I am trying to prove that in a commutative ring (with unity), every maximal ideal is prime. I wrote the the following answer which was given wrong by my tutor. I dont understand whats wrong! (he says its obvious.) Can someone help me?
Proof: Let $ab\in M$ and let if possible $a\not \in M$ and $b\not \in M$.
So $M\subsetneq <a> \neq R$ since $a\not \in M$ and $b\not\in <a>$.
Since $M$ is a maximal ideal and $<a> \neq R$, we must have $M=<a>$ and hence $a\in M$, which is a contradiction to the assumption.
Thus $M$ is a prime ideal.
Thanks in advance!
You've been answered in the comment of your misconception about "maximal ideal", and here is a possible way to prove what you wanted using your first step:
Let $\;M\le R\;$ be a maximal ideal and suppose $\;ab\in M\;$ and $\;b\notin M\;$ . This means $\;M\lneq M+bR\;$ so by maximality we get $\;M=M+bR \;$ or else $\;M+bR=R\;$ . The first option is imposible, otherwise
$$b=b\cdot1+0\in M\implies b\in M\;,\;\;\text{contradiction}$$
and thus it must be $\;M+bR=R\;$ , and this implies that there exist
$$m\in M\,,\,\,r\in R\;\;s.t.\;\;m+br=1\implies a=a\cdot 1=am +abr\in M+M=M$$
and thus $\;a\in M\;$ and we're done.