TITLE: Topologically nilpotent elements of a linearly topologized ring
QUESTION [6 upvotes]: In what follows all rings are commutative topological rings.
An element $x$ of a ring is called topologically nilpotent if $lim_{n\rightarrow \infty} x^n = 0$.
If a ring $A$ has a fundamental system of neighborhoods of $0$ consiting of ideals, $A$ is called linearly topologized.
An open ideal $I$ of a linearly topologized ring is called an ideal of definition if, for every neighborhood $V$ of $0$, there exists an integer $n > 0$ such that $I^n \subset V$.
The following proposition is Lemma 7.1.3 of Grothendieck's EGA I, Ch. 0.
Proposition
Suppose a linearly topologized ring $A$ has an ideal of definition $I$.
Suppose $x$ mod $I$ is nilpotent in $A/I$.
Then $x$ is topologically nilpotent.
The proof of EGA is as follows.
Let $V$ be a neighborhood of $0$.
There exists an integer $n > 0$ such that $I^n \subset V$.
If $x^m \in I$, $x^{mq} \in V$ for all integer $q \ge n$.
Hence $x$ is topologically nilpotent.
My question
Why $x$ is topologically nilpotent?

REPLY [2 votes]: It seems like your problem is with the last sentence. This should be just the definition of convergence of the sequence $(x^n)_{n\in \mathbb{N}}$: It convergences against $0$ iff for every neighbourhood $V$ there exists $s_0$ such that for all $s\geq s_0$ we have $x^{s}\in V$. In fact the sentence before should better read $x^s\in I^n\subset V$ for all $s\geq mn$ (since $I$ is an ideal).