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$\underline{Background}$ : Suppose $\tau$ is a preadditive category and $R$ a ring. Then one may form a new preadditive category $\tau \otimes R$ in the following way:

$\tau \otimes R$ has the same objects as $\tau$ and for objects $A$ and $B$, set $\tau \otimes R(A,B) := \tau(A,B)\otimes_{\mathbb{Z}} R$. Composition is given by $\tau(A,B)\otimes_{\mathbb{Z}} R\otimes_{\mathbb{Z}}\tau(B,C)\otimes_{\mathbb{Z}} R \cong \tau(A,B)\otimes_{\mathbb{Z}}\tau(B,C)\otimes_{\mathbb{Z}}R\otimes_{\mathbb{Z}}R \rightarrow \tau(A,C) \otimes_{\mathbb{Z}}R$, where the last arrow is given by the composition in $\tau$ and multiplication in $R$.

$\tau \otimes R$ is additve if $\tau$ is additive.

$\underline{Question}$ : Suppose $\tau$ is triangulated, is there a (canonical) structure of a triangulated category for $\tau \otimes R$?

Actually, I only need the case $R := \mathbb{Z}[\frac{1}{n}, \theta] \subset \mathbb{C}$, where $\theta$ is a primitive root of unity of order $n$ for some $n \in \mathbb{N}$, but I would also be interested in a general statement/counterexample to the general case.

Thank you for reading.

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up vote 8 down vote accepted

I would imagine it is false in general that given a triangulated category $T$ the category $T\otimes R$ is also triangulated.

The following is a concrete counterexample. Consider $D^b(\mathbb{Z})$ and let $T = D^b(\mathbb{Z})\otimes \mathbb{Z}[x]$. In order for $T$ to be triangulated the morphism $\mathbb{Z} \stackrel{x}{\to} \mathbb{Z}$ would need to be able to be completed to a distinguished triangle $$\mathbb{Z}\stackrel{x}{\to}\mathbb{Z}\to C_x \to \Sigma\mathbb{Z} $$ in $T$. Suppose that such a triangle existed and consider the exact sequence obtained by application of $Hom_T(\mathbb{Z},-)$ $$0 \to Hom_T(\mathbb{Z}, \Sigma^{-1}C_x) \to \mathbb{Z}[x] \stackrel{x}{\to} \mathbb{Z}[x] \to Hom_T(\mathbb{Z}, C_x) \to 0 $$ As multiplication by $x$ has no kernel we have $$Hom_T(\mathbb{Z}, \Sigma^{-1}C_x) = 0$$ and the object $C_x$ would have to satisfy $$\mathbb{Z} \cong Hom_T(\mathbb{Z},C_x) = Hom_{D^b(\mathbb{Z})}(\mathbb{Z}, C_x)\otimes_\mathbb{Z} \mathbb{Z}[x] $$ But decomposing $C_x$ as a sum of suspensions of free groups and torsion groups this is clearly not possible.

One situation in which something like what you ask for works is the following: let $(T,\square,\mathbf{1})$ be a tensor triangulated category (that is we have a symmetric monoidal structure on $T$ which is exact in each variable). Given a multiplicative subset of even elements $S$ of $End^*(\mathbf{1})$ one can localize the hom-sets of $T$ at $S$. This can even be realised as a Verdier quotient by a certain subcategory. This is shown in Balmer's "Spectra, spectra, spectra - Tensor triangular spectra versus Zariski spectra of endomorphism rings" section 3.

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This is a neat counterexample, thanks. The only thing I do not understand is why $C_x$ is decomposable in a free and a torsion part. But one could also argue that the assumption that T is triangulated leads to the contradiction that Z decomposes into a nontrivial direct sum of abelian groups. The situation you mention in the end handles Z[1/n]. I think this even works without the assumption that T is tensor triangulated - the morphisms which become isomorphisms after multiplication with some power of n form a system which arises from a cohomological functor (Weibel, 10.4) – Manuel Koehler Jun 2 '10 at 10:47
I should have made it clear that by $D^b(\mathbb{Z})$ I meant the bounded derived category of finitely generated abelian groups. So in fact one can split $C_x$ as a finite sum of shifts of $\mathbb{Z}$ and $\mathbb{Z}/p^n\mathbb{Z}$ for various primes. This follows from the structure theorem for finitely generated abelian groups together with the fact that $\mathbb{Z}$ has global dimension 1 which can be used to show that complexes are quasi-isomorphic to the direct sum of their cohomology groups. – Greg Stevenson Jun 2 '10 at 23:37

It seems unreasonable to expect tensoring the morphisms in a triangulated category by a ring to be a meaningful operation - after all these morphisms are typically cohomology groups of natural complexes, and we know there are universal-coefficient-theorem type corrections to tensoring cohomology with a ring as opposed to the more natural operation of tensoring a complex.

On the other hand, if we enrich our triangulated categories everything works fine. For example on the level of (pretriangulated) differential graded categories or $A_\infty$ categories, or if you prefer, stable $\infty$-categories, there's a perfectly nice operation of tensor product by a ring - in fact one can tensor by another such category (generalizing the case you're asking about, which is tensoring with R-mod). Even more, there's a structure of symmetric monoidal $\infty$-category on such categories (developed in Lurie's Derived Algebraic Geometry series (an exposition and references can be found eg here).

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Thanks for the reference, I will look into that. – Manuel Koehler Jun 2 '10 at 10:49
Manuel Koehler's example is flat, so he might not need to derive it. Of course, one still need to add cones of new morphisms, as in Greg Stevenson's example. – Ben Wieland Jun 2 '10 at 17:38
Fair enough, I can certainly believe in some contexts one can get away with triangulated constructions, but if it doesn't come easily it seems like an unnatural thing to try to push through. – David Ben-Zvi Jun 2 '10 at 23:35

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