For a maximal ideal $M$ of $R$ of a commutative ring $R$ ( not necessarily with unity ) , then is $R/M$ a simple ring ?

Question

Let $M$ be a maximal ideal of a commutative ring $R$ ( not necessarily with unity ) ; then is it true that the only ideals of $R/M$ are the trivial ones i.e. is it true that $R/M$ is a simple ring ? I know that if $R$ has unity then $R/M$ is a field ; but here unity is not assumed ; I tried to proceed as let $I$ be an ideal of $R/M$ such that $\exists M \ne a+M \in I$ , off-course $Ra+M$ is an ideal containing $M$ , then I figured out that if I could show $Ra+M=M$ , then I could show $I=R/M$ thus proving $R/M$ is simple ; but I cannot seem to be able to conclude $Ra+M=R$ without the existence of a unity in $R$ and am not even sure whether $Ra+M=R$ actually is true or not . Please help , thanks in advance

Answer 1

Let $I$ be an ideal of $R/M$. Let $J = \{j \in R: j+M \in I \subset R/M\}$.

J is an ideal of $R$: if $j+M \in I$ and $j'+M \in I$ then $j + j' + M \in I$, and if $r \in R$ then $rj + M = r(j+M) \in I$.

Therefore either $J = R$ or $J = M$. If $J = M$ then $I$ is the zero ideal in $R/M$, and if $J = R$ then $I = R/M$. So $R/M$ is simple.