# 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 $M$. Add to solve later

## 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 $M$. Add to solve later

### 1 Response

1. 05/17/2017

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

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