Ascending Chain of Submodules and Union of its Submodules

Module Theory problems and solutions

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:

  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 $M$.


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  1. 05/17/2017

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

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