For a commutative ring $R$, the <Jacobson radical> is
$$
J(R)=\bigcap_{\mathfrak m\in\operatorname{MaxSpec}R}\mathfrak m.
$$
Let $A\subseteq B$ be integral. If $\mathfrak n$ is maximal in $B$, then its contraction $\mathfrak n\cap A$ is maximal in $A$. Therefore every $a\in J(A)$ belongs to every $\mathfrak n$, and
$$
J(A)\subseteq J(B)\cap A.
$$
Conversely, the <Lying-over theorem> puts a maximal ideal $\mathfrak n$ of $B$ above every maximal ideal $\mathfrak m$ of $A$. Hence an element of $J(B)\cap A$ lies in every $\mathfrak m$, proving
$$
\boxed{J(A)=J(B)\cap A}.
$$
This is the <Jacobson radical under an integral extension> formula.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 / 2 / ii / Solution

{c}

For an <integral extension> $A\subseteq B$ of commutative rings,
$$
J(A)=J(B)\cap A.
$$
Contraction sends maximal ideals of $B$ to maximal ideals of $A$, while the <Lying-over theorem> puts a maximal ideal of $B$ over every maximal ideal of $A$.

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 Incoming links: Jacobson radical under an integral extension