# Is it really necessary to choose a basis to prove that every element of a comodule is contained in a finite-dimensional subcomodule?

Let $C$ be a coalgebra over a field and $M$ be a $C$-comodule. Then it's well-known that every element of $M$ is contained in a finite-dimensional subcomodule $M' \subset M$. This is for example an adaptation of proof given in Hovey's Model categories (Lemma 2.5.1, with notations changed a bit):

Let $\{c_i\}$ be a basis of $C$ and write the coaction $\gamma : M \to C \otimes M$ as: $$\gamma(m) = \sum_i c_i \otimes m_i,$$ where only a finite number of $m_i$ may be nonzero. For a fixed $m \in M$, let $M'$ be the finite-dimensional subspace of $M$ spanned by the nonzero $m_i$. By counitality, $M'$ contains $m$, and using coassociativity of $\gamma$, $M'$ is a $C$-comodule.

As you can see, this proof requires choosing a basis of $C$. Is it possible to rewrite the proof to avoid choosing this basis? If not, is it possible that the result becomes false in a setting where choosing a basis is not always possible (infinite dimensional $C$ and no axiom of choice)?

• So... What's a coalgebra? Mar 30, 2016 at 9:34
• @AsafKaragila Take all the axioms defining an (associative, unital) algebra and reverse all the arrows. en.wikipedia.org/wiki/Coalgebra Basically a vector space $C$ equipped with a coproduct $\Delta : C \to C \otimes C$ that satisfies coassociativity, and a counit $\epsilon : C \to \Bbbk$. Mar 30, 2016 at 9:35
• Since I don't have the time to study a whole lot about comodules and the likes, let me ask some questions which might be helpful. How strange can comodules be? In particular it is possible to have comodules which are not vector spaces over your field? Are there examples for infinitely generated coalgebra and comodule over it which both have a lot of automorphisms (so every element can be moved and the orbit of each element is rather large)? Mar 30, 2016 at 13:40
• @AsafKaragila A comodule will be a vector space over $\Bbbk$ by definition, it's a vector space $M$ equipped with a coaction $\gamma : M \to C \otimes_\Bbbk M$ that satisfies conditions dual of the axioms of a module over an algebra (i.e. $a \cdot (b \cdot m) = (ab) \cdot m$ and $1 \cdot m = m$). I don't really know for your second question -- I guess a cofree coalgebra has a ton of automorphisms, and a coalgebra is a (cofree of rank one) comodule over itself. Mar 30, 2016 at 14:00
• Coffee coal.? :) But kidding aside, that would make sense. I am asking because if you have a structure with sufficient homogeneity conditions you can "mimic it" using a symmetric extension so you have some control over what type of substructures it has. So for example if you had sufficiently homogeneous comodule, you might be able to ensure that the only finitely generated comodule is the trivial one. I don't know, this is all just vague ideas, and it would require some serious weirdness on the part of the comodule. And I really don't know enough on this to say anything substantial. [...] Mar 30, 2016 at 14:09

Theorem: Let $$k$$ be a noetherian ring, $$A$$ a $$k-$$coalgebra (coassociative counital) that is flat as $$k-$$module and $$V$$ a $$A-$$comodule. Let $$v \in V$$ and $$(v)$$ be the $$A-$$subcomodule generated by $$v$$, then $$(v)$$ is finitely generated as $$k-$$module.
Lemma: Let $$k$$ be a ring, $$A$$ a flat $$k-$$coalgebra, $$V$$ a $$A-$$comodule and $$W \subseteq V$$ a $$k-$$submodule. Let $$W^0 = \Delta_V^{-1}(W \otimes_k A) = \{v \in V \,|\, \Delta_V(v) \in W \otimes_k A\}$$ Then $$W^0$$ is a $$A-$$subcomodule and $$W^0 \subseteq W$$.