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It is a fun exercise to show that a (unital) ring $R$ is a division ring if and only if each $R$-module is free (a sketch is given here). From a categorical standpoint this characterization is very pleasing as it is completely expressible in categorical terms (the free functor of the monad on $\mathsf{Ab}$ induced by $R\otimes_{\Bbb Z} -$ is essentially surjective), especially when contrasted with other ambiguous definitions of a field in a category/topos (e.g. on the nlab).

I am aware that the characterization above relies on the law of excluded middle while the other definitions are supposed to work in a constructive setting. In fact, as mentioned in Ingo Blechschmidt's thesis, in a topos not every vector space is free (p.9) but still not not free (p.11). I think this renders the question, whether above characterization is useful in other settings than set (like that of sheafs of rings, topological rings, simplicial rings etc.) obsolete (or does it?). But still one might ask

Is there a constructive version of above characterization? Can we for example weaken the characterization to the free functor being weakly essentially surjective with respect to some notion of weak equivalence?

The problem may be that we need to know what not not isomorphic is supposed to mean. A naive guess could be having a monic + epic morphism, but unfortunately every topos is balanced, so this doesn't make sense...

As always, thank you for your time!

Jonas Linssen
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    As for the title question and first paragraph, I'd say yes it is pleasing and useful. There are many ring theoretic properties which can be characterized purely from properties of their module categories. Examples include: (semi)hereditary rings, semisimple rings, quasi-Frobenius rings, (semi)perfect rings, absolutely flat rings, etc – rschwieb Jul 21 '21 at 16:35
  • You are right, I think the title doesn’t reflect my intention enough. In my limited experience it is often very easy to see that fields have a certain ring theoretic property though (eg because they lack nontrivial ideals), without invoking the freeness of vector spaces. My question was aimed (but not phrased) towards rings and modules in other settings than sets (eg. vector bundles), but anyway thank you very much for bringing this up! – Jonas Linssen Jul 21 '21 at 16:42
  • You seem to be asking for a classification of vector space objects in a topos? There simply isn't one. – Martin Brandenburg Jul 22 '21 at 08:08
  • Why is this? Even for Grothendieck Topoi? And what about vector spaces in $\operatorname{Fun}(\mathcal{C},\mathsf{Set})$? In fact I would like to know how much linear algebra one can do in the monadic/algebraic theory setting as opposed to the internal logic/topoi setting. Knowing what a field is supposed to be would be helpful for that and generalizing above characterization would have been really nice… – Jonas Linssen Jul 22 '21 at 14:45
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    In algebraic geometry, I'd say about the closest thing we get to a field is a locally ringed space (where there exist inverses on the non-zero open sets of sections). I'm not sure off the top of my head if there's a nice categorical formulation of that criterion, aside from the usual characterization that if the space has enough points, then every point must evaluate the sheaf to a local ring. Even then, it's far from true that every sheaf of modules is locally free (or projective, which IIRC is equivalent for finitely generated sheaves). – Daniel Schepler Jul 22 '21 at 15:32
  • The theory of vector spaces is algebraic, and so there is a category $\mathbb{V}$ so that ${ \text{Vector spaces in } \mathcal{C} } \cong \mathsf{FP}(\mathbb{V}, \mathcal{C})$, the category of finite product preserving functors $\mathbb{V} \to \mathcal{C}$ (this really is a natural isomorphism of categories too, more than just an equivalence). But then it's easy to see we have... – HallaSurvivor Jul 22 '21 at 19:43
  • ${ \text{vector spaces in } \mathsf{Fun}(\mathcal{C}, \mathsf{Set}) } \cong \mathsf{FP}(\mathbb{V}, \mathsf{Fun}(\mathcal{C}, \mathsf{Set})) \cong \mathsf{Fun}(\mathcal{C}, \mathsf{FP}(\mathbb{V}, \mathsf{Set})) \cong \mathsf{Fun}(\mathcal{C}, \mathsf{Vect})$. That is, a vector space in a presheaf category is exactly a functor into $\mathsf{Vect}$. In fact, this is true for all algebraic theories. – HallaSurvivor Jul 22 '21 at 19:46
  • Fields are trickier, because the theory of fields is not algebraic. In fact, the theory of fields is coherent, which is much more delicate. – HallaSurvivor Jul 22 '21 at 19:52
  • @DanielSchepler I don’t quite understand the inverses on the non-zero open sets of sections part. You don’t mean a sheaf of fields, do you? (Considering the nullring as field gives a terminal object to $\mathsf{Field} \subseteq \mathsf{CRing}$, or alternatively forcefully adjoining a terminal object) And with evaluation at a point you mean taking stalks, right? – Jonas Linssen Jul 22 '21 at 22:29
  • @HallaSurvivor yes, this is where my question originates. The fact that vector spaces are an algebraic theory comes from the fact that modules over a ring are. So my question may be rephrased as: Can we decide, whether given a category of modules in a nice category $\mathcal C$ we have a theory of modules (= coming from a ring in $\mathcal{C}$) or a theory of vector spaces (= coming from a field in $\mathcal C$, whatever this means). In $\mathcal C = \mathsf{Set}$ we can make a distinction via above characterization, but does this generalize? – Jonas Linssen Jul 22 '21 at 22:36
  • In a locally ringed space, if you have a section $x \in \mathscr{O}X(U)$ for some open set $U$, then you can form $D(x) := { p \in U \mid x_p \notin \mathfrak{m}_p }$ which is an open subset of $U$; and then $x |{D(x)}$ is a unit of $\Gamma(D(x), \mathscr{O}_X)$. And yes, in general on a topos $\mathcal{E}$, a "point" is a functor $\mathcal{E} \to \mathbf{Set}$ satisfying certain properties; and for the case of the topos of sheaves on a topological space, taking stalks at a point is an example of such a functor. – Daniel Schepler Jul 22 '21 at 22:53
  • And the condition that a topos "has enough points" is the condition that all points, taken together, form a faithful family. (Or maybe it's the condition that a morphism is an isomorphism if and only if its image under each point is an isomorphism; I don't recall exactly.) – Daniel Schepler Jul 22 '21 at 22:55
  • @PrudiiArca -- well you have to remember how we get a theory of modules to be algebraic. We add in a unary operation for each element of our ring, and basically view modules as abelian groups with operators. This does not necessarily mean that there's a ring/field internal to $\mathcal{C}$ which acts on your internal modules in a nice way (or if it does, I'm not immediately seeing how). In the case of modules you can build a $2$-sorted theory with a ring and a module to combat this issue, but vector spaces and fields are still tricky – HallaSurvivor Jul 22 '21 at 23:28
  • @DanielSchepler thanks, my algebraic geometry is a bit rusty and I didn’t know the fact about invertibility. – Jonas Linssen Jul 23 '21 at 13:48
  • @HallaSurvivor I see, thanks. – Jonas Linssen Jul 23 '21 at 13:50

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