The submodule generated by a subset consists of the finite $R$-linear combinations of that subset

Let $L$ be the set of all sums ${\sum }_{i=1}^k{r}_i{s}_i$ with $k\in \mathbb{N}$, $r_1,\dots ,r_k\in R$ and $s_1,\dots ,s_k\in S$, the value at $k=0$ being the empty sum $0_M$; in particular $0_M\in L$, so $L$ is nonempty even when $S$ is empty.

Every $s\in S$ lies in $L$, being the one-term sum $1_{R}s$, which equals $s$ by the unitality axiom of a module; hence $S\subseteq L$.

$L$ is a submodule of $M$: it is nonempty, and for $r\in R$ and $u={\sum }_{i=1}^k{r}_i{s}_i$, $v={\sum }_{j=1}^m{t}_j{u}_j$ in $L$ the element $ru+v$ is the sum ${\sum }_{i=1}^k(r{r}_i)s_i+{\sum }_{j=1}^m{t}_j{u}_j$, again a finite $R$-linear combination of elements of $S$ with $k+m$ terms, so $ru+v\in L$ and the submodule criterion applies.

Every submodule $N\le M$ with $S\subseteq N$ contains $L$: each $s_i$ lies in $N$, so repeated use of the submodule criterion [L2] puts every $r_i{s}_i$ and every finite sum of those elements in $N$; the case $k=0$ gives $0_M\in N$.

$⟨S{⟩}_R=L$: by steps 2.1 and 2.2 the set $L$ is one of the submodules over which the intersection defining $⟨S{⟩}_R$ runs, so $⟨S{⟩}_R\subseteq L$; and $⟨S{⟩}_R$ is itself a submodule containing $S$, so step 2.3 applied to it gives $L\subseteq ⟨S{⟩}_R$.

Two consequences follow by reading the description at its extreme cases. Taking $S=\mathrm{\varnothing }$ leaves only the empty sum, so $⟨\mathrm{\varnothing }{⟩}_R=0$. Taking $M$ to be the regular module over a commutative $R$, where the submodules are the ideals, the description becomes the finite-sum description of the ideal generated by $S$, so the two agree and this lemma extends that one from ideals to modules rather than competing with it.