# 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?

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
Homework Helper
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?

Yes, it is. Satisfy yourself that the properties of det over a general ring aren't that different from over the integers.

Gold Member
Sweet! Thank you for the help! I think I can figure the rest out from here