Is the category of finite-dimensional $k[x]$-modules a comodule category?

Question

Fix a field $k$, denote by $k[x]$ the polynomial algebra. The category of finite-dimensional modules over $k[x]$ is precisely the category $\mathcal{C}$ consisting of pairs $(V, T_V: V \to V)$ of finite-dimensional vector spaces equipped with an endomorphism. Morphisms in this category are those linear maps $f: V \to W$ such that $f T_V = T_W f$. The category $\mathcal{C}$ is $k$-linear and abelian, every object is of finite length, and it has a natural fibre functor $\omega: \mathcal{C} \to \mathsf{Vect}_k$ by forgetting the endomorphism associated with a vector space.

Various things I have read say that the category $\mathcal{C}$ should be equivalent to the category of finitely-generated comodules over some coalgebra $B$, in a way which is compatible with the fiber functor $\omega$. But I am struggling to find any coalgebra which realises this. Do I have an assumption wrong? Perhaps something about generators of the category?

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Addition: Nilpotent transformations. I can solve this problem if I require all endomorphisms to be nilpotent. Let $\mathcal{N}$ be the full subcategory of $\mathcal{C}$ where each endomorphism is nilpotent. Then $\mathcal{N}$ is the category of comodules over the following coalgebra $N$:

• As a $k$-vector space, $N = \{x_0, x_1, x_2, \ldots\}$.
• The counit is $\epsilon(x_i) = \delta_{i0}$.
• The coproduct is $\Delta(x_n) = \sum_{i + j = n} x_i \otimes x_j$

This can be systematically derived by taking the dual coalgebras of the algebras $k[x]/x^n$ for increasing $n$, and then taking a limit. However, I have no idea how to do this for eigenvalues which are not zero.

Answer 1

Let me assume $k$ to be algebraically closed. Otherwise the story is more complicated.

Then finite dimensional modules over $k[x]$ can be described by Jordan normal form. In particular, the indecomposable modules are given by Jordan blocks. Moreover it is easy to see that there are no homomorphisms between indecomposable modules corresponding to different eigenvalues. Thus, the category of finite dimensional modules over $k[x]$ decomposes into 'blocks', i.e. subcategories such that there are no homomorphisms between each other.

Every block is isomorphic to the block for eigenvalue $0$ (just subtract $\lambda\operatorname{id}$ from a Jordan block with eigenvalue $\lambda$ to obtain a nilpotent Jordan block). One can check that this defines an equivalence of categories.

You already computed that the coalgebra corresponding to $\mathcal{N}$ is the tensor coalgebra $k[x]$. The category of comodules over a direct sum of coalgebras is given precisely by the "union" of the categories of comodules of the blocks (i.e. each object is just given by a direct sum of objects in the blocks). Therefore the coalgebra corresponding to finite dimensional comodules over $k[x]$ is $\bigoplus_{\lambda\in k} k[x]$.

Answer 2

Here's one way of thinking about this. First, over a field $k$, the category of coalgebras is the Ind-category of the category of finite-dimensional coalgebras. Second, the category of finite-dimensional coalgebras is equivalent to the opposite of the category of finite-dimensional algebras, by taking linear duals. Hence we have

$$\text{Coalg}(k) \cong \text{Ind}(\text{Coalg}_f(k)) \cong \text{Ind}(\text{Alg}_f(k)^{op}) \cong \text{Pro}(\text{Alg}_f(k))^{op}$$

from which we conclude that the category of coalgebras is equivalent to the opposite of the category of profinite algebras; that is, formal cofiltered limits of finite-dimensional algebras. The correspondence comes again from taking linear duals.

Furthermore, this correspondence respects modules in the following way: the category of finite-dimensional comodules over a coalgebra $C$ is equivalent to the category of finite-dimensional modules over the corresponding dual profinite algebra (where "module" means "module over a finite quotient," or equivalently "continuous module"). Finally:

Observation: The category of finite-dimensional modules over an algebra $A$ is equivalent to the category of finite-dimensional modules over its profinite completion $\widehat{A}$.

Here the profinite completion of an algebra $A$ is the cofiltered limit over all finite-dimensional quotients $A/I$.

So the question remains: what is the profinite completion of $k[x]$? Every finite quotient takes the form $k[x]/f(x)$ for some monic polynomial $f(x)$, which has some factorization $f(x) = \prod_i f_i(x)^{m_i}$ into irreducibles. The computation of the resulting cofiltered limit splits up into an independent piece for each irreducible, and we end up getting

$$\widehat{k[x]} \cong \prod_f \lim_m k[x]/f(x)^m$$

where the product runs over all monic irreducibles. In the special case that $k$ is algebraically closed, these are all linear polynomials $f(x) = x - a$, and we get

$$\widehat{k[x]} \cong \prod_{a \in k} k[[x - a]].$$

Compare to the usual computation of the profinite completion of $\mathbb{Z}$ as the product $\prod_p \mathbb{Z}_p$ over the $p$-adics. The corresponding dual coalgebra is the direct sum of the dual coalgebra of each $k[[x - a]]$ which is the thing in Julian's answer.

Answer 3

I would like to add an additional possibility that doesn't seem to have arisen yet. I assume $k$ to be a field for the sake of simplicity. Recall that if $A$ is an algebra over $k$ then we can consider its finite dual coalgebra $$A^\circ=\left\{f\in A^*\mid \ker(f)\supseteq I \text{ for some finite-codimensional ideal }I\subseteq A\right\}.$$ It can be deduced from Abe, Hopf Algebras, Chapter 3, §1.2, that there is an isomorphism of categories $$\mathsf{mod}_A\cong {^{A^\circ}\mathsf{comod}}$$ where $\mathsf{mod}_A$ are finite-dimensional right $A$-modules and ${^{A^\circ}\mathsf{comod}}$ are finite-dimensional left $A^\circ$-comodules.

In one direction, it is well-known that every comodule over $A^\circ$ is a (rational) module over $A^{\circ*}$ and, by restriction of scalars along $A\to A^{\circ*}$, this fact induces a functor ${^{A^\circ}\mathsf{Comod}}\to \mathsf{Mod}_A$ which can be (co)restricted to a functor ${^{A^\circ}\mathsf{comod}}\to \mathsf{mod}_A$ compatible with the underlying functors to $\mathsf{vec}_k$. Namely, if $(V,\delta)$ is a finite-dimensional $A^\circ$-comodule then $V$ becomes a finite-dimensional $A$-module via $$V\otimes A\to V, \qquad v\otimes a\mapsto \sum_{(v)}v_{[-1]}(a)v_{[0]}.$$

The other way around, assume that $(M,\mu)$ is a finite-dimensional $A$-module with dual basis $\sum_i e_i^*\otimes e_i$ over $k$. For every $m\in M$, set $f_m:A\to M,a\mapsto m\cdot a$. Then $$M\to A^\circ\otimes M, \qquad m\mapsto \sum_ie_i^*f_m\otimes e_i$$ is a well-defined $A^\circ$-coaction on $M$ and you may prove that these two assignments are each other inverses.

In fact, the idea is that $A^\circ$ is the universal coacting coalgebra that the Tannaka-Krein reconstruction provides out of the underlying functor $\omega:\mathsf{mod}_A\to\mathsf{vec}$.