$\underline{Background}$ : 
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.



Is the tensorproduct of a triangulated category with a ring again triangulated?
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.


