Question:
Let $R$ be a non-commutative ring with $1\neq 0$. Let $M$ be an ideal(two-sided) of $R$.
If $\frac{R}{M}$ is a field. Show that $M$ is a maximal ideal.
My Attempt:
Suppose that $M$ contains $1$.
$\Rightarrow M=R$
$\Rightarrow \frac{R}{M}=(0)$ which is not a Field.
$\Leftarrow$ Contradiction to the fact that $R/M$ is a field.
$\Rightarrow M$ doesn't contain $1$.
$\Rightarrow M \subsetneq R$
Now, Let $J$ be an ideal of $R$ such that $J\subset M\subsetneq R$
Define $\phi:\frac{R}{J}\rightarrow \frac{R}{M}$ by $\phi(r+J)=r+M$
By Third Isomorphism Theorem (Dummit and Foote, 3rd ed.), We have
$\frac{(\frac{R}{J})}{(\frac{M}{J})} \cong \frac{R}{M}$
Given that $\frac{R}{M}$ is a field and $\frac{M}{J}$ is the proper ideal of $\frac{R}{J}$, we have
$\frac{M}{J}=(0)$
$\Rightarrow M=J$
Therefore, $M$ is the maximal ideal of $R$.
Please Help me to spot my mistake and Give me hints to correct it.
We see that for any ring $R$, commutative or not, $R/M$ is a division ring only if $M$ is a maximal two-sided ideal, to wit:
If $R/M$ is a division ring, then
$a \notin M \Longrightarrow 0 \ne a + M \in R/M \Longrightarrow$ $\exists 0 \ne b + M \in R/M, \; ab + M = (a + M)(b + M) = 1 + M \in R/M; \tag 1$
thus
$\exists m \in M, \; ab + m = 1; \tag 2$
now if $M$ is not maximal in the set of two-sided ideals of $R$, there exists a two-sided ideal $J$ such that
$M \subsetneq J \subsetneq R; \tag 3$
let
$a \in J \setminus M; \tag 4$
then by what we have seen we may find $b$ such that
$1 = ab + m \in J, \; m \in M: \tag 5$
since
$a, m \in J; \tag 6$
but $1 \in J$ yields
$r \in R \Longrightarrow r = r \cdot 1 \in J \Longrightarrow J \subset R \Longrightarrow J = R, \tag 7$
in contradiction to (3); therefore no such $J$ can exists and we have $M$ maximal amongst the two-sided ideals of $R$.