# Uniqueness of infinite direct sum decomposition

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?

• What's unique is the decomposition is into isotypic summands. The decomposition into simple summands is only unique up to isomorphism.
– YCor
Dec 13 '20 at 14:14
• Yes, see Decomposition of a module — this is Azumaya theorem.
– abx
Dec 13 '20 at 14:15
• A reference is T-Y. Lam. “Lectures on modules and rings”. Graduate Texts in Math. 189. SpringerVerlag, New York, 1999.
– YCor
Dec 13 '20 at 14:15
• @abx: You sure? This is about the Krull-Schmidt theorem which I thought only applies to finite direct sum decompositions!?
– user78294
Dec 13 '20 at 14:36
• @YCor: Sorry, could you elaborate a bit more? I'm not sure I got this.
– user78294
Dec 13 '20 at 14:37

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.)