Lemma 7.6.3.label Let $R$ be a unital ring, $M$ be a left $R$-module, $\p:M\to R$ be an $R$-linear map and suppose there exists $e\in M$ such that $\p(e)=1_{R}$ then $M=\ker\p\op Re$
Proof. Let $m\in M$ then consider $m_{0}\define m-\p(m)e$
\begin{align*}\p(m)=\p(m-\p(m)e+\p(m)e)=\parens{\p(m)-\p(m)\p(e)}+\p(m)\p(e)\in \ker\p+Re\end{align*}
If $r e\in\ker\p$ for some $r\in R$ then
\begin{align*}0=\p(re)=r\p(e)=r\end{align*}
showing the sum is direct.$\square$
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