## The union of a chain of submodules is a submodule

Let $R$ be a ring with 1 and let $M$ be a left $R$-module. Let $\{M_t\}_{t \in T}$ be a nonempty chain of left $R$-submodules of $M$; that is, if $i \leq j$, then $M_i \subseteq M_j$. Prove that $\bigcup_T M_t$ is a submodule of $M$.

We use the submodule criterion.

Certainly $0 \in M_i$ for some $i$. Thus $0 \in \bigcup M_t$, and hence $\bigcup M_t \neq 0$.

Now suppose $a,b \in \bigcup M_t$ and $r \in R$. Now $a \in M_i$ and $b \in M_j$ for some $i$ and $j$. Suppose without loss of generality that $i \leq j$; then $a,b \in M_j$. Since $M_j$ is a submodule, $a+rb \in M_j$. Thus $a+rb \in \bigcup M_t$.

Hence $\bigcup M_t$ is a submodule of $M$.