# Give an example of a maximal ideal in a noncommutative ring which is not prime

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!!

• Are these left ideals, right ideals, two-sided ideals...? For two-sided ideals, what about $(0) \subset M_2(\mathbb{R})$ for example? Commented Oct 14, 2014 at 9:53
• Maximal ideal needs to be a proper subset. Two sided ideals @NajibIdrissi Commented Oct 14, 2014 at 10:01
• Nothing says $(0)$ cannot be a maximal ideal; it's $R$ that cannot be a maximal ideal. In fact, in a field $(0)$ is a maximal ideal. $(0)$ also a maximal two-sided ideal in $M_2(\mathbb{R})$ which isn't prime. Commented Oct 14, 2014 at 10:04
• @rschwieb I came upon this from the title itself.. so the title Commented Oct 14, 2014 at 10:10

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.
• @tattwamasiamrutam No, I am not saying that. It would be correct to say you gave a valid example of a prime, but not completely prime ideal of $M_n(\Bbb Z)$. Commented Oct 14, 2014 at 10:23