Show that, the ideal $M_2(2\mathbb{Z})$ is maximal ideal of $M_2(\mathbb{Z})$

Question

In addition to the title I want to say that there is an exercise given before this question which is - Let $R$ be a non-commutative ring with 1. Prove that if $M$ is an ideal of are such that every nonzero element of $R/M$ is an unit then $M$ is a maximal ideal. I have solved this problem. Then my above question is asked like - Show that, the converse of above result is false by considering the ideal $M_2(2\mathbb{Z})$ of the ring $M_2(\mathbb{Z})$. I am stuck with the second question. Can anybody give a solution to this question?

Answer 1

As far as I can tell the point of the exercise is to show that

• $M_2(2\Bbb{Z})$ is a maximal ideal. This is more or less equivalent to showing that the quotient ring $M_2(\Bbb{Z}_2)$ has only the trivial ideals.
• There are many cosets of $M_2(2\Bbb{Z})$ that don't have an inverse in the quotient ring. For example the coset of $$\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$$ is nilpotent and therefore cannot be invertible.

It is a general fact (and a nice exercise) that for a commutative ring $R$ all the ideals of $M_n(R)$ are of the form $M_n(I)$ for some ideal $I\subset R$.