The Stacks project

Lemma 47.3.9. Let $R$ be a Noetherian ring. Let $I$ be an injective $R$-module.

  1. Let $f \in R$. Then $E = \bigcup I[f^ n] = I[f^\infty ]$ is an injective submodule of $I$.

  2. Let $J \subset R$ be an ideal. Then the $J$-power torsion submodule $I[J^\infty ]$ is an injective submodule of $I$.

Proof. We will use Lemma 47.3.5 to prove (1). Suppose that $E \subset E' \subset I$ and that $E'$ is an essential extension of $E$. We will show that $E' = E$. If not, then we can find $x \in E'$ and $x \not\in E$. Let $J = \{ a \in R \mid ax \in E\} $. Since $R$ is Noetherian, we may choose $x$ so that $J$ is maximal among ideals of this form. Again since $R$ is Noetherian, we may write $J = (g_1, \ldots , g_ t)$ for some $g_ i \in R$. By definition $E$ is the set of elements of $I$ annihilated by powers of $f$, so we may choose integers $n_ i$ so that $f^{n_ i}g_ ix = 0$. Set $n = \mathrm{max}\{ n_ i \} $. Then $x' = f^ n x$ is an element of $E'$ not in $E$ and is annihilated by $J$. Then by maximality of $J$ we have $J = \{ a \in R \mid ax' \in E \} = \text{Ann}(x')$, so $Rx' \cap E = 0$. Hence $E'$ is not an essential extension of $E$, a contradiction.

To prove (2) write $J = (f_1, \ldots , f_ t)$. Then $I[J^\infty ]$ is equal to

\[ (\ldots ((I[f_1^\infty ])[f_2^\infty ])\ldots )[f_ t^\infty ] \]

and the result follows from (1) and induction. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 47.3: Injective modules

Post a comment

Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08XW. Beware of the difference between the letter 'O' and the digit '0'.