# Does projective imply flat?

Let $\mathcal C$ be an abelian category equipped with a closed symmetric monoidal structure. This implies in particular that the monoidal structure $\otimes$ is right exact in each variable. I care most about the situation where $\mathcal C$ is finite $\mathbb C$-linear in the sense of arXiv:1406.4204 (in which case the monoidal structure is closed iff it is right exact in each variable). Note that, unlike in that paper, I specifically care about monoidal structures which are not rigid. An example of the category I have in mind is the category $\mathrm{Mod}^f_A$ of finite-dimensional modules for any finite-dimensional commutative algebra $A$ (with $\otimes = \otimes_A$).

Recall that an object $P \in \mathcal C$ is projective if $\hom(P,-) : \mathcal C \to \mathrm{AbGp}$ is right exact. (It is already left exact.) Note that this has nothing to do with the monoidal structure.

An object $F \in \mathcal C$ is flat if $F \otimes : \mathcal C \to \mathcal C$ is left exact. (It is already right exact.) Note that this has everything to do with the monoidal structure.

Are projective objects necessarily flat?

Of course, in $\mathrm{Mod}_A^f$ they are. The other examples I usually use of non-rigid monoidal categories are the representation theories of non-Hopf bialgebras, but there every object is flat.

• If you have enough injectives, then yes, would you be willing to assume this? Jun 30 '15 at 16:56
• @FernandoMuro Yes, all the categories I care about have enough injectives, so I'm happy to add that as an axiom. If you have a proof available, I'll be happy to accept it as an answer. Jun 30 '15 at 17:02
• @AlexDegtyarev Do you mind spelling out your abstract nonsense? Eric Wofsey in an answer below seems to provide a counterexample. Jun 30 '15 at 21:59
• @Alex: I think you can only conclude that Tor is zero if one of the arguments is flat; to conclude this from one of the arguments being projective you already need to know that projective implies flat. Jun 30 '15 at 23:56
• @TheoJohnson-Freyd the argument I had in mind wrongly assumed that $\hom(-,I)$ had to be exact, where $\hom$ is the inner $\hom$ and $I$ is an injective object. But asking that is no different to asking projectives to be flat. Sorry for creating wrong expectations ;) Jul 1 '15 at 9:44

I believe the following is a counterexample. Let $\mathcal{A}$ and $\mathcal{B}$ be closed symmetric monoidal abelian categories such that the unit object $1\in\mathcal{B}$ is projective and let $F:\mathcal{A}\to\mathcal{B}$ be a non-exact strong symmetric monoidal functor which has a right adjoint $G:\mathcal{B}\to\mathcal{A}$. For instance, if $A$ is a commutative ring and $B$ is a commutative $A$-algebra which is not flat over $A$, you could have $\mathcal{A}=\mathrm{Mod}_A$ and $\mathcal{B}=\mathrm{Mod}_B$ and $F(M)=M\otimes_A B$. Let $\mathcal{C}=\mathcal{A}\times\mathcal{B}$, and equip it with the symmetric monoidal structure given by $$(M,V)\otimes (N,W)=(M\otimes N,F(M)\otimes W\oplus V\otimes F(N)\oplus V\otimes W).$$

The unit is $(1,0)$, and associativity follows from $F$ being strong symmetric monoidal. Furthermore, this monoidal structure is closed, with internal hom given by $$\operatorname{hom}((M,V),(N,W))=(\operatorname{hom}(M,N)\oplus G(\operatorname{hom}(V,W)),\operatorname{hom}(F(M),W)\oplus\operatorname{hom}(V,W)).$$

In this category, the object $(0,1)$ is projective by hypothesis, but it is not flat because $(M,0)\otimes (0,1)=(0,F(M))$ and $F$ is not exact.

In the example mentioned above where $\mathcal{A}=\mathrm{Mod}_A$ and $\mathcal{B}=\mathrm{Mod}_B$ and $B$ happens to be a quotient of $A$, this construction has the following intuitive explanation. The monoidal product is defined as if $(M,V)$ were secretly the $A$-module $M\oplus V$ and the tensor product is just the ordinary tensor product of $A$-modules. In particular, since $B=0\oplus B$ is not flat over $A$, the object $(0,B)$ is not flat. However, the category itself doesn't believe that $(M,V)$ is just a single $A$-module $M\oplus V$, and in particular the quotient map $A\to B$ does not exist as a map $(A,0)\to (0,B)$ that would cause $(0,B)$ to fail to be projective.

For a finite $\mathbb{C}$-linear version of this example, you can take $A$ and $B$ to be finite-dimensional $\mathbb{C}$-algebras and restrict to finitely generated modules everywhere.

• I think your $V \otimes W$ factor may be spurious? Doesn't it spoil associativity? I'm pretty sure the square-zero extension version of what you're saying works. Beautiful example.
– thel
Jun 30 '15 at 20:46
• @ThelSeraphim: I don't see how the $V\otimes W$ term causes any trouble; it is natural to include it in the special case described in the penultimate paragraph. I think you're right that the "square-zero" version without it also works, though. Jun 30 '15 at 21:01
• yeah, book-keeping problem on my side, sorry. Both ways work.
– thel
Jun 30 '15 at 21:10
• Well, grumble. And this gives an example where I don't even have flat resolutions, since by the same logic $(N,1)$ fails to be flat for every $N$. (Tensoring with $(M,0)$ and then projecting onto the $\mathcal B$ factor still gives the functor $F$.) Jun 30 '15 at 22:06
• @TheoJohnson-Freyd: A projective non-flat object can never have a flat resolution, since by projectivity the first map in the resolution will split and a direct summand of a flat object is flat. Jun 30 '15 at 22:14

The paper:

When projective does not imply flat, and other homological anomalies, Theory and Applications of Categories, Vol 5, pp. 202-250, 1999, available here

by Gaunce Lewis shows that this behaviour is quite common in categories of Mackey functors for compact Lie groups. These categories arise very naturally in equivariant stable homotopy category. Here is a quote from the abstract: "These examples were not fabricated to illustrate the abstract possibility of misbehavior. Rather, they are drawn from the literature."

• Nice! (On a different note, did you do that thing you emailed me about? :-P) Jul 1 '15 at 1:24