Over a ring with a central idempotent, every module is an internal direct sum

Let $R$ be a ring with 1 and let $M$ be a left unital $R$-module. Suppose there exists $e \in R$ such that $e^2 = e$ and $e \in Z(R)$. Prove that $M = eM \oplus (1-e)M$.

First, recall by this previous exercise that $eM$ and $(1-e)M$ are both submodules of $M$. Suppose $m \in M$. Then $m = e \cdot m + 1 \cdot m - e \cdot m \in eM + (1-e)M$, so that $M = eM + (1-e)M$. Finally, suppose $m \in eM \cap (1-e)M$; say $m = e \cdot a = (1-e) \cdot b$. Now $e \cdot a = 1 \cdot b - e \cdot b$, so that $e \cdot (a+b) = b$. Now $e \cdot b = e^2 \cdot (a+b) = e \cdot (a + b)$ $= e \cdot a + e \cdot b$, so that $e \cdot a = 0$. Thus $m = 0$, and we have $eM \cap (1-e)M = 0$. So $M = eM \oplus (1-e)M$.

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