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Call a finite dimensional (acyclic) quiver $K$-algebra A derived indecomposable in case $A$ is not derived equivalent to an algebra of the form $B \otimes_K C$.

For example when the number of simples or the vector space dimension of $A$ is a prime number $A$ is derived indecomposble. But $K D_4$ is derived equivalent to $K A_2 \otimes_K KA_2$ and thus not derived indecomposable. $KE_6$ is derived equivalent to $K A_2 \otimes_K K A_3$. $KE_8$ is derived equivalent to $K A_2 \otimes K A_4$.

Question 1: Is there an (easy) criterion when a given algebra is derived indecomposable?

Easy could mean that it can be decided by some discrete data that are computable for example using QPA.

Question 2: Is a derived equivalence like $K E_6$ to $K A_2 \otimes_K K A_3$ pure coincidence or is there more behind it? Can we also "build" $E_6$ up to some equivalences from $A_2$ and $A_3$ in other situations like Lie algebras?

Question 3: For which acyclic graphs $Q$ is the path algebra $KQ$ derived indecomposable?

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    $\begingroup$ In singularity theory, one can make the simple singularity E6 ($x^3+y^4$) from A2 ($x^3$) and A3 ($y^4$). This sum-with-disjoint-variables is named the Thom–Sebastiani sum. $\endgroup$
    – F. C.
    Aug 31, 2020 at 16:22
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    $\begingroup$ There is a necessary condition, that the Coxeter polynomial is a tensor product of two polynomials. This can be checked for instance on the set of roots. $\endgroup$
    – F. C.
    Sep 8, 2020 at 18:38

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