The Ring $A$ is commutative or not. I'm trying to understand the proof of the theorem of unicity of simple factorization of semisimple modules. I already know Shur's lemma, and that, as a corollary of isomorphism theorems of direct sum and product of modules, we have $$ \operatorname{Hom_A}(M_\alpha, \bigoplus_{B \in B}\Lambda_{\beta} N_{\beta}) \cong \bigoplus_{B \in B}\Lambda_{\beta}\left(M_\alpha, N_{\beta}\right)$$ but I can't understand why if we have $M_\alpha \cong N_\beta$ (isomorphism) then $$ \operatorname{Hom_A}(M_\alpha, \bigoplus_{B \in B}\Lambda_{\beta} N_{\beta}) \cong \Lambda_{\beta} \left(M_\alpha, N_{\beta}\right)$$ for a unique $\beta$. I have two questions:
- why they omitted the direct sum?
- why this is true for a unique $\beta$.
Please click here to see the picture: The theorem and the begining of the proof with the difficulty highlited
And click please here for the notation used in the theorem
The answer is that $M_\alpha$ can't be isomorphic to more than one $N_\beta$ because $N_\beta$s are pairwise distincts. Ans by shur's lemma, we can conslude that the other homomorphism will be trivial, beacause of the simplicity of the modules.