# Submodule Consists of Elements Annihilated by Some Power of an Ideal ## Problem 417

Let $R$ be a ring with $1$ and let $M$ be an $R$-module. Let $I$ be an ideal of $R$.
Let $M’$ be the subset of elements $a$ of $M$ that are annihilated by some power $I^k$ of the ideal $I$, where the power $k$ may depend on $a$.
Prove that $M’$ is a submodule of $M$. Add to solve later

## Proof.

Let us define the subset of $M$ by
$N_i=:\{a\in M \mid sa=0 \text{ for all } s\in I^i\}.$ That is, $N_i$ consists of elements of $M$ that are annihilated by the power $I^i$.

We claim that:

1. the subset $N_i$ is a submodule of $M$ for each integer $i$, and
2. we have the ascending chain
$N_1 \subset N_2 \subset \cdots,$ and
3. $M’=\cup_{i=1}^{\infty} N_i$.

Once we prove these claims, the result follows from the previous problem.

Let us prove claim 1. Let $a, b\in N_i$ and let $r\in R$.
For any $s\in I^i$ we have
\begin{align*}
s(a+b)&=sa+sb=0
\end{align*}
because $a, b$ are annihilated by $s\in I^i$.
Also, we have
\begin{align*}
s(ra)=(sr)a=0
\end{align*}
since $sr\in I$ as $I$ is an ideal.
Thus, $N_i$ is a submodule of $M$.

To prove claim 2, we note the inclusion
$I^{i+1}=I^i\cdot I\subset I^{i}.$ Thus each $a\in N_i$ is annihilated by elements in $I^{i+1}$.
Hence $N_i\subset N_{i+1}$ for any $i$, and this proves claim 2.

The claim 3 follows from the definition of the subset $M’$.

Since the union of submodules in an ascending chain of submodules is a submodule, we conclude that $M’$ is a submodule of $M$.

(For a proof of this fact, see the post “Ascending chain of submodules and union of its submodules“.) Add to solve later

### More from my site

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

###### More in Module Theory ##### Ascending Chain of Submodules and Union of its Submodules

Let $R$ be a ring with $1$. Let $M$ be an $R$-module. Consider an ascending chain \[N_1 \subset N_2 \subset...

Close