Fix a ground field $k$. By a linear category I will mean an Abelian category which is compatibly enriched over $k$-vector spaces. A linear category is called finite if it satisfies the following four conditions:

  1. Every hom vector space is finite dimensional;
  2. There are finitely many simple objects, up to isomorphism;
  3. There are enough projective objects (i.e. for every object $x$ there is a projective object $p$ and a surjection $p \to x$; and
  4. Every object has finite length, i.e. for each object $x$, any descending chain of proper subobjects $$ x \supsetneq x_0 \supsetneq x_1 \supsetneq x_2 \supseteq \cdots $$ is necessarily finite.

It is well known that a linear category is finite if and only if it is equivalent to $Rep(A)$ the category of finite dimensional modules of a finite dimensional $k$-algebra $A$. See for example Prop. 1.4 of arXiv:1406.4204 for a proof of this fact.

The proof given in this reference uses all four of these properties. But one can still wonder if each of these assumptions is absolutely necessary? Perhaps some of these conditions imply the others? The following examples show that this is not the case for the first three properties:

Example 1: Let $K$ be an infinite field extension of $k$. Then the category $Vect_K$ of finite dimensional $K$-vector spaces (finite $K$-dimension, that is) is a $k$-linear category which satisfies (2), (3), and (4), but not (1).

For the next examples if $A$ is a $k$-algebra, not necessarily finite, we will let $Rep(A)$ denote the category of finite dimensional $A$-modules, i.e. finite dimensional vector spaces equipped with an action of $A$.

Example 2: Let $A = \oplus_\infty k$, the countable direct sum of copies of the algebra $k$. Then $Rep(A)$ satisfies (1), (3), and (4), but not (2).

Example 3: Let $A = k[[x]]$ be the formal power series algebra on one generator. Then $Rep(A)$ satisfies (1), (2), and (4), but not (3).

So that leaves the finial property. My question is:

Question: Is there an example of a linear category which satisfies the above properties (1), (2), and (3), but which fails to satisfy property (4)?

It is easy to construct examples which fail to satisfy property (4), but everything I have come up with so far also fails to satisfy one of the other properties as well. One of the reasons I am interested in this question is to better understand the Deligne tensor product. It is usually defined for finite linear categories, and I would like to understand what kinds of categories we would be dealing with if various assumptions were dropped.


Consider the category of functors from the poset $(\mathbb{Q}_{\geq0},\leq)$ to finite-dimensional vector spaces, and let $\mathcal{A}$ be the full subcategory consisting of functors $F$ for which there is a finite sequence $$0=x_0<x_1<\dots<x_n$$ in $\mathbb{Q}_{\geq0}$ such that $F(y)\to F(z)$ is an isomorphism whenever $x_i\leq y<z<x_{i+1}$ for some $i$ or when $x_n\leq y<z$. Call $\{x_0,\dots,x_n\}$ an exceptional set of $F$.

Then $\mathcal{A}$ is an exact abelian subcategory of the functor category: if $F$ and $G$ are in $\mathcal{A}$ with exceptional sets $S_F$ and $S_G$, then the kernel and cokernel of a morphism $F\to G$ are in $\mathcal{A}$ with exceptional set $S_F\cup S_G$.

(1) If $F$ has exceptional set $\{x_0,\dots,x_n\}$ then a morphism $F\to G$ is determined by the maps $F(x_i)\to G(x_i)$, so Hom-spaces are finite-dimensional.

(2) There are finitely many simple objects because there are no simple objects at all.

(3) For $x\in\mathbb{Q}_{\geq 0}$ let $P_x$ be the object where $$P_x(y)= \begin{cases}0&\mbox{if }y<x\\ k&\mbox{if }y\geq x \end{cases}$$ and $P_x(y)\to P_x(z)$ is an isomorphism for $x\leq y<z$. Then $P_x$ is projective, and if $F$ has exceptional set $\{x_0,\dots,x_n\}$ then $F$ is a quotient of $$\left(P_{x_0}\otimes_kF(x_0)\right)\oplus\dots\oplus\left(P_{x_n}\otimes_kF(x_n)\right),$$ so $\mathcal{A}$ has enough projectives.

(4) If $y_0<y_1<\dots$ is an increasing sequence in $\mathbb{Q}_{\geq0}$, then $$P_{y_0}\supsetneq P_{y_1}\supsetneq\dots,$$ so $\mathcal{A}$ has infinite descending chains of proper subobjects.

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  • $\begingroup$ This is a very clever example, thank you! $\endgroup$ – Chris Schommer-Pries Oct 24 '14 at 9:14

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