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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?

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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).

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    Indeed, the sentence before the last one is misleading.2012-11-02