# Gabriel's theorem over a commutative ring

Is Gabriel's theorem on the indecomposables of representations of quivers of finite type true over a commutative ring, i.e. not necessarily a field?

• Could you give a precise statement of the theorem? – Dag Oskar Madsen Jan 11 '15 at 7:14
• The representations of the simplest quiver with only one vertex and no edge over a ring $R$, are the same as $R$ modules. So this quiver is of finite type iff $R$ has finitely many indecomposible modules. A good questions is to ask whether the classification of finite type quivers for such rings is the same as Gabriel's theorem. – Mostafa Jan 11 '15 at 7:38
• One could try to state the theorem in such a way that we allow for infinitely many indecomposables but the infinity "comes from R" instead of the quiver. Another way to make sense of such a theorem in a more general form would be to consider subcategories of the full module category like the category of $R\Lambda$-modules that are f.g. projective as $R$-modules. – Johannes Hahn Jan 11 '15 at 15:18
• Good point, Mostafa. I'd be happy for any result given any hypotheses you like on the ring, as I'm sure my intended application will satisfy them. For Dag Oskar Madsen, I mean the theorem that a quiver has finitely many isomorphism classes of indecomposable representations iff its underlying graph is a simply-laced Dynkin diagram, and that these are in correspondence with positive roots. – Eric Zaslow Jan 11 '15 at 18:02
• @Mostafa The short answer is no. Already the quiver $A_2$ creates trouble. Representations over $A_2$ are equivalent to modules over the 2x2 triangular matrix ring over $R$. For $R=\mathbb Z/(p^n)$ wtih $n \geq 4$ this is of infinite representation type, see blms.oxfordjournals.org/content/3/3/333.extract – Dag Oskar Madsen Jan 11 '15 at 18:07

Certainly it doesn't generalize in a really obvious way. One way to think about this is the following: Gabriel's theorem uses in really deep way that the category of quiver representations over a field is hereditary (in particular, $\mathrm{Ext}^2(M,N)=0$ for any representations). Over an arbitrary commutative ring, this is false, and so lots of things go out the window. For example, my preferred proof of Gabriel's theorem is from Crawley-Boevey's notes (http://www1.maths.leeds.ac.uk/~pmtwc/quivlecs.pdf); perhaps the most important lemma there is that a Dynkin quiver can't have any indecomposable quiver representations that have automorphisms other than scalars, or any self-extensions. Of course, this is totally false over a non-semisimple ring.

• Thanks. And if I include semi-simplicity? As I said, I'm willing to impose whatever hypotheses on my ring to get some kind of positive statement. – Eric Zaslow Jan 12 '15 at 0:34
• semisimple is most probably not a condition you're willing to impose, since a semisimple ring is a direct sum of fields. – Vivek Shende Jan 12 '15 at 4:22
• Confession: I only want it for the A_n case and for representations for which the cone on each map is a free module of rank one. – Eric Zaslow Jan 12 '15 at 7:10