# Infinite Krull-Schmidt categories?

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"?

• 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. – Sam Gunningham Jan 5 at 13:55
• 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. – Ivan Di Liberti Jan 5 at 14:22

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.)

• 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. – Jeremy Rickard Jan 9 at 22:51
• 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. – Nicholas Kuhn Jan 10 at 14:28
• Yes, I think that’s right. – Jeremy Rickard Jan 10 at 15:17