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 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 $J$. Set $J' = \{ a \in R \mid ax' \in E \}$ so $J \subset J'$. Conversely, we have $a \in J'$ if and only if $ax' \in E$ if and only if $f^ m a x' = 0$ for some $m \geq 0$. But then $f^ m a x' = f^{m + n} a x$ implies $ax \in E$, i.e., $a \in J$. Hence $J = J'$. Thus $J = J' = \text{Ann}(x')$, 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 (2)

Comment #6237 by Owen on

I don't think it's necessary to take $J$ maximal among ideals of this form. Indeed, the inclusion $J:=\{a\in R|ax\in E\}\subset\{a\in R|ax'\in E\}$ is automatic, and the reverse inclusion is also automatic, since $ax'\in E$ iff $f^max'=0$ for some $m$, but $f^max'=f^{m+n}ax$, so $ax\in E$. Clearly $\{a\in R|ax'\in E\}\supset\operatorname{Ann}(x')$, but it is just shown that $J\subset\operatorname{Ann}(x')$. So I don't see where maximality of $J$ figures into the argument.

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