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Endomorphism ring of a module

  1. Jan 30, 2012 #1
    Let R be a finite dimensional C-algebra (C=Complex numbers) and S a simple R-module. Why does it follow that [itex]End_{R}(S)[/itex] is also finite dimensional (as C-vector spaces, I'm guessing)? I'm not really sure how to construct a basis for it using one of S, and there's probably another reason for it (is end(S) embedded in S or something?)
     
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  3. Jan 30, 2012 #2

    morphism

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    Is S finitely generated over R (hence C)?
     
  4. Jan 30, 2012 #3
    yes; I have this result
    "For a finite dimensional C-algebra R, there are only finitely many isomorphism classes of simple R-modules and they are finite dimensional"
     
  5. Jan 30, 2012 #4

    morphism

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    Perfect. So S is a finite-dimensional C-vector space and End_R(S) is a subspace of End_C(S), hence is finite-dimensional over C. In fact, you can use Schur's lemma to show that End_R(S) is one-dimensional.
     
  6. Jan 30, 2012 #5
    ah of course, I didn't think of the fact End_{R}(S) is a subspace of End_{C}(S).
    I'm actually going through Schur's Lemma's proof to show End_R(S) is isomorphic to C but this was the detail I wasn't understanding.

    Thank you, morphism!
     
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