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Show a matrix ring is simple

  1. Feb 10, 2008 #1
    [SOLVED] show a matrix ring is simple

    1. The problem statement, all variables and given/known data
    Show that the matrix ring M_2(Z_2) is a simple ring; that is, M_2(Z_2) has no proper nontrivial ideals.

    In my book M_2(Z_2) is the ring of 2 by 2 matrices whose elements are in Z_2.

    2. Relevant equations

    3. The attempt at a solution
    I wrote down all of the elements of the ring. I stared at them for awhile and I couldn't think of anything to do. We want to show that if H is a nontrivial additive subgroup of the ring, then M_2(Z_2) H must equal M_2(Z_2). But how...
  2. jcsd
  3. Feb 10, 2008 #2


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    Hint: Think about the "standard bases matrices eij" of M_2(Z_2); here eij is the matrix with a '1' in the (i,j)th position and zeros elsewhere. An ideal is closed under left and right multiplication from the ring, so in particular if a is in your ideal, then so are eij*a and a*eij.

    In general, the ideals of R correspond to the ideals of M_n(R) via the order-preserving bijection I <-> M_n(I). Consequently, M_n(R) is simple iff R is.
  4. Feb 10, 2008 #3
    Is an order-preserving bijection the same thing as an isomorphism?
  5. Feb 10, 2008 #4


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    No. Well, it depends on what context the word "isomorphism" is being used in. I'm just setting up a bijection between the ideals of two rings; order-preserving here means if I is a subset of J [as ideals of R], then M_n(I) is a subset of M_n(J) [as ideals of M_n(R)]. This does NOT mean that R and M_n(R) are isomorphic. The fact that M_n(R) shares simplicity with R is a by-product of their Morita equivalence -- this means that these two rings are very similar but not necessarily isomorphic. But this is neither here nor there.

    Another irrelevant remark is that order-preserving bijections can be considered "isomorphisms" if we're working in the 'category' (I'm using this term loosely) of ordered sets, where the underlying structure is the order on the sets. This is consistent with thinking about "isomorphisms" as structure-preserving maps.
  6. Feb 10, 2008 #5
    Hmmm. So let I be a nontrivial ideal of M_2(Z_2). We know that there are at least two elements in I and there is one that is not 0, call it a. a must have some component that is not 0. Say that component is (i.j). Obviously you cannot get the whole ring just by applying the basis matrices on the right and on the left since that only gives 8 matrices. But you can apply the elementary matrices in succession and apparently that will give you the whole ring. But how to prove that? Again, I wrote out the 8 products that result from applying the basis matrices on the right and I didn't see the proof :(
  7. Feb 10, 2008 #6


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    Well, suppose we have a nonzero element a in the ideal, whose (1,2)th entry is nonzero. Then e11*a*e21 is the matrix with the (1,2)th entry of a in the (1,1)th position, and e21*a*e22 is the matrix with the (1,2)th entry of a in the (2,2)th position. Both of these are in the ideal, and so is their sum, which is the identity matrix. But if an ideal contains a unit (i.e. an invertible element), it must be the entire ring.

    Try to generalize this. You'll have to get your hands dirty to see how the matrices eij behave.
    Last edited: Feb 10, 2008
  8. Feb 10, 2008 #7


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    Well, there aren't many nontrivial additive subgroups, are there? And you only need to consider the minimal ones anyways....
  9. Feb 11, 2008 #8
    There are tons of additive subgroups; every nonzero element of the ring has additive order 2.
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