While trying to find an example, I came up with this:
Since if $J$ is an ideal of a the ring $M_n(R)$, where $R$ is a commutative ring, then $J=M_n(I)$ for some ideal $I$ of $R$. IF I could show that for every maximal ideal $J$ of $R$, $M_n(J)$ is a maximal ideal of $M_n(R)$, then I could take $R=\mathbb{Z}$ and $J=2\mathbb{Z}$, then $M_n(2\mathbb{Z})$ will be a maximal ideal and I could cook up an example using $M_2(2\mathbb{Z})$ and the matrix $$
M=
\left[ {\begin{array}{cc}
1 & 1 \\
1 & 1 \\
\end{array} } \right]
$$
Clearly $M^2\in M_n(2\mathbb{Z})$ but $M \not \in M_n(2\mathbb{Z})$
Here is how I tried to prove: for every maximal ideal $I$ of $R$, $M_n(I)$ is a maximal ideal of $M_n(R)$.
Suppose $$M_n(I) \subseteq M_n(K) \subseteq M_n(R).$$ If $M_n(I) \ne M_n(K)$, then there exists an element $k_1 \in M_n(K)$ such that $k_1 \not \in M_n(I) $. let $$
K_1=
\left[ {\begin{array}{cc}
a_{11} & a_{12} & . &. &a_{1n} \\
a_{21} & a_{22} & . &. &a_{2n} \\
. & . & . & .& . \\
a_{n1} & a_{n2} & . &. &a_{nn} \\
\end{array} } \right]
$$
Then there exists at least one $a_{ij} \in K$ such that $a_{ij} \not \in I$. But since $M_{n}(I) \subset M_n(K)$, $I \subset K$. BUt $I$ is maximal. Hence $K=R$. Thus $M_n(K)=M_n(R)$. So $M_n(I)$ is maximal.
IS is alright of a proof??
Thanks for the help!!
Best Answer
Yes, your reasoning is valid that if $M_n(I)\subsetneq M_n(K) \subseteq M_n(R)$, then $I\subsetneq K\subseteq R$, and so if $I$ is maximal, $M_n(I)$ is maximal.
As for the title question, a maximal ideal in a noncommutative ring is always prime, unless you mean to apply the commutative definition of prime ideals to noncommutative rings. The general definition of "prime ideal" is "$AB\subseteq P\implies A\subseteq P \text{ or } B\subseteq P$."
In noncommutative algebra, an ideal that satisfies $ab\in I\implies a\in I \text{ or } b\in I$ is called a completely prime ideal, and the zero ideal of a matrix ring over a field is an example of a maximal ideal that isn't completely prime.
The example you gave is a maximal and prime ideal of $M_n(\Bbb Z)$ which is not a completely prime ideal.