One-step submodule test

Nonempty and closed under $rm+n$ already means submodule.Suggest a slogan

Let $R$ be a ring, let $M$ be a left $R$-module, and let $N\subseteq M$ be a nonempty subset. If $rm+n\in N$ for all $r\in R$ and $m,n\in N$, then $N$ is a submodule of $M$.

Choose $n\in N$; then $0=(-1)n+n\in N$. For $r\in R$ and $m\in N$, $rm=rm+0\in N$; in particular $-m=(-1)m\in N$. For $m,n\in N$, $m+n=1\cdot m+n\in N$. So $N$ is an additive subgroup closed under scalar multiplication.