In a Krull--Schmidt 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"?

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    $\begingroup$ Some googling brought me to the book: "Module Theory: Endomorphism rings and direct sum decompositions in some classes of modules", by Alberto Facchini, which seems to contain a detailed account of this sort of thing. $\endgroup$ Commented Jan 5, 2019 at 13:55
  • $\begingroup$ The best that I know of is Theorem 1.3 in Abelian Categories by Popescu. I am not a specialist of Lie Algebras, thus I do not know if it applies to your case. $\endgroup$ Commented Jan 5, 2019 at 14:22

2 Answers 2


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, Compos. Math. 54, 63-78 (1985). ZBL0565.16012

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 $P\oplus U(\mathfrak{g})\cong U(\mathfrak{g})\oplus U(\mathfrak{g})$.


Theorem 1 of Chapter 1 of Pierre Gabriel's famous paper [Des catégories abéliennes, Bull. Soc. Math. France 90 (1962), 323–448] says that one gets a nice Krull-Schmidt theorem for arbitrary direct sums of objects in an abelian category $\mathcal A$ if $\mathcal A$ has a set of generators and exact "inductive" limits, and we are considering direct sums of indecomposable objects having local endomorphism rings. I am guessing that this applies to your situation. (I don't know much about the endomorphism rings of infinite dimensional indecomposables in your category.)

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    $\begingroup$ There are infinite dimensional indecomposable $\mathfrak{g}$-modules which don’t have local endomorphism rings. For example, the universal enveloping algebra with the regular action. $\endgroup$ Commented Jan 9, 2019 at 22:51
  • $\begingroup$ So the problem is with infinite dimensional indecomposables. The other thing asked about - if one can have a Krull Schmidt theorem for infinite direct sums - is fine, as long if the summands are all finite dimensional. $\endgroup$ Commented Jan 10, 2019 at 14:28
  • $\begingroup$ Yes, I think that’s right. $\endgroup$ Commented Jan 10, 2019 at 15:17

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