Ascending Chain of Submodules and Union of its Submodules
Problem 416
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 $M$.
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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:
- For any $x, y\in U$, we have $x+y\in U$, and
- 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 $M$.
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