If $R$ is a ring and $M$ be any maximal ideal with $a\notin M$, then there exists $q\in M$ and $s\in R$ such that forall $r\in R$ $r=sar+qr$

Question

If $R$ is a ring and $M$ be any maximal ideal with $a\notin M$, then there exists $q\in M$ and $s\in R$ such that forall $r\in R$ we have $r=sar+qr$. Also show that this is true only when $M$ is maximal ideal. I am assuming $R$ is commutative and $1\in R$. Please help me. I am unable to proceed.

Answer 1

Since $b\notin M\implies M\subsetneqq \langle b\rangle+M. $ As $\langle b \rangle+M$ is an ideal of $R$ and $M\subsetneqq \langle b\rangle +M\subseteq R $. Since $M$ is maximal, $\langle b\rangle +M=R$. Since $1\in R\implies \exists\ s\in R$ and $q\in M$ such that $$ 1=bs+q\implies a=sba+qa $$