In a [Krull--Schmidt category](https://en.wikipedia.org/wiki/Krull%E2%80%93Schmidt_category), if 
$$
X\_{1}\oplus X\_{2}\oplus \cdots \oplus X\_{r}\cong Y\_{1}\oplus Y\_{2}\oplus \cdots \oplus Y\_{s},
$$
where the $X\_{i}$ and $Y\_j$ are all indecomposable, then $r = s$, and there exists a permutation $\pi$ such that $X\_{\pi (i)}\cong Y\_{i}$, for all $i$.


I believe this works for the abelian category of not-necessarily finite-dimensional modules over a simple Lie algebra $\frak{g}$, where we no longer require that the number of summands is finite, but I can't seem to prove it. Does somebody know of a "nice" proof?


Also, those there exist a notion of an "infinite Krull-Schmidt category" abstracting these properties? If so, when does is a general abelian of ""infinite Krull-Schmidt type"?



Infinite Krull-Schmidt categories?
The statement about simple Lie algebras is not true.


A (finitely generated right) module $P$ for a ring $R$ is **stably free** if $P\oplus R^m\cong R^n$ for some integers $m,n$. 


Suppose $R$ has a non-free stably free module $P$, as above. If also $R$ is a (right) Noetherian domain, then the regular module $R$ is indecomposable and $P$ is a finite direct sum of indecomposable modules. So $R^n$ has two distinct decompositions into finitely many indecomposable summands, contradicting the Krull-Schmidt property.


If $\mathfrak{g}$ is any finite dimensional Lie algebra, then its universal enveloping algebra $U(\mathfrak{g})$ is a Noetherian domain.


Probably there are particular examples that predate this, but Theorem 2.6 of


*Stafford, J. T.*, [**Stably free, projective right ideals**](http://www.numdam.org/item?id=CM_1985__54_1_63_0), Compos. Math. 54, 63-78 (1985). [ZBL0565.16012](https://zbmath.org/?q=an:0565.16012)


shows that if $\mathfrak{g}$ is a finite dimensional non-abelian Lie algebra, then $U(\mathfrak{g})$ always has a non-free (finitely generated) stably free module.


In fact, I think that the construction gives an indecomposable non-free $P$ with $P\oplus U(\mathfrak{g})\cong U(\mathfrak{g})\oplus U(\mathfrak{g})$. 


