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 $M$.

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 $M$.

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 […]

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 […]

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