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Let me first recall some basic well-known definitions: Let $R$ be a ring (as always commutative). A (cocomplete) abelian tensor category is defined to be a symmetric monoidal category, whose underlying category is also a $R$-linear (cocomplete) abelian category, such that the tensor product is right exact (cocontinuous) and $R$-linear in each variable. An object $X$ of such a category $C$ is called flat if $X \otimes -$ is an exact functor. Also recall that an object $X$ of a symmetric monoidal category is called dualizable with dual $X^\*$ if it comes equipped with morphisms $1_C \to X \otimes X^\*$ and $X^\* \otimes X \to 1_C$ satisfying the two triangular identities.

Now these notions are connected in the case of $C = \text{Mod}(R)$: The dualizable objects are precisely the finitely generated projective $R$-modules (equivalently, locally free of finite rank). Thus, every dualizable object is flat, and Lazard's Theorem implies that every flat module is a directed colimit of locally free modules. Now I wonder if these statements remain true in the general case. I've already this concerning Lazard's theorem here for the category of quasi-coherent modules on a nice scheme (without answer). Thus my question is:

Question: Is every dualizable object in a (cocomplete) abelian tensor category flat?

Remark that we cannot insert "projective" as an intermediate step. It is easy to see that $1$ is projective iff every dualizable object is projective. But $1$ is not projective in $\text{Qcoh}(S)$ forr some non-affine concentrated scheme $S$, whereas the question in this case is answered with "yes" since every locally free quasi-coherent sheaf is locally flat and thus (globally) flat.

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Thank you Nacho and Todd for your answers. I would like to accept them both ;) –  Martin Brandenburg Jun 24 '11 at 12:45

2 Answers 2

up vote 14 down vote accepted

A dual pair as you describe above induces an adjunction $(X^*\otimes-)\dashv(X\otimes-)$, and since right adjoints preserve limits, $X$ is flat.

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Yes of course! I was, again, blind. –  Martin Brandenburg Jun 24 '11 at 12:44
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Indeed, this proves something much more than flatness for $X$: namely, $X\otimes$ preserves all limits, not just the finite ones. (I've usually seen the word "exact" meaning only preservation of finite limits and colimits.) –  Theo Johnson-Freyd Jun 24 '11 at 13:49

In a symmetric monoidal category, $X$ and $X^\ast$ are mutually dual. (We can compose with the symmetry to obtain a unit $1_C \to X^\ast \otimes X$ and counit $X \otimes X^\ast \to 1_C$ which realize $X$ as the dual of $X^\ast$.) This means we have both $X \otimes - \dashv X^\ast \otimes -$ and $X^\ast \otimes - \dashv X \otimes -$. Since $X \otimes -$ is both a left and right adjoint, it is exact.

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