Characterizing Units in M_n(R) for Commutative Rings with 1

[jgens]

Homework Statement

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

Homework Equations

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]

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?

 

[jgens]

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?

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

 

[jgens]

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