Proposition 6

*Let $A$ be a commutative Noetherian ring and $m$ an ideal of $A$. The following properties are equivalent: (a) $m$ is contained in the Jacobson radical of $A$. (b) Every finitely generated $A$-module is Hausdorff with the $m$-adic topology. (c) For every finitely generated $A$-module $E$, every submodule of $E$ is closed with respect to the $m$-adic topology on $E$. (d) Every maximal ideal of $A$ is closed with respect to the $m$-adic topology.

Let us show that (a) implies (b). Suppose that $m$ is contained in the Jacobson radical of $A$ and let $E$ be a finitely generated $A$-module. If $x \in E$ and $m \in m$ are such that $(1 - m)x = 0$, then $x = 0$, for $1 - m$ is invertible in $A$. Then (no. 2, Proposition 5) $E$ is Hausdorff with the $m$-adic topology.

Let us prove that (b) implies (c). Suppose (b) holds. Let $E$ be a finitely generated $A$-module and $F$ a submodule of $E$. Then $E/F$ is Hausdorff with the $m$-adic topology, which is the quotient topology of the $m$-adic topology on $E$; then $F$ is closed in $E$.

Clearly (c) implies (d). Let us show finally that (d) implies (a). It follows from (d) that, for every maximal ideal $a$ of $A$, the $A$-module $A/a$ is Hausdorff with the $m$-adic topology. This implies $m(A/a) \neq A/a$, unless the $m$-adic topology on $A/a$ were the coarsest topology and $A/a$ were reduced to 0, which is absurd since $A/a$ is ajield. The canonical image of $m$ in $A/a$ is therefore an ideal of $A/a$ distinct from $A/a$ and hence reduced to 0; then $m \subset a$, which proves that $m$ is contained in the Jacobson radical of $A$.