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Units in a matrix ring

  • Thread starter jgens
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  • #1
jgens
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Homework Statement



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

Homework Equations



N/A

The Attempt at a Solution



If [itex]R[/itex] 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 [itex]R[/itex] is not even an integral domain necessarily, I could use some help. Is there any nice way to characterize the units?
 

Answers and Replies

  • #2
Dick
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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?
 
  • #3
jgens
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If we take [itex]R=\mathbb{Z}[/itex], then it is pretty easy to show that any matrix [itex](a_{ij}) \in M_n(R)[/itex] has an inverse if and only if [itex]\det(a_{ij})=\pm 1[/itex]. 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: [itex](a_{ij}) \in M_n(R)[/itex] is a unit if and only if [itex]\det(a_{ij})=x[/itex] where [itex]x[/itex] is a unit in [itex]R[/itex]. I have not checked to see if this is actually true, but is a condition like this what you were getting at?
 
  • #4
Dick
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If we take [itex]R=\mathbb{Z}[/itex], then it is pretty easy to show that any matrix [itex](a_{ij}) \in M_n(R)[/itex] has an inverse if and only if [itex]\det(a_{ij})=\pm 1[/itex]. 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: [itex](a_{ij}) \in M_n(R)[/itex] is a unit if and only if [itex]\det(a_{ij})=x[/itex] where [itex]x[/itex] is a unit in [itex]R[/itex]. 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.
 
  • #5
jgens
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Sweet! Thank you for the help! I think I can figure the rest out from here
 

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