## Two facts about unital modules

Let $R$ be a ring with 1 and let $M$ be a left unital $R$-module. Prove that for all $m \in M$, $0 \cdot m = 0$ and $(-1) \cdot m = -m$.

Let $m \in M$. Now $0 \cdot m + 1 \cdot m = (0+1) \cdot m = 1 \cdot m$ by the definition of module. Since $M$ is an abelian group, subtracting $1 \cdot m$ from both sides we have $0 \cdot m = 0$.

Now note that $1 \cdot m + (-1) \cdot m = (1 - 1) \cdot m = 0 \cdot m = 0$. By the uniqueness of inverses in a group, $-m = -(1 \cdot m) = (-1) \cdot m$.