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One variant of Schur's Lemma states that $$ \text{Hom}(S,T) \cong \left\{ \begin{matrix} 0 & \text{if } S \neq T \\ \mathbb{C} & \text{if } S = T \end{matrix} \right. $$ when $S,T$ are simple modules over a $\mathbb{C}$-algebra $A$.

I would like a similar result for $\text{Hom}(\bigoplus_{i=1}^n S_i,S)$. Will it be isomorphic to $ \bigoplus_{i=1}^n \text{Hom}(S_i,S) $ or is this just wishful thinking?

EDIT: and the same question for $ \text{Hom}(\bigoplus S_i,\bigoplus T_j) $

EDIT 2: OK i know where i was confused, for some reason i thought $\text{Hom}(U,V)$ was an algebra when it's actually a module

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    $\begingroup$ For your deleted question: $\Bbb C\oplus\Bbb C$ and $\Bbb C[\varepsilon]/(\varepsilon^2)$ are nonisomorphic two-dimensional $\Bbb C$-algebras. | Yes $\hom$ distributes over (arbitrary) direct sums in both arguments. In fact there is a canonical isomorphism. Can you tell what it is? $\endgroup$ – anon Dec 27 '14 at 21:32
  • $\begingroup$ I'm good now, thanks for all the help! $\endgroup$ – JC574 Dec 27 '14 at 21:45
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    $\begingroup$ @JC574 For your deleted question (I think you should undelete it), the answer is yes if the algebras are commutative and semisimple. $\endgroup$ – egreg Dec 27 '14 at 21:52
  • $\begingroup$ @egreg If I get time later tonight I'll reask it and tag you. Thanks for your help. $\endgroup$ – JC574 Dec 27 '14 at 22:14
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If $S, S_i, T, T_i$ are simple modules, then $$ \dim\text{Hom}(\bigoplus_{i=1}^n S_i,S) = \sum_{i=1}^n \dim \text{Hom} (S_i, S). $$ is the number of $S_i$ isomorphic to $S$.

For $$ \dim \text{Hom}(\bigoplus S_i,\bigoplus T_j) $$ you get the "same thing".

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  • $\begingroup$ so we have dimensions, but no algebra isomorphism necessarily? $\endgroup$ – JC574 Dec 27 '14 at 20:17
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    $\begingroup$ @JC574: Sorry, yes, we have isomorphisms. I just tend to think in terms of dimensions. You just get copies of $\mathbb{C}$. $\endgroup$ – Thomas Dec 27 '14 at 20:19

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