Prove that for every $a\in A$ $\exists$ $q\in M,s\in A$ s.t. $a=sba+qa$ where $M$ is maximal ideal in $A$.

Question

Let A be a commutative ring with unity ring, M be a maximal ideal and $b\notin M$ . Show that there exists $q \in M, s\in A$ such that for every $a \in A$ we have $a=sba+qa$. Show that this is true only when $M$ is a maximal ideal.

Suppose if we take $A=2\mathbb Z$ and $M=4\mathbb Z$, $a=6.$ Then by the given statement $\exists$ $s\in A$, $q\in M$ s.t. $6=6(2s+q)$ $\implies$$s=\frac{1-q}{2}\implies q \text { is odd}$. But $q \in 4\mathbb Z.$

Please someone give some hints. how can solve this.

Thank you..

Answer 1

The set $W=\{sb +q : q\in M , s\in A\}$ is an ideal containing $M.$ Since $M$ is maximal thus $1\in W.$ Therefore the assertion is true.