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 $M$.

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“.)

The post Submodule Consists of Elements Annihilated by Some Power of an Ideal first appeared on Problems in Mathematics.

Let $R$ be a ring with $1$. Let $M$ be an $R$-module. Consider an ascending chain
\[N_1 \subset N_2 \subset \cdots\] of submodules of $M$.
Prove that the union
\[\cup_{i=1}^{\infty} N_i\] is a submodule of $M$.

Proof.

To simplify the notation, let us put
\[U=\cup_{i=1}^{\infty} N_i.\] Prove that $U$ is a submodule of $M$, it suffices to show the following two conditions:

1. For any $x, y\in U$, we have $x+y\in U$, and
2. For any $x\in U$ and $r\in R$, we have $rx\in U$.

To check condition 1, let $x, y\in U$.
Since $x$ lies in the union $U=\cup_{i=1}^{\infty} N_i$, there is an integer $n$ such that
\[x\in N_n.\] Similarly, we have
\[y \in N_m\] for some integer $m$.

Since $N_n\subset N_{\max(n, m)}$ and $N_m\subset N_{\max(n, m)}$, we have
\[x, y \in N_{\max(n, m)}.\] As $N_{\max(n, m)}$ is a submodule of $M$, it is closed under addition.

It follows that
\[x+y \in N_{\max(n, m)}\subset U.\] Hence condition 1 is met.

Next, we consider condition 2.
Let $x \in U$ and $r\in R$.
Since $x$ is in the union $U$, there exists an integer $n$ such that $x\in N_n$.

Since $N_n$ is a submodule of $M$, it is closed under scalar multiplication.
Thus we have
\[rx\in N_n \subset U.\] Therefore, condition 2 is satisfied, and so $U$ is a submodule of $M$.

The post Ascending Chain of Submodules and Union of its Submodules first appeared on Problems in Mathematics.