$M$ maximal iff $\bar{M}$ is maximal

Question

We have a ring $R$ and $I$ an ideal of $R$. Let $M$ be an ideal of $R$ containing $I$. Let $\overline{M}$ be $M/I$ and $\overline{R}$ be $R/I$. Prove that $M$ is maximal if and only if $\overline{M}$ is maximal.

I think I get the general idea, that $M$ not maximal means there exists a bigger $M'$ containing $M$ and so $M'/I$ contains $M/I$. How do I formalize this idea? And how do I show the other direction?

Answer 1

By the third isomorphy theorem you have $$R/M \cong (R/I) / (M/I).$$ Each of this quotients is a field if and only if $M$ is maximal, and if and only if $M/I$ is maximal.

Answer 2

Let $M$ be maximal and $I \subseteq M$. We show that $M/I$ is maximal in $R/I$.

So let, $J$ be an ideal in $R/I$ with $J \supseteq M/I$. By one of the isomorphism theorems, we can find an ideal $N$ of $R$ with $I \subseteq N$ and $J = N/I$. Thus $M/I \subseteq N/I$ and consequently $M \subseteq N$ (why?). Since $M$ is maximal, we obtain either $N=M$ or $N=R$. Thus $J = M/I$ or $J=R/I$. Hence, $M/I$ is maximal.

The converse is similar.

Answer 3

Let $R$ be an arbitrary ring and $I$ be an ideal. From the lattice isomorphism theorem, the lattice of ideals of $R$ containing $I$ is isomorphic to the lattice of ideals of $R/I$ under the map $J \leftrightarrow \overline{J}$. Then, of course, an ideal $M$ of $R$ containing $I$ is maximal if, and only if $\overline{M}$ is maximal. Maximality is just a property of the lattice which is preserved by the lattice isomorphism.