Let $K$ be a field. Suppose $A$ and $B$ are $K$-algebra and there is a derived equivalence $F:D^b(A)\cong D^b(B)$ between their bounded derived categories. If we assume that $A$ has a tensor decomposition $A\cong A\_1\otimes\_K A\_2$ for two $K$-algebras $A\_1$ and $A\_2$.


Does $B$ also have a tensor decomposition $B\cong B\_1\otimes\_K B\_2$ such that $D^b(B\_1)\cong D^b(A\_1)$ and $D^b(B\_2)\cong D^b(A\_2)$.



Tensor decomposition under derived equivalence
Not in general.


For an easy example, let $C$ and $D$ be derived equivalent algebras. Then $A=C\times C$ and $B=C\times D$ are derived equivalent, and $A=C\otimes\_K(K\times K)$ has a tensor decomposition, but $B$ will usually not.


There are also connected examples. For example, $KA\_2\otimes\_K KA\_2$ is derived equivalent to $KD\_4$ (where $KA\_2$ and $KD\_4$ are path algebras of quivers of the indicated Dynkin type).


There are more examples in [this question](https://mathoverflow.net/questions/370535/when-is-an-algebra-derived-indecomposable). But also, examples are very easy to construct. If you take an algebra of the form $A=A\_1\otimes\_KA\_2$, then it will usually have many tilting complexes, and unless you choose one of the form $T\_1\otimes\_KT\_2$, there is no particular reason to expect its endomorphism algebra (which is derived equivalent to $A$) to have a tensor decomposition.


