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Homework Help: Artin-Wedderburn theorem contradiction? Mind = Blown

  1. Nov 23, 2013 #1
    1. The problem statement, all variables and given/known data

    My question basically wants me to write the direct product of rings [itex] R = \mathbb{Z}_3 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5 [/itex] as a direct product of matrix rings over division rings.

    2. Relevant equations

    Relevant theorems

    http://img713.imageshack.us/img713/8471/g7pn.png [Broken]

    3. The attempt at a solution

    So I found two ways of doing this which seemed contradictory to the above Theorem.

    First, [itex] R [/itex] is the direct sum of the simple ideals [itex] (\mathbb{Z}_3, 0 , 0) = S_1 [/itex], [itex] (0, \mathbb{Z}_5, 0) = S_2 [/itex] and [itex] (0, 0, \mathbb{Z}_5) = S_3 [/itex]. These are simple since the [itex] \mathbb{Z}_3, \mathbb{Z}_5 [/itex] are fields (and have no proper non-zero ideals).

    But then treat [itex] R [/itex] as a left [itex] R [/itex] module. Then these three simple ideals above become three simple submodules of [itex] R [/itex] and

    [itex] R = S_1 + S_2 + S_3 [/itex]

    As left [itex] R [/itex] modules, [itex] S_2 \cong S_3 [/itex] whereas [itex] S_1 [/itex] is not isomorphic to either of the other two (different cardinalities).

    Then [itex] R \cong S_1 \oplus S_2 \oplus S_3 \cong S_1 \oplus {S_2}^2 [/itex]

    [itex] End_R(M) [/itex] is the set of R-module homomorphisms from [itex] M [/itex] to [itex] M [/itex]. Another theorem previously shows as rings [itex] R [/itex] is isomorphic to [itex] End_R(R) [/itex]. But since this is the case, Theorem 4.24 tells me;

    [itex] R \cong End_R(R) \cong End_R(S_1 \oplus {S_2}^2) \cong M_1(D_1) \oplus M_2(D_2) [/itex]

    Where [itex] D_1 = End_R(S_1), D_2 = End_R(S_2) [/itex] are division rings.

    So thats one direct product of matrix rings over division rings. But also I could've simply said.

    [itex] \mathbb{Z_3} \cong M_1(\mathbb{Z}_3), \mathbb{Z_5} \cong M_1(\mathbb{Z}_5) [/itex] as rings.

    [itex] R \cong M_1(\mathbb{Z}_3) \oplus M_1(\mathbb{Z}_5) \oplus M_1(\mathbb{Z}_5) [/itex]

    Also a direct product of matrix rings over fields (thus division rings)

    But this surely contradicts what is said at the very bottom of my image above. Since one is a direct product over 2 matrix rings, another is a direct product over 3? So can anyone tell me where I'm wrong?
    Last edited by a moderator: May 6, 2017
  2. jcsd
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