An exercise about the direct sum of almost R-modules

Question

Let $R$ beacommutative ring.Call an (additive) abelian group $M$ an almost $R$-module if there is a function $R \times M \to M$ satisfying all the axioms of an $R$-module except axiom (iv):We do not assume that $1m=m$ for all $m$ $\in$ $M$.

Prove that

$M=M_1 \oplus M_0$, where $M_1=${$m \in M:1m=m$} and $M_0=${$m \in M:1m=0$ for all r $\in R$}.

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My question: It suffices to prove that $M \subseteq M_1+M_0$,in other words,every element of M has the form $x+y$,where $x \in M_1$ and $y \in M_0$,but I don't know how to prove it. By the way,I have found that $m \in M_0$ iff $1m=0$,is the fact helpful?

I am new to modern algebra,so please forgive me asking people naive questions here and there...

Answer 1

Hint : Given $m\in M$, look at $m_1 := 1m$ and $m_0 := m-m_1$.