Lemma 15.88.4. Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$.

1. For any $R$-module $M$ we have $(M/M[I^\infty ])[I] = 0$.

2. An extension of $I$-power torsion modules is $I$-power torsion.

Proof. Let $m \in M$. If $m$ maps to an element of $(M/M[I^\infty ])[I]$ then $Im \subset M[I^\infty ]$. Write $I = (f_1, \ldots , f_ t)$. Then we see that $f_ i m \in M[I^\infty ]$, i.e., $I^{n_ i}f_ i m = 0$ for some $n_ i > 0$. Thus we see that $I^ Nm = 0$ with $N = \sum n_ i + 2$. Hence $m$ maps to zero in $(M/M[I^\infty ])$ which proves the first statement of the lemma.

For the second, suppose that $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of modules with $M'$ and $M''$ both $I$-power torsion modules. Then $M[I^\infty ] \supset M'$ and hence $M/M[I^\infty ]$ is a quotient of $M''$ and therefore $I$-power torsion. Combined with the first statement and Lemma 15.88.3 this implies that it is zero. $\square$

Comment #8387 by Peng Du on

Need to add "." in the end of Proof.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).