I want to prove the statement: "The ideal $\mathcal m \subset \mathcal R $ is maximal $\iff $ the zero ideal $0 \subset \mathcal R/m$ is maximal, where $\mathcal R$ is a commutative ring with unity", by using the correspondence result: There is a one-to-one order preserving correspondence between the ideals $\mathcal b $ of $\mathcal R$ which contain $\mathcal m $, and the ideals $\bar b $ of $\mathcal R/m:$ $$ \pi: \mathcal R \rightarrow \mathcal R/m, \,\,\,\mathcal R \supset b \mapsto \pi(b)\subset \mathcal R/m.$$
$"\implies ":$ Let $m$ be a maximal ideal, i.e. there is no ideal $b$ satisfying $m \subset b \subset \mathcal R.$ If this is the case, then there is no argument left for the canonical ring homomorphism $\pi.$ I thus considered the situation $m \subset b \subseteq \mathcal R,$ but I dont know if it is correct. Thus I can find at most only one ideal containing $m,$ which is $\mathcal R.$ At the right side, $\mathcal R/m,\,$ $m$ is an ideal of $\mathcal R/m,$ since $\bar c m=(c+m)m=m, \forall \bar c \in \mathcal R/m.$ Actually $m$ is the zero ideal of $\mathcal R/m.$ Are there other ideals in $\mathcal R/m$ ? I know that there must be in total 2: the zero ideal, which is $m$ and $(1)=\mathcal R/m.$ Let $n$ be an ideal of $\mathcal R/m,$ such that $m \in n \in \mathcal R/m.$ An element of $n$ will have the form $\bar c \in \mathcal R/m,$ thus $mn=m.$ Since $n$ is also an ideal, we must also have $mn=n,$ and so $m=n.$ I am somehow not getting the desired ideal $(1)=\mathcal R/m$. But even if I did, it would contradict the correspondence theorem, since I could find at most only one ideal at the left side of the canonical mapping. And I am blocked. I also need to prove the other implication.
Can somebody help me by pointing out my mistakes and proposing something ? Many thanks.
There is a one-to-one order preserving correspondence between the ideals of $R/m$ and the ideals of $R$ containing $m$.
If $m$ is maximal, there are exactly two ideals of $R$ containing $m$, namely $R$ and $m$. Thus $R/m$ has exactly two ideals and therefore the zero ideal is maximal.
If the zero ideal in $R/m$ is maximal, $R/m$ has exactly two ideals. So there are exactly two ideals of $R$ containing $m$. Since $m$ and $R$ do contain $m$, we conclude that there is no ideal properly between $m$ and $R$, therefore $m$ is maximal.