Show that $0\cdot m = 0$

Question

If $R$ is a ring with $1$ and $M$ is a left $R$-module, how could I prove that $0_R m = 0_M$? I am using the following axioms

1. $(M,+)$ is an abelien group.
2. $(r+s)\cdot m = r\cdot m + s\cdot m$ for all $r,s\in R, m\in M$.
3. $(r\times s) \cdot m = r\times (s\cdot m)$ for all $r,s\in R, m \in M$.
4. $r\cdot(m+n) = r\cdot m + r\cdot n$ for all $r\in R,m,n\in M$
5. $1\cdot m = m$ for all $m\in R$

I actually did $$0_R m = (r + (-r)) m=rm + (-r)m=rm - rm = 0$$

But then I realized that I made a circular reasoning, because I used this conclusion to show that

$$m + (-1)m=1\cdot m + (-1)\cdot m = (1 + (-1))\cdot m =0 \cdot m = 0$$

, which I used in the first proof.

Answer 1

Notice that by the ring axioms we have that $$0_R=0_R+0_R$$ and by abelian groups we have that $$a+a=a$$ implies that $a=0_M$ and as such with $$0_Rm=(0_R+0_R)m=0_Rm+0_Rm\in M$$ and the rest follows.