$R$ is a communitive without an identity, $M$ is the maximal ideal of $R$ and $R/M$ is not a field.How to prove $(R/M)^{2}=0?$

Question

$R$ is a communitive without an identity, $M$ is the maximal ideal of $R$ and $R/M$ is not a field.How to prove $(R/M)^{2}=0?$

I have known an example for a this context.Take $R=(2)$ and $M=(2p)$ then $M$ is maximal but $R/M$ is not a field.Maybe the example can give some hint but I cannot work it out.

Answer 1

Suppose that $M$ is a proper ideal in $R$ which is maximal with respect to the inclusion. Pick $S = R/M$. Then $S$ is a nontrivial ring (possibly without unit) and $S$ has no ideals except $0$ and $S$. Since $S\cdot S\subseteq S$ is an ideal, we have two possibilities.

1. $S\cdot S = 0$ and then we are done.
2. $S\cdot S = S$ then there exists $x\in S$, $x\neq 0$ such that $S\cdot x =\{yx\in S\,|y\in S\}\neq 0$. Thus $S\cdot x\subseteq S$ is a nontrivial ideal and we deduce that $S\cdot x= S$. In particular, there is $e\in S$ such that $e\cdot x = x$. Clearly $e\neq 0$. Now define a morphism $f:S\rightarrow S$ by formula $$f(s) = e\cdot s - s$$ This is clearly a morphism of $S$-modules and $x\in \mathrm{ker}(f)$. Thus $\mathrm{ker}(f)\neq 0$ and since the kernel is also an ideal, we derive that $\mathrm{ker}(f) = S$. Hence $f\equiv 0$ and $e\cdot s = s$ for all $s\in S$. Therefore, $e$ is a unit in $S$ and $S$ is a commutative ring with unit such that its family of ideals is $\{0,S\}$. Thus $S$ is a field.