Two decreasing <filtrations of a module> $(M_n)$ and $(M'_n)$ are equivalent when each contains a fixed shift of the other: there are $a,b\ge0$ such that
$$
M_{n+a}\subseteq M'_n,
\qquad
M'_{n+b}\subseteq M_n
$$
for every $n\ge0$.

For an $I$-filtration, $IM_n\subseteq M_{n+1}$, so $I^nM_0\subseteq M_n$. If it is stable from $s$ onward, then
$$
M_{s+n}=I^nM_s\subseteq I^nM_0.
$$
It is therefore equivalent to the <I-adic filtration>. Any two stable $I$-filtrations are consequently equivalent to each other.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 101 / 4 / b / Solution

{c}
{title2=$(I^nM)$}
{wiki=I-adic_topology}

For an ideal $I$, the $I$-adic filtration of a module $M$ is $M\supseteq IM\supseteq I^2M\supseteq\cdots$.


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 Incoming links: I-adic filtration