A module $M$ over a ring $R$ is called *semisimple* if it admits a direct sum decomposition into simple modules. If $M$ admits a *finite* decomposition $M=\bigoplus\_{i=1}^n S\_i$ into simple $R$-modules $S\_i$, then this decomposition is unique up to isomorphism and permutation of the factors because from such a decomposition one can cook up a composition series and then use the Jordan–Hölder theorem.


I would like to know: are *infinite* decompositions unique as well?


E.g. if $R$ is semisimple, then *all* $R$-modules are semisimple. Do they decompose uniquely?



Uniqueness of infinite direct sum decomposition
Yes. A semisimple module $M$ is *canonically* isomorphic to


$$M \cong \bigoplus\_i \text{Hom}\_R(S\_i, M) \otimes\_{\text{End}(S\_i)} S\_i$$


where $\text{Hom}\_R(S\_i, M)$ is what you might call the multiplicity space of the simple module $S\_i$. It is naturally a module over the division ring $\text{End}(S\_i)$, and from this expression it follows that the multiplicity of $S\_i$ in any direct sum decomposition of $M$ into simples must be $\dim\_{\text{End}(S\_i)} \text{Hom}\_R(S\_i, M)$. (For this we need to know that $\text{Hom}\_R(S\_i, -)$ preserves infinite coproducts, but this follows from the fact that simple modules are cyclic and so in particular finitely generated.)


