsThis uses Zariski lemma and elementary integrality, not a claim that a field merely finitely generated as a field extension must be algebraic.

A nilpotent element belongs to every prime ideal and therefore to every maximal ideal, soFor the reverse containment, let $r$ be nonnilpotent. The localization of a ring $S=R[1/r]$ is nonzero: if $1=0$ there, some power of $r$ would annihilate $1$ in $R$. This is still a finite-type integer algebra, because it can be presented as $R[T]/(rT-1)$.

Choose a maximal ideal $\mathfrak{n}$ of $S$. Its residue field $L=S/\mathfrak{n}$ is a finite-type integer algebra, hence finite by part (a). Consider the image $F$ of $R$ in $L$. It is a finite integral domain containing $1$, so it is a field: multiplication by a nonzero element is injective on the finite set $F$, hence surjective, and therefore has an inverse in $F$.

The kernel $\mathfrak{m}$ of $R\to F$ is consequently maximal. The image of $r$ is nonzero, since $r$ became a unit in $S$ and remains a unit in its nonzero residue field. Thus $r\notin \mathfrak{m}$. Every nonnilpotent element can therefore be avoided by a maximal ideal, givingThis radical equality for finitely generated integer algebras uses part (a) to ensure that the contraction of the localized maximal ideal is maximal. Such contraction is not generally maximal for arbitrary localization of a ring; the finite-field image is the decisive additional step. The zero ring is immediate, with the intersection of its empty set of maximal ideals understood as the whole ring. No general theorem that integer algebras are Jacobson rings is being quoted.