I have a monoidal (not symmetric) triangulated category $(A,\otimes, 1)$ with unit 1.
Define $C$ the localizing subcategory of $A$ generated by the unit 1.


* is $(C, \otimes, 1) $ a symmetric monoidal triangulated category?
* Do we have a natural isomorphism $c\otimes a\cong a\otimes c$ for any $a\in A$ and any $c\in C$ ?



Monoidal triangulated categories
The answer to question 2, even with the assumption that $a\in C$, is no in general. It follows that the answer to question 1 is also no.


For a counterexample, one can consider for example a (derived) category of bimodules. Let $k$ be a base commutative ring, and $R$ a (say) flat $k$-algebra. Consider the derived category $A= D(R\otimes\_k R^{op})$ of $(R,R)$-bimodules over $k$. This is monoidal under $\otimes^\mathbb L\_R$, with unit $R$ viewed as an $R$-bimodule in the canonical way.


An explicit example is given by the polynomial ring $R= k[t]$. In this case, $R\otimes\_k R^{op} = k[t\_0,t\_1]$.


There is $k[t\_0,t\_1]$-linear extension $0\to k[t]\to k[t\_0,t\_1]/(t\_0-t\_1)^2\to k[t]\to 0$, where the rightmost map is the algebra map $t\_0,t\_1\mapsto t$, while the leftmost map is $P\mapsto (t\_0-t\_1)P(t\_0,t\_0)$. One easily checks that this is an extension: first, the rightmost map is clearly a bimodule map. For the leftmost map, one checks that $(t\_0-t\_1)t\_1P(t\_0,t\_0) = (t\_0-t\_1)t\_0P(t\_0,t\_0)$. Indeed it suffices to check it when $P = 1$ and then the relation is $t\_0t\_1-t\_1^2 = t\_0^2-t\_1t\_0$ which is a rephrasing of the relation $(t\_0-t\_1)^2 = 0$.


Second, suppose $P(t,t) = 0$. Then $P$ is divisible by $(t\_0-t\_1)$, to $P = (t\_0-t\_1)Q$ for some $Q$, and now $(t\_0-t\_1)Q = (t\_0-t\_1)Q(t\_0,t\_0)$ modulo $(t\_0-t\_1)^2$, because $Q-Q(t\_0,t\_0)$ is divisible by $(t\_0-t\_1)$. It follows that $P$ is in the image of the leftmost map.


This extension corresponds to a morphism $k[t]\to \Sigma k[t]$ in $A$, with fiber $E= k[t\_0,t\_1]/(t\_0-t\_1)^2$. In particular, $E$ belongs to $C$.


Another object in $C$ is $k$ where $t\_0,t\_1$ act trivially. Indeed, it is the mapping cone of $k[t]\xrightarrow{t}k[t]$ (this is a bimodule map, as $t$ is central).


I claim that $E\otimes^\mathbb L\_{k[t]}k \not\simeq k\otimes\_{k[t]}^\mathbb L E$.


To see this, we first observe that these derived tensor products are underived. The key lemma for this is:


*Lemma*: Suppose $(t\_0-t\_1)$ divides $t\_0 P$. Then $(t\_0-t\_1)$ divides $P$.
*Proof*: The assumption implies that $tP(t,t) = 0$ in $k[t]$. But $t$ is not a zero divisor, so $P(t,t) = 0$ and the conclusion follows.


Now we use this lemma twice : suppose $t\_0 P = 0$ in $E$. This means $(t\_0-t\_1)^2\mid t\_0P$ in $k[t\_0,t\_1]$. It follows that $(t\_0-t\_1)$ divides $t\_0 P$, and thus by the lemma, it divides $P$. Write $P= (t\_0-t\_1)Q$, and it follows easily that $(t\_0-t\_1) \mid t\_0 Q$, so that by the lemma, $(t\_0-t\_1)$ divides $Q$, and so $(t\_0-t\_1)^2$ divides $P$. So $P = 0$ in $E$.


By symmetry, the same holds for $t\_1$, and so the action of $t\_0$ and $t\_1$ on $E$ is injective, which proves the claim that the tensor products are underived.


Now, $E\otimes\_{k[t]}k$ is therefore $E/t\_1 E\cong k[t\_0,t\_1]/((t\_0-t\_1)^2, t\_1) = k[t\_0,t\_1]/(t\_0^2,t\_1)\cong k[t\_0]/(t\_0)^2$, on which $t\_0$ acts nontrivially while $t\_1$ acts trivially. Conversely, $k\otimes\_{k[t]}E\cong k[t\_1]/(t\_1^2)$, on which $t\_0$ acts trivially but $t\_1$ nontrivially. It follows that they are not isomorphic as bimodules, hence not isomorphic in the derived category either.


