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Let $R$ be a ring and $M$ an $R$-module. Suppose that $\{M_i\}_{i\in I}$ is a (possibly infinite) collection of simple submodules of $M$, which are pairwise isomorphic. Suppose that $M$ is the direct sum of $\{M_i\}_{i\in I}$. Suppose further that $M$ is also the direct sum of $\{K_j\}_{j\in J}$, where each $K_j$ is a simple submodule of $M$.

I can prove that each $K_j$ is isomorphic to each $M_i$. However, is it true that $I$ and $J$ are of the same cardinality? If so, I'd appreciate a direct proof (using basic equivalent definitions of a semisimple module is fine).

Chris A
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  • It's clearly true if $I$ is finite, so we can restrict to $|I|\geq\aleph_0$. I have some things I think are true when $I$ is infinite, but it will take a better set-theorist than myself to justify it. The intuition I have is that if $|I|\leq |M_i|$, then $|\oplus_{i\in I}M_i|=|M_i|$ and that if $|I|> |M_i|$, $|\oplus_{i\in I}M_i|=|I|$. Please take this with a grain of salt, because I am not 100% sure both these facts are true. – rschwieb Mar 20 '19 at 13:46
  • I would assume it is, and I would proof it by testing it with a copy of $M_j \cong K_i$, since you know precisely the form of those morphism spaces ($k J$ and $k I$ where $k X$ is the free vectorspace on $X$). – Felix Mar 20 '19 at 13:53
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    If your ring is commutative, it follows from uniqueness of cardinality of bases over vector spaces, right ? @rschwieb : your cardinality computations are correct, in fact in general $|\oplus_i M_i| = \displaystyle\sum_n \sum_{E\subset I, |E| = n} |M_i|^n$, so when $|M_i|$ is finite this is just the cardinality of the set of finite subsets of $I$, which is just $|I|$, and $|M_i|$ is infinite, this is just $|I||M_i| = \max {|I|, |M_i|}$ – Maxime Ramzi Mar 20 '19 at 13:57
  • We can get even more precise by noting that $\hom(M,M_i) = \hom (K_j, M_i)^J = \hom(M_i, M_i)^I$, so if $\kappa = |\hom (M_i, M_i)|$ then $\kappa^I = \kappa^J$. Of course exponentiation of cardinals is a terrible thing, but if $|I|\leq |M_i|$ and $\kappa$ is small enough this might bring some information – Maxime Ramzi Mar 20 '19 at 14:07
  • Also note $|\hom(M_i, M)|= \sum_{E\subset I, \mathrm{finite}} |\hom (M_i, M_i^E)|= \sum_{E\subset I, \mathrm{finite}} \kappa^E = |I|$ if $\kappa$ is finite, $=\kappa |I|$ otherwise. So this is more precise : if $\kappa \leq |I|$ then we get $|I|=|J|$ too (morally, this can happen if $M_i$ has very few endomorphisms, so for instance if $Z(R)$ is very small) – Maxime Ramzi Mar 20 '19 at 14:13
  • @Max Yes, I forgot we were talking about simple modules, so a dimensionality argument would make sense after identifying the endomorphism ring with the full linear ring over a division ring. Index sets with different cardinalities ought to correspond to the endomorphism ring having different dimensions over the base division ring. I think the hardest part is convincing oneself that the isomorphism of the two semisimple modules results in a ring isomorphism that's an algebra isomorphism (over the center of the division ring.) – rschwieb Mar 20 '19 at 15:16
  • @rschwieb : is there actually a theory of dimension over division rings ? (I don't know much about the noncommutative world) – Maxime Ramzi Mar 20 '19 at 15:18
  • @max Sure. All arguments about dimension go through identically without commutativity for a division ring. But again as I just realized we can probably argue the dimensions are different over the center of the division ring to dispense with the problem (Thank goodness the division ring has to be finite dimensional over its center.) – rschwieb Mar 20 '19 at 15:19
  • Or maybe we don't care about dimension? Maybe just the cardinality of the two matrix rings is enough – rschwieb Mar 20 '19 at 15:23
  • @rschwieb : thank you for confirming that. Why does the division ring have to be finite dimensional over its center though ? (+ come to think of it I'm not sure I undertsand your argument, taking $H= \hom(M_i, M_i)$, there are two $H$-module structures on $M$, one induced by $M_i$'s and one by $K_j$'s, how do you relate them ? ) – Maxime Ramzi Mar 20 '19 at 15:26
  • @user3533 : why is it true though ? It doesn't seem clear to me (I have already identified $M_i$ and $K_j$ so there is no post/pre composition with anything) – Maxime Ramzi Mar 20 '19 at 15:33
  • @Max I don't immediately recall the argument, but the proposition that given a simple module $S_R$, $End(S_R)$ is a division ring that's finite dimensional over its center is usually a result appearing near the Artin-Wedderburn theorems in any noncommutative text. – rschwieb Mar 20 '19 at 15:46

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This is a special case of the general Krull-Schmidt-Remak-Azumaya theorem.

Theorem 2.12 (Krull-Schmidt-Remak-Azumaya Theorem) Let $M$ be a module that is a direct sum of modules with local endomorphism rings. Then any two direct sum decompositions of $M$ into indecomposable direct summands are isomorphic.

There is an elementary proof of the special case based on a generalization of the concept of dimension of vector spaces, which you can find, for instance, in Jacobson's “Basic Algebra II”.

egreg
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