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Let $A$, $B$ be two DG algebras and $D(A)$, $D(B)$ be derived categories of DG-modules of $A$, $B$, respectively. We call $A$ and $B$ are dg Morita equivalent if there is an $A$-$B$ bimodule $T$ with the following properties

  1. $H^i(A)\cong \text{Hom}_{D(B)}(T,T[i])$ for any $i\in \mathbb{Z}$;
  2. $T$ defines a compact object in $D(B)$;
  3. For an object $N\in D(B)$,$\text{Hom}_{D(B)}(T,N[i])=0$ for any $i\in \mathbb{Z}$ implies $N=0$.

We know that $D(A)\simeq D(B)$ does not imply $A$ and $B$ are dg Morita equivalent. A counterexample could be found in On Differential Graded Categories Page 166.

However, we notice that the DG algebras in this counterexample have char$=p$. If we consider DG algebras $A$, $B$ over a field $k$ with char$k=0$, does $D(A)\simeq D(B)$ implies $A$ and $B$ are dg Morita equivalent? If not, is there any counterexample in char $0$?

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    $\begingroup$ Are you sure this is what you mean to ask? The link you give seems to be giving a counterexample to a rather different statement. There are plenty of examples, even for ordinary algebras (i.e., DG algebras concentrated in degree zero) where $A$ and $B$ are derived equivalent but not quasi-isomorphic. For example, path algebras of different orientations of a finite quiver without cycles. $\endgroup$
    – Jeremy Rickard
    Commented Jul 30, 2016 at 10:11
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    $\begingroup$ The way the question is posed, it suffices to consider the field $k$ and the matrix algebra $Mat_2(k)$. The abelian categories of $k$-modules/vector spaces and $Mat_2(k)$-modules are equivalent (by Morita), hence their derived categories are also equivalent. Still, the algebras $k$ and $Mat_2(k)$ are not isomorphic, hence not quasi-isomorphic when viewed as DG-algebras. $\endgroup$
    – Leonid Positselski
    Commented Jul 30, 2016 at 11:04
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    $\begingroup$ @JeremyRickard Thank you very much for pointing out the problem in my original question. What I want to ask is whether they are dg Morita equivalent instead of quasi-isomorphic. I have edited my question and maybe it makes more sense now. $\endgroup$
    – Zhaoting Wei
    Commented Jul 30, 2016 at 14:16
  • $\begingroup$ @LeonidPositselski Thank you very much for your comment. What I want to ask is whether they are dg Morita equivalent instead of quasi-isomorphic. I have edited my question and maybe it makes more sense now. $\endgroup$
    – Zhaoting Wei
    Commented Jul 30, 2016 at 14:18

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