Flat Modules are Filtered Colimits of Free Modules

Question

A result by Wraith and Blass states that every flat module is a filtered colimit of free modules (see nLab, Thm 1).

I am wondering if this is simply a corollary of Yoneda's density theorem which states that every presheaf is a colimit of representables, or more precisely given a presheaf $F\colon C^{op}\to \mathbf{Set}$ we have

$$F\cong colim_{(C,x)\in el(F)}Y(C)$$

where $el(F)$ is the category of elements (see nLab page) and $Y\colon C\to [C^{op},\mathbf{Set}]$ is the Yoneda embedding.

To connect these ideas, we note that the category of $R$-modules is equivalent (as additive categories) with the category of additive functors $[\mathbb{R},\mathbf{Ab}]_{add}$ where we are interpreting $\mathbb{R}$ as the additive category with one object $*$ and homset $\mathbb{R}(*,*)=R$ with obvious addition of morphisms. There is a forgetful functor

$$ U\colon [\mathbb{R},\mathbf{Ab}]_{add}\to [\mathbb{R},\mathbf{Set}]$$

which forgets the additive structures. Now if $U(N)\colon \mathbb{R}\to \mathbf{Set}$ is flat (as a functor) for a module $N$ its category of elements is filtered and we have

$$ U(N)\cong colim_{(*,n)\in(el(U(N)))}Y(*) $$

which shows that as sets we have $$ U(N)\cong colim_{(*,n)\in el(U(N)}R $$ since $Y(*)(*)=\mathbb{R}(*,*)=R$ and colimits of presheaves can be computed pointwise (which in this case is only one point $*$). Thus $N$ is a filtered colimit of the free modules of rank 1, namely $R$, as sets. We also know that the forgetful functor $G\colon \mathbf{Ab}\to \mathbf{Set}$ creates filtered colimits (all algebraic categories share this property). So shouldn't the isomorphism lift to an isomorphism of abelian groups.

An assumption I am making but am not entirely convinced of is that a module $N$ is flat iff the presheaf $U(N)\colon \mathbb{R}\to \mathbf{Set}$ is flat.

Ideally, I am looking for the answer to the following:

Can the left Kan extension $N\otimes_{\mathbb{R}}-\colon [\mathbb{R}^{op},\mathbf{Ab}]\to \mathbf{Ab}$ along the $\mathbf{Ab}$-enriched Yoneda embedding $Y\colon \mathbb{R}\to [\mathbb{R}^{op},\mathbf{Ab}]_{add}$ of a flat module $N\colon \mathbb{R}\to \mathbf{Ab}$ be computed as the left Kan extension of presheaves with values in $\mathbf{Set}$?

Answer 1

a module $N$ is flat iff the presheaf $U(N)\colon \mathbb{R}\to \mathbf{Set}$ is flat.

This is true, if I'm not wrong.

Notice that there are embeddings of $\text{Mod}_R$ and ${}_R\text{Mod}$ in the bicategory of $\bf Ab$-valued profunctors, respectively as ${\bf Prof}({\bf 1}, \mathbb R)$ and ${\bf Prof}(\mathbb R,{\bf 1})$ (in your notation, $\mathbb R$ is the ring $R$ regarded as a category).

This entails, in particular, that tensor product $\otimes_R$ of modules $M\in{}_R\text{Mod}, N\in\text{Mod}_R$ corresponds to the composition of profunctors $\star$.

This motivates the following chain of equivalences:

1. $N$ is a flat module iff $-\otimes_R N$ commutes with finite limits;
2. iff, regarded as a profunctor, postcomposition $-\star N$ commutes with finite limits
3. (6.3 here) iff the presheaf $\tilde N$ corresponding to $N$ is flat
4. iff the category of elements of $\tilde N$ is cofiltered.

Answer 2

[...] a module $N$ is flat if and only if the functor $U(N)\colon \mathbb R \to \mathbf{Set}$ is flat.

I think this is false.

I change a bit the notations.

Let $\mathbb A = (A, +_A, \cdot, 1_A)$ be a commutative unitary ring and $\mathbb M = (M, +, 0_M)$ a $\mathbb A$-module. I denote by $\mathbf A$ the category (monoid) with a single object $\ast$ and with $A$ as set of morphisms, where $\cdot$ is the composition.

Consider the functor $\mathbf M$ from $\mathbf A$ to $\mathbf{Set}$ that maps the unique object of $\mathbf A$ to $M$ and, for each morphism $a$ of $\mathbf A$, the function $\mathbf M_a\colon M \to M$ is the action of $a$, i.e. $\mathbf M_a(m) = am$. The functor $\mathbf M$ is flat if and only if the category $\mathbf A^{\text op}/\mathbf M$ of elements of $\mathbf M$, regarded as a presheaf over $\mathbf A^{\text op}$, is filtered. This is spelled out by the following three conditions:

1. the only object $\ast$ is not mapped to the empty-set (this is always satisfied);

2. for any couple of elements $m, m'$ in $M$, there exist $n$ in $M$ and $a, a'$ in $A$ such that $an = m$ and $a'n = m'$;

3. for any element $m$ in $M$ and any couple of elements $a, a'$ in $A$ such that $am = a'm$ there exist $m'$ in $M$ and $b$ in $A$ such that $ab = a'b$ and $bm' = m$.

It's rather immediatele to notice that, if $\mathbb A = \mathbb Z$ and $\mathbb M = \mathbb Z/2\mathbb Z$, then the relative functor $\mathbf M$ is flat. Indeed, the category $\mathbf Z^{\text op}/\mathbf M$ is non-empty, it has $(\ast, 1)$ as weakly terminal object and the axiom of the couple of parallel arrows is trivial. (This example is wrong: see the comments)

The assumption of the OP fails on one side. For consider $\mathbb A = \mathbb Z$ and $\mathbb M = \mathbb Z\oplus \mathbb Z$, the latter with its canonical structure of free $\mathbb Z$-module. Further, consider the objects $(\ast, (1, 0))$ and $(\ast, (0, 1))$ of $\mathbf Z^{\text op}/\mathbf M$. Then it does not exist any object $(\ast, (m, n))$ and any couple of morphisms $a, a'$ from $(\ast, (1, 0))$ and $(\ast, (0, 1))$, resp., to $(\ast, (m, n))$ in $\mathbf Z^{\text op}/\mathbf M$. In fact, if they existed we should have on one hand $am= 1$ and $an= 0$, thus $m\neq 0$ e $n=0$, and on the other hand $a'm= 0$ and $a'n=1$, which is a contradiction.

Added later.

It seems that if $\mathbf M$ is flat as a functor, then $\mathbb M$ is flat as an $\mathbb A$-module.

Suppose now $\mathbf M$ is a flat functor and $\mathbb M$ a finitely generated $\mathbb A$-module. By induction, condition 2 above implies that the $\mathbb A$-module $\mathbb M$ is generated by a single element $n$. In this case, the $\mathbb A$-module $\mathbb M$ is flat if and only if whenever $an = 0_M$, for some $a$ in $A$, there exists $c$ in $A$ such that $cn = n$ and $ac = 0_A$ (see proposition 1 here ).

The flatness of the functor $\mathbf M$ implies then the flatness of the module $\mathbb M$. Indeed, consider condition 3 above with $m = n$ and $a' = 0$, i.e. suppose $an = 0_M$; hence there exist $b$ in $A$ and $m'$ in $M$ such that $bm' = n$ and $ab = 0_A$ and since $An = M$ we have that $m' = b'n$, for some $b'$ in $A$. Thus, posing $bb' = c$, we deduce the equalities $cn = n$ and $ac = 0_A$.

For the general case, a very similar argument should work. Suppose $\sum_{i=1}^k a_i m_i$, where $a_i$ in $A$ and $m_i$ in $M$ for all $1\leq i \leq k$. By condition 2 we find an $n'$ in $M$ which generates all the $m_i$, say $m_i = b_i n'$, so that we can rewrite the finite sum as $an' = 0_M$, where $a = \sum_i a_ib_i$.

Condition 3 implies then that we can find $n$ in $M$ and $b'$ in $A$ such that $n' = b'n$ and $ab' = 0$. Hence we have a collection $(b_ib')$ of elements of $A$ such that $m_i = b_ib' n$, for every $1\leq i \leq k$, and furthermore $\sum_i a_ib_ib' = ab' = 0_A$, which implies that $\mathbb M$ is flat as $\mathbb A$-module.