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Let $A$ be an associative algebra, And let $V$ be a representation of $A$. By $End_{S}(V)$ one denotes the algebra of homomorphisms of representations $V \to V$

Show that $End_{A}(A)=A^{op}$ the algebra with opposite multiplication.

john
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1 Answers1

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Hint: fix $a\in A$ and denote by $R_a:x\mapsto xa$ the right multiplication by $a$. Then $End_A(A)=\{ R_a \mid a \in A\}$. Show that this is isomorphic to $A^{op}$.

Dietrich Burde
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  • Why don't you use left multiplication ? I think that by Schur's lemma , if $f$ is a homomorphism between two representations $A \to A$ then $f=aId$ where $a$ is in $A$ right ? – john Jun 07 '15 at 15:44
  • For Schur's lemma you need irreducible representations. – Dietrich Burde Jun 07 '15 at 15:48
  • Thanks. I thought this is an irreducible representation.I forgot about ideals. – john Jun 07 '15 at 15:52