I was reading the following from Liu's Algebraic Geometry and Arithmetic Curves on page 18. In this book all rings are commutative and with unit.

Let $A$ be a ring endowed with the $I$-adic topology. We denote the completion of $A$ by $\hat A$. We also call $\hat A$ the formal completion of $A$ for the $I$-adic topology. As $A$ is a ring, so is $\hat A$. Let $M$ be an $A$-module. "Then any submodule filtration $(M_n)$ of $M$ defines the structure of a topological $A$-module on $M$."

But Liu wrote on errata that the last sentence in the quote should be deleted. So what would be an example of a commutative ring $A$ with unit, its ideal $I$, and a submodule filtration $(M_n)$ of $M$ that does not define the structure of a topological $A$-module on $M$?


1 Answer 1


One would usually want to insist that the filtration on $M$ is compatible with the $I$-adic filtration on $R$, i.e. for all $n$, there exists $k$ such that $I^k M_n \subset M_{n+1}$ (or just $R_i M_j \subset M_{i+j}$, if we are working with a general filtration on $R$).

So, for example, if no power of $I$ is $0$, we can take the filtration $M\supset 0 \supset 0 \cdots$ of $M \cong R$, which does not induce an $I$-adic structure on $M$.

  • $\begingroup$ For your example to work, one needs that no power of $I$ be zero, no? $\endgroup$ Feb 4, 2015 at 22:53
  • $\begingroup$ @MarianoSuárez-Alvarez Yes, I think that's right. I've updated with a slightly more general definition. $\endgroup$
    – Slade
    Feb 4, 2015 at 22:55

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