Finite Dimensionality of Endomorphism Ring in Simple Modules?

  • Context: Graduate 
  • Thread starter Thread starter Bleys
  • Start date Start date
  • Tags Tags
    module Ring
Click For Summary

Discussion Overview

The discussion centers on the properties of the endomorphism ring of simple modules over finite dimensional complex algebras. Participants explore the dimensionality of the endomorphism ring, specifically End_{R}(S), where R is a finite dimensional C-algebra and S is a simple R-module. The conversation touches on concepts such as basis construction, finite generation, and Schur's lemma.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • One participant questions why End_{R}(S) is finite dimensional, suggesting a need for a basis construction and possible embedding considerations.
  • Another participant confirms that S is finitely generated over R and C.
  • A third participant states that for a finite dimensional C-algebra R, there are finitely many isomorphism classes of simple R-modules, which are also finite dimensional.
  • It is noted that since S is a finite-dimensional C-vector space, End_{R}(S) is a subspace of End_{C}(S), leading to the conclusion that it is finite-dimensional over C.
  • One participant acknowledges the relationship between End_{R}(S) and End_{C}(S) and expresses a desire to understand the proof of Schur's lemma to show that End_{R}(S) is isomorphic to C.

Areas of Agreement / Disagreement

Participants generally agree on the finite dimensionality of End_{R}(S) and the implications of Schur's lemma, but there is still some uncertainty regarding the construction of a basis and the details of the proof.

Contextual Notes

Some assumptions about the properties of finite dimensional algebras and modules are present, but not all details are fully resolved, particularly regarding the basis construction for End_{R}(S).

Bleys
Messages
74
Reaction score
0
Let R be a finite dimensional C-algebra (C=Complex numbers) and S a simple R-module. Why does it follow that End_{R}(S) 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?)
 
Physics news on Phys.org
Is S finitely generated over R (hence C)?
 
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"
 
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.
 
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!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K