# Tame abelian tensor categories

In the article "Tannaka duality for geometric stacks" (arxiv, see nlab for a summary) Jacob Lurie introduced the notion of a tame abelian tensor category. An abelian tensor category is called tame if every short exact sequence $0 \to M' \to M \to M'' \to 0$ with $M''$ flat remains exact after tensoring with an arbitrary object. Basically this means that flat objects behave in the usual way. For example it is easy to check that in a tame abelian tensor category every extension of flat objects is flat (Lemma 5.4).

Question 1: Do you know any further literature about tame abelian tensor categories?

Question 2: Can you give an explicit example of an abelian tensor category which is not tame? Jacob Lurie remarks that every abelian tensor category with enough flat objects is tame, because then we can define $Tor_1$ and use symmetry. So in particular, a counterexample does not have enough flat objects. [Tom Goodwillie has given an example below]

Question 3: Can you give an example of a cocontinuous tensor functor between tame abelian tensor categories which is not tame?

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What's a flat object? –  Qfwfq Mar 7 '11 at 11:22
Flat module: en.wikipedia.org/wiki/Flat_module –  David Roberts Mar 7 '11 at 12:07
Well it's an object $M$ such that the functor $M \otimes -$ is exact. See also Lurie's article for more definitions and background. –  Martin Brandenburg Mar 7 '11 at 14:00

I don't know the definition: is the category of torsion abelian groups an example of an abelian tensor category? With $\otimes$ having its usual meaning? If so, then it seems that $\mathbb Q/\mathbb Z$ is flat (tensor product with anything is zero!) but the sequence

$0\to \mathbb Z/p \to \mathbb Q/\mathbb Z\to \mathbb Q/\mathbb Z\to 0$

does not remain exact when you tensor it with $\mathbb Z/p$.

EDIT: For a true example I need the $\otimes$ operation to have a unit object. You can artificially introduce a unit, but it won't help with this example, so first let me alter the example. Choose a domain $R$ that is a $\mathbb Q$-algebra and not a field. Let $K$ be the field of fractions. In the category of torsion $R$-modules $M\otimes (K/R)$ is always zero, and the sequence

$0\to R/fR \to K/fR\to K/R\to 0$

does not remain exact when tensored with $R/fR$. Now, to introduce a unit, consider the product of the category of $\mathbb Z$-modules and the category of torsion $R$-modules. Define

$(A,M)\otimes (A',M')=(A\otimes_{\mathbb Z}A',A\otimes_{\mathbb Z}M'\oplus A'\otimes_{\mathbb Z}M\oplus M\otimes_RM')$.

The object $(0,K/R)$ is flat (this uses that $K/R$ is a flat $\mathbb Z$-module), and the sequence

$0\to (0,R/fR) \to (0,K/fR)\to (0,K/R)\to 0$

does not remain exact when tensored with $(0,R/fR)$.

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Thanks! This answers 2). [The definition of abelian tensor categories is given in the nlab article which I have linked, for example] –  Martin Brandenburg Jun 23 '11 at 14:40
Hmm ... My example doesn't qualify because there is no unit object for $\otimes$. –  Tom Goodwillie Jun 23 '11 at 15:34
I didn't see your edit. This example works :). –  Martin Brandenburg Jul 8 '11 at 9:36