Lemma 89.4.7. Let $R$ be an object of $\widehat{\mathcal{C}}_\Lambda$. Let $(J_ n)$ be a decreasing sequence of ideals such that $\mathfrak m_ R^ n \subset J_ n$. Set $J = \bigcap J_ n$. Then the sequence $(J_ n/J)$ defines the $\mathfrak m_{R/J}$-adic topology on $R/J$.

Proof. It is clear that $\mathfrak m_{R/J}^ n \subset J_ n/J$. Thus it suffices to show that for every $n$ there exists an $N$ such that $J_ N/J \subset \mathfrak m_{R/J}^ n$. This is equivalent to $J_ N \subset \mathfrak m_ R^ n + J$. For each $n$ the ring $R/\mathfrak m_ R^ n$ is Artinian, hence there exists a $N_ n$ such that

$J_{N_ n} + \mathfrak m_ R^ n = J_{N_ n + 1} + \mathfrak m_ R^ n = \ldots$

Set $E_ n = (J_{N_ n} + \mathfrak m_ R^ n)/\mathfrak m_ R^ n$. Set $E = \mathop{\mathrm{lim}}\nolimits E_ n \subset \mathop{\mathrm{lim}}\nolimits R/\mathfrak m_ R^ n = R$. Note that $E \subset J$ as for any $f \in E$ and any $m$ we have $f \in J_ m + \mathfrak m_ R^ n$ for all $n \gg 0$, so $f \in J_ m$ by Artin-Rees, see Algebra, Lemma 10.51.4. Since the transition maps $E_ n \to E_{n - 1}$ are all surjective, we see that $J$ surjects onto $E_ n$. Hence for $N = N_ n$ works. $\square$

Comment #7732 by Mingchen on

I understand that Krull's intersection theorem is a simple consequence of Artin-Rees, but here what you really need is Krull's intersection theorem, so it might be better to say $f\in J_m$ by Krull's intersection theorem.

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