# Units in a matrix ring

Gold Member

## Homework Statement

Let $R$ be a commutative ring with $1$. What are the units of $M_n(R)$?

N/A

## The Attempt at a Solution

If $R$ is a field, then I know that we can characterize the units as those matrices with non-zero determinant (since those are the invertible matrices). But since $R$ is not even an integral domain necessarily, I could use some help. Is there any nice way to characterize the units?

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper
Think about an integer matrix M. If det(M)=1 or -1 then its inverse is an integer matrix. Can you show that? Now suppose |det(M)| is not 1. Can you show that the inverse isn't an integer matrix?

Gold Member
If we take $R=\mathbb{Z}$, then it is pretty easy to show that any matrix $(a_{ij}) \in M_n(R)$ has an inverse if and only if $\det(a_{ij})=\pm 1$. But the integers have a good deal more structure than a general commutative ring with 1. The determinant condition here is also different than the one for fields.

But then maybe the condition is: $(a_{ij}) \in M_n(R)$ is a unit if and only if $\det(a_{ij})=x$ where $x$ is a unit in $R$. I have not checked to see if this is actually true, but is a condition like this what you were getting at?

Dick
If we take $R=\mathbb{Z}$, then it is pretty easy to show that any matrix $(a_{ij}) \in M_n(R)$ has an inverse if and only if $\det(a_{ij})=\pm 1$. But the integers have a good deal more structure than a general commutative ring with 1. The determinant condition here is also different than the one for fields.
But then maybe the condition is: $(a_{ij}) \in M_n(R)$ is a unit if and only if $\det(a_{ij})=x$ where $x$ is a unit in $R$. I have not checked to see if this is actually true, but is a condition like this what you were getting at?