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Let $Q_1, Q_2$ be (connected) acyclic quivers and $I_1, I_2$ admissible ideals (in which the relations have only coefficients 1 or -1). Let $K$ and $F$ be two fields.

Question 1: Is $KQ_1/I_1$ derived equivalent to $K Q_2/I_2$ iff $FQ_1/I_1$ is derived equivalent to $F Q_2/I_2$? (is this true at least in case both $F$ and $K$ have characteristic different from 2 or even characteristic 0?)

That is, is being derived equivalent independet of the field? This is not true when the quiver are not acyclic.

Question 2: Is $KQ_1/I_1$ derived equivalent to its opposite algebra?

(it feels like I asked myself this question before, but I dont remember what the answer was)

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    $\begingroup$ How do you regard $I_1$ as an ideal of both $FQ_1$ and $KQ_1$? Do you want your admissible ideals to be defined over the integers in some way? $\endgroup$
    – Jeremy Rickard
    Commented Dec 29, 2019 at 14:55
  • $\begingroup$ @JeremyRickard Thanks, I added that the coefficients are only 1 or -1 so that they are defined over any field. (I never met acyclic quiver algebras in practise having relations that do not fullfill this or can be normalised in that way) $\endgroup$
    – Mare
    Commented Dec 29, 2019 at 15:00

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