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Math Help - Ideal in matrix ring

  1. #1
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    Smile Ideal in matrix ring

    Hi,
    I need help with the following problem:
    consider the ring of 3x3 matrices with entries in the ring Z36.How many matrices are there in the two sided ideal generated by the matrix diag(o,-6,18)?
    Attempt at a solution:
    I computed the general matrix in the two sided ideal,but counting was complicated since different products of the parameters may be equal in Z36.
    I will be gratefull for any help.
    Hedi
    Last edited by hedi; October 24th 2012 at 09:48 AM.
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  2. #2
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    Re: Ideal in matrix ring

    Perhaps there's a more clever way, but the method you proposed should work. Remember that you have 18 free parameters.
    To simplify things, I'd factor out a -6: A diag{0, -6, 18 }B = (-6) A diag {0, 1, -3} B. To get you started:

     \left( \begin{matrix} a_{1, 1} & a_{1, 2} & a_{1, 3} \\ a_{2, 1} & a_{2, 2} & a_{2, 3} \\ a_{3, 1} & a_{3, 2} & a_{3, 3} \end{matrix} \right) \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -3 \end{matrix} \right) \left( \begin{matrix} b_{1, 1} & b_{1, 2} & b_{1, 3} \\ b_{2, 1} & b_{2, 2} & b_{2, 3} \\ b_{3, 1} & b_{3, 2} & b_{3, 3} \end{matrix} \right) = \left (\begin{matrix} a_{1, 1} & a_{1, 2} & a_{1, 3} \\ a_{2, 1} & a_{2, 2} & a_{2, 3} \\ a_{3, 1} & a_{3, 2} & a_{3, 3} \end{matrix} \right) \left( \begin{matrix} 0 & 0 & 0 \\ b_{2, 1} & b_{2, 2} & b_ {2, 3} \\ -3b_{3, 1} & -3b_{3, 2} & -3b_{3, 3} \end{matrix} \right)

    = \left(\begin{matrix} (a_{1, 2}b_{2,1} -3a_{1,3}b_{3, 1}) & (a_{1, 2}b_{2,2} -3a_{1,3}b_{3, 2}) & (a_{1, 2}b_{2,3} -3a_{1,3}b_{3, 3}) \\ (a_{2, 2}b_{2,1} -3a_{2,3}b_{3, 1}) & (a_{2, 2}b_{2,2} -3a_{2,3}b_{3, 2}) & (a_{2, 2}b_{2,3} -3a_{2,3}b_{3, 3}) \\ (a_{3,2}b_{2,1} -3a_{3,3}b_{3, 1}) & (a_{3, 2}b_{2,2} -3a_{3,3}b_{3, 2}) & (a_{3, 2}b_{2,3} -3a_{3,3}b_{3, 3}) \end{matrix} \right)

    If you can choose (I don't know that you can, but it seems possible at first glance) combinations of a's and b's so that the resulting matrix is anything

    in R = M_{3, 3}\left(\mathbb{Z}_{36}\right), then your ideal I would equal (-6)R, whose number you can count.

    \text{Suppose } A = \left( \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right), B = \left( \begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) \text{ so } a_{1,2} = b_{2,1} = 1, \text{ and all entries 0 otherwise.}

    \text{Then } A \text{ diag}\{0, 1, -3 \} \ B = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) = E_{1, 1}, \text{ so } -6E_{1, 1} \in I.

    What of the other elementary matricies?
    Last edited by johnsomeone; October 24th 2012 at 09:54 PM.
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    Re: Ideal in matrix ring

    This ideal cannot be the whole ring because it's matrices are not invertible.From this reason it cannot contain all possible
    matrices with zero entries except one 1.
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    Re: Ideal in matrix ring

    In second thought there is no meaning for rank for matrices over rings.
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    Re: Ideal in matrix ring

    it appears that I = (-6)R = 6R. in Z36, <6> = <-6> (since 6 and -6 are additive inverses, and both generate the same additive subgroup of (Z36,+)).

    to prove this, it suffices to show that we can always pick 4 elements a,b,c,d in Z36 with:

    ab - 3cd = 1. a = b = 2, and c = d = 1 will serve.

    thus (2Ei2+Ei3)(diag(0,1,-3))(2E2j+E3j) =

    (2Ei2-3Ei3)(2E2j+E3j) =

    (4Ei2E2j) + (2Ei2E3j) + (2Ei32E2j) - (3Ei3E3j)

    = 4Eij + 0 + 0 - 3Eij = Eij.

    for example, with i = 3, j = 1:

    \begin{bmatrix}0&0&0\\0&0&0\\0&2&1 \end{bmatrix} \begin{bmatrix}0&0&0\\0&1&0\\0&0&-3 \end{bmatrix} \begin{bmatrix}0&0&0\\2&0&0\\1&0&0 \end{bmatrix} =

    \begin{bmatrix}0&0&0\\0&0&0\\0&2&-3 \end{bmatrix}\begin{bmatrix}0&0&0\\2&0&0\\1&0&0 \end{bmatrix} = \begin{bmatrix}0&0&0\\0&0&0\\1&0&0 \end{bmatrix}.

    yes, we can have "duplicates", because in Z36, ab = ac doesn't imply b = c. but the relevant fact here is that the map:

    k --> 6k is a surjective homomorphism from Z36 to <6>.

    since 6Eij is in our ideal for any choice of i and j, then any Z36-linear combination of the 6Eij are also in the ideal (by closure), and so I = 6R.

    this gives 10,077,696 elements of I (i think).
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    Re: Ideal in matrix ring

    Thank you very much for your help,It was very helpfull.
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    Re: Ideal in matrix ring

    There's an observation that Deveno and I relied on that you should understand make an argument for.

    \text{Since you can show that }6E_{i, j} \in I \ \forall \ i, j \in \{1, 2, 3 \},

    \text{ it follows that } 6R \subset I.

    \text{But how do you then justify that }I = 6R \ ?

    \text{Perhaps }I = 2R\supsetneq 6R? \text{ How do you know that can't happen?}
    Last edited by johnsomeone; October 25th 2012 at 01:58 PM.
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  8. #8
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    Re: Ideal in matrix ring

    I is the two sided ideal in the ring of matrices over Z36 generated by the matrix 6diag(0,1,-3),so the entries of all matrices in I are elements of Z36 that are multiples of 6.
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  9. #9
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    Re: Ideal in matrix ring

    Right. Since X_0 = \text{ diag} \{0, -6, 18 \} = 6 \text{ diag} \{0, -1, 3 \}\subset 6R, have that I = <X_0>  \ \subset \ <6R>  \ = \ 6R.

    So once you've also shown that 6R \subset I, you'll have that I = 6R.
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