Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 4

An element $x\in B$ is integral over $A$ when it satisfies a monic polynomial with coefficients in $A$. Equivalently, $A[x]$ is a finite $A$-module.

If $x,y$ are integral over $A$, then $A[x,y]$ is finite over $A$: it is generated by finitely many monomials $x^i{y}^j$. Multiplication by $x+y$, $x-y$, or $xy$ is an endomorphism of this finite module, so the determinant trick gives a monic annihilating polynomial. Hence the integral elements form a subring $C$ containing $A$.

The integral closure of $A$ in $B$ is this subring $C$. The ring $A$ is integrally closed in $B$ when $C=A$, and $B$ is integral over $A$ when $C=B$.

For $x\in A$, considerThis is monic, and every $\tau \in G$ permutes its factors, so all coefficients lie in $A^G$. Since $f_x(x)=0$, every $x\in A$ is integral over $A^G$. Thus

The Going-up theorem states: if $B$ is integral over $A$, ${\mathfrak{p}}_1\subseteq {\mathfrak{p}}_2$ are primes of $A$, and ${\mathfrak{q}}_1$ lies over ${\mathfrak{p}}_1$, then some prime ${\mathfrak{q}}_2\supseteq {\mathfrak{q}}_1$ lies over ${\mathfrak{p}}_2$.

Pass to $A/{\mathfrak{p}}_1\subseteq B/{\mathfrak{q}}_1$, which remains integral, and localize at the complement of ${\mathfrak{p}}_2/{\mathfrak{p}}_1$. The lying-over theorem supplies a prime of the localized upper ring over the maximal ideal of the localized lower ring. Contracting it to $B/{\mathfrak{q}}_1$, and then pulling it back to $B$, gives the required ${\mathfrak{q}}_2$.

If $x\in A$ is a unit in $B$, then $x^{-1}$ is integral over $A$:Multiplication by $x^{n-1}$ expresses $x^{-1}$ as an element of $A$, so $x$ is a unit in $A$.

Use the characterization $a\in J(A)$ exactly when $1-ra$ is a unit for every $r\in A$. If $a\in J(B)\cap A$, then $1-ra$ is a unit in $B$ and hence in $A$, proving $a\in J(A)$. Conversely, if $a\in J(A)$, every maximal ideal $\mathfrak{n}$ of $B$ contracts under the integral extension to a maximal ideal of $A$, which contains $a$. Thus every $\mathfrak{n}$ contains $a$, so $a\in J(B)$. Therefore

Put $S=A\setminus \mathfrak{p}$. Primes of $B_{\mathfrak{p}}=S^{-1}B$ correspond to primes $\mathfrak{q}$ of $B$ whose contractions are contained in $\mathfrak{p}$. By going up, each such $\mathfrak{q}$ is contained in a prime lying over $\mathfrak{p}$, and this prime is uniquely ${\mathfrak{p}}^{\prime }$. Hence ${\mathfrak{p}}^{\prime }{B}_{\mathfrak{p}}$ is the unique maximal ideal of $B_{\mathfrak{p}}$.

Localizing this already local ring at its unique maximal ideal changes nothing, soFinally, localization preserves integral extensions; therefore $B_{{\mathfrak{p}}^{\prime }}=B_{\mathfrak{p}}$ is integral over $A_{\mathfrak{p}}$.