Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

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".

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.