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Let $R$ be a ring with a descending filtration by ideals

$$R = I^{(0)} \supset I^{(1)} \supset I^{(2)} \supset \dots$$

such that $I^{(j)} I^{(k)} \subset I^{(j+k)}$, with equality whenever $k$ is sufficiently large. Is the topology induced by this filtration necessarily the same as the $\mathfrak a$-adic topology for some ideal $\mathfrak a \subset R$?

isekaijin
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    If you take on $\mathbb{Z}$ the flitration defined by $I^0=(2)$, $I^1=(2.3)$...$I^n=(2.3...p_n)$ where $p_n$ is the n-th prime number. This topology won't be an $\mathfrak{a}$-adic topology for no ideal $\mathfrak{a}$. – Ahr Sep 04 '19 at 07:54
  • @Ahr: I don't think your filtration has some instant $n \in \mathbb N$ such that $I^{(j)} I^{(k)} = I^{(j+k)}$ for all $k \ge n$. – isekaijin Sep 04 '19 at 07:57
  • Oh, right. I didn't see you wanted equality for some sufficiently large $k$. Sorry. – Ahr Sep 04 '19 at 07:58

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