## A restriction on left invertability in a ring in terms of its left modules

Let $R$ be a ring with 1 and let $M$ be a left unital $R$-module. Suppose $r \in R$ and $m \in M$, $m \neq 0$, such that $r \cdot m = 0$. Prove that $r$ does not have a left inverse in $R$.

We begin with a lemma.

Lemma: Let $M$ be a unital left $R$-module. The for all $r \in R$, $r \cdot 0 = 0$. Proof: Let $m \in M$. Note that $r \cdot 0 + r \cdot m = r \cdot (0+m) = r \cdot m$. Thus $r \cdot 0 = 0_M$. $\square$

Suppose there exists $s \in R$ such that $sr = 1$. Now $m = 1 \cdot m$ $= (sr) \cdot m$ $= s \cdot (r \cdot m)$ $= s \cdot 0$ $= 0$, a contradiction; hence no such left inverse exists.