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Quotients of direct sums of modules

  1. Jan 21, 2012 #1
    Hi, I keep seeing indirect uses of a result which I think would be stated as follows:

    If a module [itex]M[/itex] over the unital associative algebra [itex]A[/itex] is written
    [itex]M\cong S_1\oplus\cdots\oplus S_r[/itex] (where the [itex]S_i[/itex] are simple modules), then in any comosition series of [itex]M[/itex], the composition factors are, up to order and isomorphism, [itex]S_1,\ldots,S_r[/itex]. Perhaps this statement is not correct, but it's the best I can do when I can't find an explicit statement anywhere.

    I think I would use the Jordan Holder theorem to prove this. I would argue as follows: a composition series for [itex]S_1\oplus\cdots\oplus S_r[/itex] is given by
    [itex]S_1\oplus\cdots\oplus S_r>S_1\oplus\cdots\oplus S_{r-1}>S_1\oplus\cdots\oplus S_{r-2}>\cdots>S_1>\{0\}.[/itex]
    Now, [itex](S_1/\{0\})\cong S_1,\ (S_1\oplus S_2)/S_1\cong S_2,\ (S_1\oplus S_2\oplus S_3)/(S_1\oplus S_2)\cong S_3,\ldots, (S_1\oplus\cdots\oplus S_r)/(S_1\oplus\cdots\oplus S_{r-1})\cong S_r[/itex] and so by the Jordan Holder theorem, in any comosition series of [itex]M[/itex], the composition factors are, up to order and isomorphism, [itex]S_1,\ldots,S_r[/itex].

    I have a feeling that I am going wrong somewhere, perhaps in my cancelling when I do things like [itex] (S_1\oplus S_2)/S_1\cong S_2[/itex]. The terminology of "direct sum" would suggest that this is not allowed.
     
    Last edited: Jan 21, 2012
  2. jcsd
  3. Jan 21, 2012 #2

    morphism

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    Assuming the S_i's are simple modules, what you've written is correct.

    To justify canceling off a factor in the direct sums, just use the first isomorphism theorem: e.g., the map [itex]S_1 \oplus S_2 \to S_2[/itex] defined by [itex](s_1,s_2) \mapsto s_2[/itex] is surjective and has kernel equal to [itex]S_1\oplus 0[/itex].
     
  4. Jan 21, 2012 #3
    Yes, they are simple, I've edited that in. Thanks a million for your help.
     
  5. Jan 21, 2012 #4

    mathwonk

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    this looks fishy.
     
  6. Jan 21, 2012 #5

    morphism

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    Is there anything specific that's making you skeptical?
     
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