Lemma 87.4.10. Let $B$ be a linearly topologized ring. The set of topologically nilpotent elements of $B$ is a closed, radical ideal of $B$. Let $\varphi : A \to B$ be a continuous map of linearly topologized rings.

1. If $f \in A$ is topologically nilpotent, then $\varphi (f)$ is topologically nilpotent.

2. If $I \subset A$ consists of topologically nilpotent elements, then the closure of $\varphi (I)B$ consists of topologically nilpotent elements.

Proof. Let $\mathfrak b \subset B$ be the set of topologically nilpotent elements. We omit the proof of the fact that $\mathfrak b$ is a radical ideal (good exercise in the definitions). Let $g$ be an element of the closure of $\mathfrak b$. Our goal is to show that $g$ is topologically nilpotent. Let $J \subset B$ be an open ideal. We have to show $g^ e \in J$ for some $e \geq 1$. We have $g \in \mathfrak b + J$ by Lemma 87.4.2. Hence $g = f + h$ for some $f \in \mathfrak b$ and $h \in J$. Pick $m \geq 1$ such that $f^ m \in J$. Then $g^{m + 1} \in J$ as desired.

Let $\varphi : A \to B$ be as in the statement of the lemma. Assertion (1) is clear and assertion (2) follows from this and the fact that $\mathfrak b$ is a closed ideal. $\square$

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