Let $I, M$ be proper idelas of a ring $R$. Prove the following:

Question

1) $M$ is maximal ideal iff $R/M$ is a simple ring,

2) $R$ has a maximal ideal that contains $I$,

3) $R$ has at least one maximal ideal.

I'll post how far I've got in a while, but now I'm leaving the city. I basically need some hints, because I'm stuck at one point for about two hours.

Answer 1

Hints

1) If $M$ is maximal, then $R/M$ is a field. Conversely, if $R/M$ is simple, consider $\pi: R\to R/M$ the natural projection and let $I$ an ideal of $R$. Why can you say about the ideal generated by $\pi(I)$ ?

2) Let $\{I_i\}$ a chain of ideal that contain $I$ (i.e. $I_{i+1}\supset I_i$ and $I\subset I_i$ for all $i$). Consider $\bigcup_{i}I_i$ and use Zorn lemma.

3) $(0)$ is an ideal. Using 2) allow you to conclude.