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Math Help - Simple C*-algebra

  1. #1
    Member Mauritzvdworm's Avatar
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    Simple C*-algebra

    Suppose we have a simple C*-algebra, say A. Show that M_n(A) is also simple.

    My first guess would be to try and obtain some contradiction by assuming M_n(A) has some closed ideal J other than 0 or M_n(A) and then find that A will then have some closed ideal other than 0 or A which will conclude the proof.

    However, I'm having some trouble with the particulars...

    We can suppose that there is some sequence \{T^{(m)}\}\in J that converges to T\in J. Then we have RT^{(m)},T^{(m)}R\in J and also RT,TR\in J for every R\in M_n(A).

    How can we now use this to construct a closed ideal in A? (If it is at all possible)
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  2. #2
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    Let e_{ij}\ (1\leqslant i,j\leqslant n) denote the matrix units in M_n(A). Show that e_{11}Je_{11} is a closed ideal in A (where A is identified with e_{11}M_n(A)e_{11}). Hence e_{11}Je_{11} = e_{11}M_n(A)e_{11} (unless e_{11}Je_{11} = \{0\}, in which case show that J=\{0\}). Deduce that e_{ij}Je_{ij} = e_{ij}M_n(A)e_{ij} for all i,j, and so J = M_n(A).
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  3. #3
    Member Mauritzvdworm's Avatar
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    I can show the first part, but I'm not sure how to attack the part where I need to show that e_{ij}M_n(A)e_{ij}=e_{ij}Je_{ij}?
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  4. #4
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    Quote Originally Posted by Mauritzvdworm View Post
    I can show the first part, but I'm not sure how to attack the part where I need to show that e_{ij}M_n(A)e_{ij}=e_{ij}Je_{ij}?
    The (1,1)-element of the matrix x is the (i,j)-element of the matrix e_{i1}xe_{1j}. If an element x\in J can have an arbitrary element of A in its (1,1)-position, then e_{i1}xe_{1j} (which is also an element of J) can have an arbitrary element of A in its (i,j)-position.
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