Why does $0_R\cdot m=0_M$ and $r\cdot 0_M=0_M$ in a module?

Question

According to Wikipedia an R-module is define as:

Suppose that $R$ is a ring and $1_R$ is its multiplicative identity. A left R-module $M$ consists of an abelian group $(M, +)$ and an operation $⋅ : R × M → M$ such that for all $r, s \in R$ and $x, y \in M$, we have:

$$r\cdot (x+y)=r\cdot x+r\cdot y $$ $$(r+s)\cdot x=r\cdot x+s\cdot x $$ $$(rs)\cdot x=r\cdot (s\cdot x) $$ $$1_{R}\cdot x=x. $$

My question is why we do not require that $0_R \cdot x = 0_M$ and $r \cdot 0_M = 0_M$. I think these conditions are necessary but I do not see how these follow from the above-given axioms.

Answer 1

Hint :

$0_R \cdot x$ $= (0_R +0_R)\cdot x$

By distributivity in definition of module,

$=0_R \cdot x+0_R \cdot x$.

Can you take it over from here?

Answer 2

$$0x = (0+0)x = 0x + 0x \implies 0x = 0$$ $$r0 =r(0+0) = r0 + r0 \implies r0=0$$