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Units in a matrix ring

  1. Jan 16, 2012 #1

    jgens

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    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    N/A

    3. 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?
     
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  3. Jan 16, 2012 #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?
     
  4. Jan 16, 2012 #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?
     
  5. Jan 16, 2012 #4

    Dick

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    Yes, it is. Satisfy yourself that the properties of det over a general ring aren't that different from over the integers.
     
  6. Jan 16, 2012 #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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