k-linear abelian categories which are not categories of modules

Question

According to Joyal, Street ("An Introduction to Tannaka Duality and Quantum Groups"), any $k$-linear abelian category $\mathcal{C}$ admitting a faithful, exact functor $U: \mathcal{C} \rightarrow \mathcal{V}ect_{k}$ into finite-dimensional $k$-vector spaces arises as a category of finite-dimensional comodules over some coalgebra. In fact, the coalgebra in question is the "predual" $End^{pre}(U)$ of the endomorphism algebra of $U$.

Do all such abelian categories also arise as categories of finite-dimensional modules over some algebra? If not, what's the easiest example of a $k$-linear abelian category admitting a faithful, exact functor into finite-dimensional vector spaces which is not equivalent to a category of modules?

Edit: As pointed out below, one should also assume that $\mathcal{C}$ is essentially small so that the endomorphism coalgebra is a well-defined.

Answer 1

If $A$ is a $k$-algebra, and $M$,$N$ are finite-dimensional $A$-modules, then $$\operatorname{Ext}^i_A(M,N)\cong\operatorname{Tor}^A_i(M,N^*)^*$$ (where $*$ denotes $k$-dual).

So $\operatorname{Ext}^i_A(M,N)$ must be the dual of a vector space, and so in particular its dimension can't be countably infinite.

For $i=1$ it makes no difference whether we take $\operatorname{Ext}^1(M,N)$ in the category of finite-dimensional modules or the category of all modules, as it can be described in terms of equivalence classes of extensions $0\to N\to L\to M\to0$.

Let $\mathcal{C}$ be the category whose objects are finite-dimensional vector spaces over $k$ with a set of endomorphisms, all but finitely many zero, indexed by a countable set. Let $k$ be the one-dimensional object with all the endomorphisms zero. Then $\operatorname{Ext}^1(k,k)$ has countably infinite dimension, and so $\mathcal{C}$ can't be equivalent to the category of finite-dimensional modules for a $k$-algebra.

I think that if you translate this example into a question about profinite $k$-algebras as in Qiaochu's answer, it comes down to the fact that a countable product of copies of $k$ is not the profinite completion of any vector space over $k$.

Slightly changing the example (requiring the composition of any two of the endomorphisms to be zero) gives a simple example of a profinite $k$-algebra that is not a profinite completion:

Let $V$ be the direct product of countably many copies of $k$ with the product topology, and let $A=k\oplus V$ where $uv=0$ for all elements $u,v$ of $V$. The subalgebras of $A$ are all of the form $k\oplus U$ for some subspace $U\leq V$. The proper ideals of $k\oplus U$ are of the form $0\oplus U'$ for subspaces $U'\leq U$, so the profinite completion, as an algebra, of $k\oplus U$ is $k\oplus\hat{U}$, where $\hat{U}$ is the profinite completion, as a vector space, of $U$. But the profinite completion of any vector space $U$ is its double dual, which is larger than $V$ for any infinite dimensional $U$. So $A$ is not the profinite completion of any subalgebra.

Answer 2

This is not an answer. Below "finite" means "finite-dimensional over $k$," so "profinite" means "pro-finite-dimensional" and so forth.

The category of coalgebras is the ind-category of the category of finite coalgebras, and hence its opposite is the pro-category of the category of finite algebras, or in other words the category of profinite algebras. Moreover, the category of finite comodules over a coalgebra is equivalent, as a $k$-linear category with fiber functor, to the category of finite continuous modules over the corresponding profinite algebra, where "continuous" means that the action of the algebra factors through one of its distinguished finite quotients. (Here I am not sure where to insert "left" vs. "right," but it doesn't matter for our purposes.)

If $A$ is an algebra, then the profinite algebra corresponding to its category $\text{Mod}_f(A)$ of finite modules is its profinite completion (which is $\text{End}(U)$ where $U$ is the forgetful functor $\text{Mod}_f(A) \to \text{Vect}$, as suggested by Julian Rosen in the comments). If in the question everything is required to be essentially small and "equivalent" is taken to mean "equivalent as $k$-linear categories with fiber functors," then the question is equivalent to the following:

Is every profinite algebra a profinite completion?

This question is harder than I thought it was, and in particular something I really expected to be a counterexample isn't. Fix $k = \mathbb{F}_2$ and consider

$$B = \prod_{i \in \mathbb{N}} \mathbb{F}_2$$

regarded as the cofiltered limit of the projections $B \to \prod_{i \in S} \mathbb{F}_2$ where $S$ runs over all finite subsets of $\mathbb{N}$. The category of finite continuous $B$-modules is the category of finite $\mathbb{N}$-graded vector spaces over $\mathbb{F}_2$, so this is a somewhat simpler version of Todd's proposal in the comments.

Heuristically, the category of finite $\mathbb{N}$-graded vector spaces is the category of finite modules over the algebra generated by countably many commuting orthogonal idempotents $e_i, i \in \mathbb{N}$ (corresponding to projection onto the $i^{th}$ graded component) satisfying the additional relation

$$\sum_{i \in \mathbb{N}} e_i = 1.$$

The problem, of course, is that it is impossible to state this relation. So one might hope to use some kind of compactness argument to prove that $B$ is not a profinite completion. In addition, $B$ is not its own profinite completion: it admits maps $B \to \mathbb{F}_2$ which don't factor through any of its distinguished finite quotients coming from non-principal ultrafilters on $\mathbb{N}$.

Nevertheless:

$B$ is a profinite completion.

To motivate the construction, suppose $A$ is an algebra whose profinite completion is $B$ and $f : A \to B$ is the natural map. Since $A$ and $\text{im}(f)$ have the same finite quotients, we may assume WLOG that $f$ is injective, or equivalently that $A$ is residually finite. Now, $B$ is a Boolean ring: every element of it is idempotent. Hence $A$ is also a Boolean ring.

Every Boolean ring $A$ is the ring of continuous $\mathbb{F}_2$-valued functions on a profinite space $X$, namely its space of $\mathbb{F}_2$-valued points $\text{Hom}(A, \mathbb{F}_2)$. By hypothesis we know what the finite quotients of $A$ are, and hence we know that

$$\text{Hom}(A, \mathbb{F}_2) \cong \mathbb{N}$$

as a set. Hence $A$ must be the ring of continuous $\mathbb{F}_2$-valued functions on $\mathbb{N}$ where $\mathbb{N}$ has been equipped with a profinite (compact, Hausdorff, totally disconnected) topology, and so to construct $A$ it suffices to construct a profinite topology on $\mathbb{N}$.

But this is straightforward: we can use the topology coming from thinking of $\mathbb{N}$ as the one-point compactification of $\mathbb{N} \setminus \{ 1 \}$. With this topology, $A$ is the subalgebra of $B$ consisting of sequences $a_i \in \mathbb{F}_2$ such that $\lim_{i \to \infty} a_i = a_1$.

This construction shows more generally that every profinite Boolean ring is the profinite completion of some Boolean ring, as follows. The category of profinite Boolean rings is the pro-category of the category of finite Boolean rings, and hence its opposite is the ind-category of the category of finite sets, or in other words the category of sets. The profinite completion functor from Boolean rings to profinite Boolean rings is given on opposite categories by taking the underlying set, and so it suffices to show that every set admits a profinite topology, which the one-point compactification construction accomplishes.

On the other hand, I don't see how to adapt this construction to

$$B' = \prod_{i \in \mathbb{N}} M_i(\mathbb{F}_2)$$

regarded as the cofiltered limit of the obvious finite projections as before. The category of finite continuous $B'$-modules is again the category of finite $\mathbb{N}$-graded vector spaces, but with a different fiber functor.