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Ideal in matrix ring

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the ring of 3x3 matrices over the ring Z36.How many different matrices are there in the two sided ideal generated by the matrix diag(0,-6,18)?


    2. Relevant equations



    3. The attempt at a solution
    I computed a general matrix in the two sided ideal,but counting is complicated because different products of parameters may be equal in Z36.
     
    Last edited: Oct 24, 2012
  2. jcsd
  3. Oct 24, 2012 #2

    haruspex

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    Can you characterise the general form of a member of the ideal? (in each matrix position, think what factors it would have.) Caution: there are two broad cases, with overlap.
     
  4. Oct 24, 2012 #3
    The general form is attached.
     

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  5. Oct 24, 2012 #4

    haruspex

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    OK. Now discard that factor 6 and consider the resulting matrix M mod 3. That gives you 9 elements generated from 6 (b, e, h, D, E, F), each of which takes only 3 values. So there are at most 3^6 possibilities. Can they all be realised? Having answered that, you can go back to M and discard the remainder mod 3 and factor out the 3. Since a factor 6 was removed at the start, and we've now taken a factor 3 out, there are only two interesting values in each matrix position. Again, there may be limits on which of the potential 2^9 values arise, but now all 12 variables in M contribute.
     
  6. Oct 24, 2012 #5
    it is mod36 not mod3.2x6=2x24mod36 (for example) so i can't just multiply all possible values of the parameters.
     
  7. Oct 24, 2012 #6

    haruspex

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    Checking again, I don't think your general form is general enough. An ideal also has to be closed under addition. In consequence, the approach I outlined above is not going to work.
    What we can say easily is that the ideal cannot have more than 6^9 elements (agreed?). I suspect that is in fact the answer, but I can't yet see how to prove it.
     
  8. Oct 25, 2012 #7

    haruspex

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    OK, I think I have it... but now it seems almost too easy.
    Having reduced the problem to Z6 and the generating matrix M = diag(0 -1 3), consider post-multiplying by
    0 0 0
    -1 0 0
    0 0 0
    to get
    0 0 0
    1 0 0
    0 0 0
    Similarly, premultiplying and postmultiplying by suitable matrices we can shuffle the 1 around to any position. Then adding combinations of these we can generate all 3x3 matrices over Z6.
    Does that look right?
     
  9. Oct 25, 2012 #8
    Why the ideal cannot have more than 6^9 elements,and how did you reduced the problem to z6?This ideal cannot be the whole ring because it does not contain invertible matrices(from rank considerations).
     
  10. Oct 25, 2012 #9
    In second thought there is no meaning for rank for matrices over rings.
     
  11. Oct 26, 2012 #10

    haruspex

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    Every element in the generating matrix is divisible by 6, so the same is true of every matrix in the ideal. Therefore each element is one of the values 0, 6, 12, 18, 24, 30 (mod 36). So factor out the 6 from the generating matrix and the modulo base.
     
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