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

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.

• These all have answers in the site. Just search. – rschwieb Feb 11 '17 at 23:26
• I'll try tommorow, but I wanted just some hints at first, so I could try to find the solutions by myself – Pan Miroslav Feb 11 '17 at 23:29
• I'm sure you're perfectly capable of reading only part of the solution to give you a hint, or you could ask a friend to make a hint out of the solution. The problem is that we can't really encourage the pattern you're following. If we accepted everyone's "give me a hint - I won't look at existing questions and answers" question, then we would wind up with unacceptably many copies of the same thing. – rschwieb Feb 11 '17 at 23:44

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.