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There's a proposition that states that for an $R$-module $E=E_1^{(n_1)}\oplus\cdots\oplus E_r^{(n_r)}$, the $E_i$ being nonisomorphic and each $E_i$ being repeated $n_i$ times, then the $E_i$ are uniquely determined up to isomorphism, and the multiplicities are uniquely determined.

The proof starts by assuming there is an isomorphism between direct sum decompositions into simple modules $$ E_1^{(n_1)}\oplus\cdots\oplus E_r^{(n_r)}\to F_1^{(m_1)}\oplus\cdots\oplus F_s^{(m_s)} $$ with $E_i$ nonisomorphic, and $F_j$ nonisomorphic. It then states from Schur's lemma, (that every nonzero homomorphism between simple modules is an isomorphism), that we conclude that each $E_i$ is isomorphic to some $F_j$, and conversely. I don't see how it applies here. How is Schur's lemma used to get that conclusion in the proof?

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A homomorphism from a direct sum $A_1\oplus\cdots \oplus A_n$ to a module $M$ is equivalent to a family of homomorphism $f_i\colon A_i\to M$ by the universal property of the direct sum. So a homomorphism $$f\colon E_1^{(n_1)}\oplus\cdots\oplus E_r^{(n_r)}\to F_1^{(m_1)}\oplus\cdots\oplus F_s^{(m_s)}$$ is equivalent to a family of homomorphism from each $E_i$ to the direct sum.

Composing such a homomorphism with the projections on the $F_j$ gives you a family of maps from each $E_i$ to each $F_j$. For a fixed $E_i$ not all those homomorphisms can be equal to $0$ (since that would mean the original map has nontrivial kernel), so at least one of them must be an isomorphism by Schur's Lemma. A symmetric argument using $f^{-1}$ gives the "conversely".

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Thanks Arturo! This makes perfect sense now. – Nastassja Apr 7 '12 at 21:47
@Nastassja: You're welcome. – Arturo Magidin Apr 7 '12 at 21:48

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