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Math Help - Semisimple algebra

  1. #1
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    Semisimple algebra

    Show that an algebra A is semisimple if and only if every A-module is completely reducible.

    I know that if A is semisimple then every irreducible A-module is isomorphic to a submodule of A^{*} ( A^{*} is A viewed as a module over itself). Then the submodules of A are direct summands (there exists another submodule such that the direct sum equals A).
    Is it true that if a module is completely reducible, then all of its submodules are as well?

    For the other direction I don't have any good ideas. How about you guys?
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  2. #2
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    Quote Originally Posted by robeuler View Post
    Show that an algebra A is semisimple if and only if every A-module is completely reducible.
    Definition. The algebra A is semisimple if every nonzero A-module is semisimple.

    If your definition of the semisimple algebra is here, then I think your question is probably,

    Show that an algebra A is semisimple if and only if the A-module A is completely reducible (semisimple).

    Proof.
    Suppose that the A-module A is semisimple, and let M be an arbitrary A-module generated by \{b_1, \cdots , b_m\}. We shall show that M is a semisimple A-module. Let A^m denote the direct sum of m copies of A. Let \phi:A^m \rightarrow M be a map, which sends (x_1, \cdots, x_m) to x_1b_1 + \cdots +x_mb_m. We see that \phi is an A-module epimorphism. It follows that M is isomorphic with a quotient module of the semisimple module A^m. Thus it is semisimple by Lemma 1 below. We conclude that A is a semisimple algebra by the definition given above.

    I know that if A is semisimple then every irreducible A-module is isomorphic to a submodule of A^{*} ( A^{*} is A viewed as a module over itself). Then the submodules of A are direct summands (there exists another submodule such that the direct sum equals A).
    Is it true that if a module is completely reducible, then all of its submodules are as well?
    For the other direction I don't have any good ideas. How about you guys?
    Lemma 1. Quotient modules and submodules of semisimple modules are also semisimple.

    Let M be a semisimple A-module. Then every submodule of M is isomorphic with a quotient module of M by the first isomorphism theorem. It remains to show that the quotient modules of M are semisimple. Since M is a semisimple A-module, M has the following form:

    M=S_1 \oplus \cdots \oplus S_n.

    The quotient module of M has the form M/N, where N is an A-module. Let \phi be an A-module epimorphism such that \phi:M \rightarrow M/N. Then,

    M/N=\phi(M)=\phi(S_1) \oplus \cdots \oplus \phi(S_n).

    Can you conclude that M/N is semisimple, which implies that a quotient module of a semisimple module is semisimple?
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