# Flat Modules are Filtered Colimits of Free Modules

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 $\mathbf{Set}$?

• It's really not nice to use $\Bbb R$ so freely. – Pedro Tamaroff Apr 3 '16 at 12:40
• @PedroTamaroff Noted. Since there is an answer I will leave it unchanged. But I agree. Perhaps $\mathbb{B}R$ should be used instead? – Rachmaninoff Apr 5 '16 at 8:05
• The fact that (flat = filtered colimit of free) is certainly not originally due to Blass and Wraith. This is a classic theorem of commutative algebra, which is proved in any text on the subject. I want to say it's due to Lazard. – tcamps Sep 23 '17 at 14:18

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.
• Between 2 and 3 you pass from $\mathbf{Ab}$-valued profunctors to $\mathbf{Set}$-valued profunctors. Why is this allowed? – Zhen Lin Apr 3 '16 at 13:25
• Thanks for the link to Distributors. Are you making the assumption 2 to 3 since $\mathbf{Ab}\to \mathbf{Set}$ creates filtered colimits? – Rachmaninoff Apr 5 '16 at 8:44

[...] 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.

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.

• I am not convinced that $\mathbb Z/2\mathbb Z$ yields a flat presheaf. The pair of arrows corresponding to "multiply by 2" and "multiply by 4" on $(*,1)\to (*,0)$ is parallel in the category of elements without a coequalizing arrow. – Rachmaninoff Apr 11 '16 at 1:22
• Also, the presheaf corresponding to the module $\mathbb Z\oplus \mathbb Z$ is flat. The condition you attempted to use to derive a contradiction is wrong. The correct condition is that for any two objects in the category there is a third object and two arrows to it. You flipped this around and started with the third object $(*,(m,n))$. There is an arrow, namely "multiply by 0" which does connect $(*,(0,1))$ and $(*,(1,0)$. Perhaps you are standing on your head and looking at the dual instead? – Rachmaninoff Apr 11 '16 at 1:33
• A functor $F \colon A \to \mathbf{Set}$ is flat iff the opposite of the comma category $(e\downarrow F)$ is filtered (by $e$ I denote a terminal object in $\mathbf{Cat}$). Equivalently, regarding $F$ as a presheaf over $A^{\text op}$, iff its category of elements (à la Grothendieck), which I denoted $A^{\text op}/F$, is filtered. So one actually must check that $(e\downarrow F)$ is co-filtered. – Andrea Gagna Apr 11 '16 at 7:55
• @Rachmaninoff : I added a sketch of proof of "flat as presheaf implies flat as module". The proof uses the most explicit description of flat module that I'm aware of (and so it is not very elegant or "formal"). – Andrea Gagna Apr 12 '16 at 10:48
• Looks good. I also came across a result in Martin Brandenburg's thesis regarding the left kan extensions in R-linear tensor categories which is relevant, see the Prelim chapter of arxiv.org/pdf/1410.1716v1.pdf – Rachmaninoff Apr 12 '16 at 12:29