Prove that an ideal $ \mathfrak{m} $ of a commutative ring $ R $ is maximal iff $ R/\mathfrak{m} $ is simple. Could someone give me a hint on whether I’m on the right track or not? For sufficiency, I tried the following:
Suppose that $ \mathfrak{m} $ is a maximal ideal. With the quotient map, we get $ R/\mathfrak{m} $, where any $ x \in R/\mathfrak{m} $ is of the form $ a + \mathfrak{m} $, with $ a \in R $. Suppose that $ J $ is a non-zero ideal of $ R/\mathfrak{m} $, and let $ y \in J $. As $ J $ is an ideal, we have $ x y \in J $, so $ x y = a y + \mathfrak{m} $. Given that $ a \in R $, $ y \in J $ and $ J \subseteq R $, we get $ y \in R $, so $ a y \in R $. This implies that $ a y + \mathfrak{m} \in R/\mathfrak{m} $. Hence, $ x y \in J $ and $ x y \in R/\mathfrak{m} $, and so $ J \subseteq R/\mathfrak{m} $.
Now, pick any $ a b + \mathfrak{m} \in R/\mathfrak{m} $, where $ b \in J $, with $ J $ an ideal of $ R/\mathfrak{m} $. We know that
$$
a b + \mathfrak{m} = (a + \mathfrak{m}) b \in R/\mathfrak{m},
$$
and as $ b \in J $, we have $ (a + \mathfrak{m}) b \in J $. Hence, $ R/\mathfrak{m} \subseteq J $. We have therefore shown that $ J = R/\mathfrak{m} $ if $ J $ is not the zero ideal.
We thus have only two cases, (i) when $ J $ is not the zero ideal, which we showed implies $ J = R/\mathfrak{m} $, and (ii) when $ J $ is the zero ideal. Therefore, these are the only ideals of $ R/\mathfrak{m} $, and so $ R/\mathfrak{m} $ is simple.
Is this proof on the right path? Also, does anyone have a hint for necessity?

This is wrong because you can't multiply an element of $R/M$ by an element of $J$. They're separate rings. The proof of this I'm familiar with supposes $M$ is maximal and constructs the ideal 
$$J=\{m+rx:m\in M,r\in R\}$$
where $x$ is an element such that $x+M\neq M$ (or $x\notin M$). Then this ideal contains $M$ by letting $r=0$, and it contains $x$ by letting $m=0,r=1$. Therefore $M\subset J\implies J=R$. Then we know $1\in J$. So $1=m+rx$ for $m\in M,r\in R$.
I'll let you try to figure out the rest
