# 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 $M$.

## 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 $M$.

### More from my site

• Submodule Consists of Elements Annihilated by Some Power of an Ideal 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 […]
• Torsion Submodule, Integral Domain, and Zero Divisors Let $R$ be a ring with $1$. An element of the $R$-module $M$ is called a torsion element if $rm=0$ for some nonzero element $r\in R$. The set of torsion elements is denoted $\Tor(M)=\{m \in M \mid rm=0 \text{ for some nonzero} r\in R\}.$ (a) Prove that if $R$ is an […]
• Annihilator of a Submodule is a 2-Sided Ideal of a Ring Let $R$ be a ring with $1$ and let $M$ be a left $R$-module. Let $S$ be a subset of $M$. The annihilator of $S$ in $R$ is the subset of the ring $R$ defined to be $\Ann_R(S)=\{ r\in R\mid rx=0 \text{ for all } x\in S\}.$ (If $rx=0, r\in R, x\in S$, then we say $r$ annihilates […]
• A Module $M$ is Irreducible if and only if $M$ is isomorphic to $R/I$ for a Maximal Ideal $I$. Let $R$ be a commutative ring with $1$ and let $M$ be an $R$-module. Prove that the $R$-module $M$ is irreducible if and only if $M$ is isomorphic to $R/I$, where $I$ is a maximal ideal of $R$, as an $R$-module.     Definition (Irreducible module). An […]
• A Module is Irreducible if and only if It is a Cyclic Module With Any Nonzero Element as Generator Let $R$ be a ring with $1$. A nonzero $R$-module $M$ is called irreducible if $0$ and $M$ are the only submodules of $M$. (It is also called a simple module.) (a) Prove that a nonzero $R$-module $M$ is irreducible if and only if $M$ is a cyclic module with any nonzero element […]
• Basic Exercise Problems in Module Theory Let $R$ be a ring with $1$ and $M$ be a left $R$-module. (a) Prove that $0_Rm=0_M$ for all $m \in M$. Here $0_R$ is the zero element in the ring $R$ and $0_M$ is the zero element in the module $M$, that is, the identity element of the additive group $M$. To simplify the […]
• Short Exact Sequence and Finitely Generated Modules Let $R$ be a ring with $1$. Let $0\to M\xrightarrow{f} M' \xrightarrow{g} M^{\prime\prime} \to 0 \tag{*}$ be an exact sequence of left $R$-modules. Prove that if $M$ and $M^{\prime\prime}$ are finitely generated, then $M'$ is also finitely generated.   […]
• Linearly Dependent Module Elements / Module Homomorphism and Linearly Independency (a) Let $R$ be a commutative ring. If we regard $R$ as a left $R$-module, then prove that any two distinct elements of the module $R$ are linearly dependent. (b) Let $f: M\to M'$ be a left $R$-module homomorphism. Let $\{x_1, \dots, x_n\}$ be a subset in $M$. Prove that if the set […]

### 1 Response

1. 05/17/2017

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

##### Linearly Dependent Module Elements / Module Homomorphism and Linearly Independency

(a) Let $R$ be a commutative ring. If we regard $R$ as a left $R$-module, then prove that any two...

Close