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

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?

• For the modified question see for example this and threads linked to it. The quotient ring $M_2(\Bbb{Z})/M_2(2\Bbb{Z})\simeq M_2(\Bbb{Z}_2)$. Because $\Bbb{Z}_2$ is a field the ring $M_2(\Bbb{Z}_2)$ has only trivial ideals. Then apply correspondence (or the first homomorphism theorem). Commented Apr 2, 2018 at 15:43

• $M_2(2\Bbb{Z})$ 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 $I\subset R$.
• Jyrki, in the question we have asked to show that the converse of the above result(result is mention at the beginning of my description) is fasle considering the ideal $M_2(2\mathbb{Z})$ of the ring(non commutative ring with unity) $M_2(\mathbb{Z})$. I have noticed that the 'if' condition fails here(I.e. every non zero element of the quotient ring $M_2(\mathbb{Z})/M_2(2\mathbb{Z})$ is not unit). So the ideal $M_2(2\mathbb{Z})$ will not be maximal. Commented Apr 1, 2018 at 8:02
• @BiswarupSaha But the ideal $M_2(2\Bbb{Z})$ is maximal. The way I read it is that the converse would claim: If an ideal (of a non-commutative ring) is maximal, then all the non-zero cosets of the quotient ring are invertible. The converse is false as shown by the example hinted at in your exercise. Commented Apr 1, 2018 at 8:55
• More precisely: the converse of if p then q is if q then p. Here p says all the non-zero cosets of the quotient ring are invertible, and q says the ideal is maximal. So to disprove the converse we need to give an example of a maximal ideal such that not all the non-zero elements of the quotient ring are invertible. That's why $M_2(2\Bbb{Z})$ was suggested as a hint. Commented Apr 1, 2018 at 9:04