Lemma 15.88.4. Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$.
For any $R$-module $M$ we have $(M/M[I^\infty ])[I] = 0$.
An extension of $I$-power torsion modules is $I$-power torsion.
Lemma 15.88.4. Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$.
For any $R$-module $M$ we have $(M/M[I^\infty ])[I] = 0$.
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. $\square$
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (1)
Comment #8387 by Peng Du on