# Strategies for proving a category is Noetherian?

Let $C$ be a small linear category over a commutative ring $R$. A representation of $C$ is an $R$-linear functor $C \to \mathrm{Mod}(R)$. For example, for each $c\in C$, there is a representation $[c] = \hom_C(c,-)$. These representations are called generators, because every representation can be written as a quotient $\bigoplus_{i\in I} [c_i] / \bigoplus_{j\in J} [c'_j]$ for some indexing sets $I,J$ and objects $c_i,c'_j \in C$.

A representation $M$ is finitely generated if it can be written in such a way with $|I|<\infty$. Equivalently, $M$ is finitely generated if there is a surjection $\bigoplus_{i=1}^n [c_i] \to M$ for some finite list $\{c_i\}$ of objects in $C$. A map of representations is a surjection iff it evaluates to a surjection in $\mathrm{Mod}(R)$ when evaluated at each object in $C$.

The category $C$ is Noetherian if every subrepresentation of every generator is finitely generated. This is the a natural generalization of the case of algebras (which correspond to categories with only one object, hence subrepresentations are ideals).

What strategies are there to prove that some given category is Noetherian?

For example, is there a notion of "Ore extension" of categories for which Noetherian-ness is preserved?

In my case, I have a few categories that I would like to be Noetherian. They are constructed from smaller, known-to-be-Noetherian categories by some process that looks a bit like Ore extension, and then a bit not.

I'd rather not reinvent the wheel (or the stethoscope or the transistor or...) if I can help it.

• There are lots of equivalent definitions of Noetherian rings, each of those suggests a generalization to linear categories, and I don't know if they're equivalent. For example you could've chosen 1) submodules of f.g. modules are f.g. (maybe equivalent to your definition?) or 2) ideals (collections of arrows closed under composition with arrows in $C$ on either side) satisfy the ascending chain condition (doubt this is equivalent to your definition if $C$ has infinitely many objects), etc. May 5 '15 at 2:22
• @QiaochuYuan Condition 1) is equivalent. I don't know about 2), but it is less likely to be the one that I want for my application. But I agree that the word "the" in the second sentence of the third paragraph is perhaps should be "a". May 5 '15 at 4:30

One way to prove a ring is Noetherian is to use Groebner theory. Recent work of Sam and Snowden generalizes Groebner theory to certain small categories enriched in finite sets (so-called hom-finite categories). Any representation of such a category factors through the universal R-linearization category, and so their work handles any case where $C$ is a quotient of the universal R-linearization of some hom-finite category admitting Groebner bases.

The paper is excellent. http://arxiv.org/abs/1409.1670

Unfortunately, a practical criterion for checking if a category is Noetherian seems well beyond current technology, even assuming that the category is the linearization of a hom-finite category.

If you get those Ore extensions working, I'd love to hear about it :)

---Edit---

I have just posted a paper http://arxiv.org/abs/1508.04107 explaining when the representation theory of a category is simultaneously Noetherian and Artinian. In Corollary 1.8, I give a new way to show that a category is Noetherian: if it has what I call a "homological modulus." A homological modulus is an explicit gadget that you can look for computationally. For example, my REU student Andrew Gitlin has recently shown that the category of finite sets with relations https://en.wikipedia.org/wiki/Category_of_relations admits a homological modulus over $\mathbb{Q}$. If $R$ is a Noetherian $\mathbb{Q}$-algebra, we may conclude that finitely generated functors from $\mathrm{Rel}$ to $R$-modules are Noetherian. If $R$ is an Artinian $\mathbb{Q}$-algebra then these functors are actually finite length. I do not know if this example could also be completed using Sam-Snowden Groebner theory, but Gitlin's proof is sufficiently elegant that I'm sure his way is easier in this case.

• Thank you! I will take a look there. I realize that even for rings it's generally a very hard problem to decide if it is Noetherian, so a sharp practical criterion is definitely too much to ask for. This paper certainly provides some strategies I will attempt. May 5 '15 at 14:29

Proving that a category is Noetherian is tricky. Historically, there are two main ways of doing this: Realize your ring as the quotient of something noetherian or realize your ring as the (Ore) localization of something Noetherian. Both of these can be captured under the following slogan: Sometimes, Noetherianity can be transferred along a left adjoint. This is made precise in a note I wrote with Phil Tosteson.

Obviously, this is not the only way to prove that a category is Noetherian. As John pointed out, if you have monomials and divisibility is a WQO, then you get Noetherianity.

Nagpal, Sam and Snowden and have used the theory of twisted commutative algebras to prove that the category of finite sets with injections and a perfect matching on the complement is Noetherian. See here and here. Currently, it is not known if you can use Grobner methods to prove this result.

Also, in FI-modules over Noetherian rings, Church, Ellenberg, Farb and Nagpal proved that FI is Noetherian over any Noetherian ring. They do not use twisted commutative algebras or Grobner methods. If you are only interested in the Noetherianity of FI over a field, then you can use Grobner theory or TCAs to prove that FI is Noetherian.

• Comment on last sentence: the Grobner methods can be used to prove categories like FI are noetherian over any noetherian base. May 5 '15 at 17:23
• Regarding the last sentence and Steven's comment: to prove that FI is Noetherian, the Grobner methods (using that OI -> FI is finite) feel more modern than our original proof in Church-Ellenberg-Farb-Nagpal. (For this question, though, CEFN could still be valuable just for providing another approach to Noetherianness.) In any case, these methods certainly work over Z. In contrast, I believe the result of Nagpal-Sam-Snowden in your third paragraph does not apply except over C; see Question 1.5(3) and the remark following 5.3. Perhaps one of the authors can clarify this? Aug 18 '15 at 21:15
• The methods of the Nagpal-Sam-Snowden paper only work over a field of characteristic 0 (whether or not it's C is irrelevant, we just adopt that for simplicity of notation). Aug 18 '15 at 21:25