The first Tachikawa conjecture states that for a finite dimensional algebra $A$, $Ext_A^i(D(A),A)=0$ for all $i \geq 1$ implies that $A$ is selfinjective.

Question: In case $A$ has the property that $Ext_A^i(D(A),A)=0$ for all $i \geq 1$ and $B$ is derived equivalent to $A$, can we conclude that also $Ext_B^i(D(B),B)=0$ for all $i \geq 1$?

In case this is true, the first Tachikawa conjecture would imply that selfinjective algebras are closed under derived equivalences. I think this is only known for algebras over algebraically closed fields at the moment? Is it true for general abelian categories with injectives=projectives?.

Note that $Ext_A^i(D(A),A) \cong Ext_{A^e}^i(A,A^e)$ when $A^e$ is the enveloping algebra of $A$. A derived equivalence $F$ between $A$ and $B$ induces a derived equivalence between $A^e$ and $B^e$ that sends $A$ to $B$ and $D(A)$ to $D(B)$, so maybe this can be used to prove the question somehow?

  • 4
    $\begingroup$ There's a proof that selfinjective algebras are closed under derived equivalences in a paper of mine and Rouquier's that I don't think needs an algebraically closed field (Corollary 3.12 in doi.org/10.1016/j.jalgebra.2016.05.018). $\endgroup$ Oct 15, 2019 at 12:39
  • $\begingroup$ @JeremyRickard Is it also known for general Artin algebras (or abelian categories where injectives=projectives)? $\endgroup$
    – Mare
    Oct 17, 2019 at 11:16
  • 2
    $\begingroup$ I would guess that the proof would generalize to Artin algebras, although I haven't checked. I don't know about more general abelian categories. $\endgroup$ Oct 17, 2019 at 11:55
  • 1
    $\begingroup$ In the category of graded modules for $k[x]/(x^2)$, where $x$ has degree one, injectives$=$projectives, and I think I see how to produce a derived equivalence with a category where injectives$\neq$projectives. $\endgroup$ Oct 17, 2019 at 12:18
  • $\begingroup$ @JeremyRickard Interesting, I added the question to this thread question. So if you want you might post it as an answer. $\endgroup$
    – Mare
    Oct 17, 2019 at 12:38

1 Answer 1


This is not an answer to the main question, but to the subsidiary question of whether abelian categories with projectives coinciding with injectives are closed under derived equivalence. The answer is no.

The categories I'll describe also have enough projectives and injectives, whch was probably meant to be assumed.

Let $k$ be a field, and $\mathcal{A}$ the category of cochain complexes of $k$-vector spaces, and let $P_i$ be the object that is $\dots\to0\to k\stackrel{\sim}{\to}k\to0\to\dots$ with the two nonzero terms in degrees $i$ and ${i+1}$. Then $P_i$ is projective and injective, and every projective or injective object is a direct sum of copies of the $P_i$ (i.e., a contractible complex). Note that the only nonzero maps between the objects $P_i$, up to multiplication by a scalar, are the identity maps and maps $\require{AMScd}$ \begin{CD} \cdots@>>>P_2@>>>P_1@>>>P_0@>>>P_{-1}@>>>P_{-2}@>>>\cdots \end{CD} where the composition of two such maps is zero.

Let $$X_i=\begin{cases} 0\to P_i&\text{ if }i<0\\ P_1\to P_0&\text{ if }i=0\\ P_i\to0&\text{ if }i>0, \end{cases}$$ considered as complexes over the category $\mathcal{A}$ with the displayed terms in degrees $0$ and $1$, and zero in all other degrees.

Let $\mathcal{D}(\mathcal{A})$ be the derived category of $\mathcal{A}$. It is easy to check that $\text{Hom}_{\mathcal{D}(\mathcal{A})}(X_i,X_j[t])=0$ whenever $t\neq0$, and that the objects $X_i$ generated the same subcategory of $\mathcal{D}(\mathcal{A})$ as the objects $P_i$. It follows from the Morita theory of derived categories ("Deriving DG categories" by Bernhard Keller contains a sufficiently general formulation) that there is another abelian category $\mathcal{B}$ and an equivalence of derived categories $\mathcal{D}(\mathcal{A})\approx\mathcal{D}(\mathcal{B})$ sending the objects $X_i$ to a set of projective generators of $\mathcal{B}$. Let $Q_i$ be the object that is the image of $X_i$ in $\mathcal{D}(\mathcal{B})$.

The only nonzero maps between the $Q_i$ (or equivalently between the $X_i$), up to multiplication by a scalar, are the identity maps and maps

\begin{CD} \cdots@>>>Q_3@>>>Q_2@>>>Q_0@>>>Q_{-1}@>>>Q_{-2}@>>>\cdots\\ @.@.@.@VVV\\ @.@.@.Q_1 \end{CD} where the composition of two such maps is zero, except for the composition $Q_2\to Q_0\to Q_1$.

The map $Q_0\to Q_1$ is a monomorphism in $\mathcal{B}$ since $\text{Hom}_\mathcal{B}(Q_i,Q_0)\to\text{Hom}_\mathcal{B}(Q_i,Q_1)$ is injective for every $i$, and the $Q_i$ are a set of projective generators. However, it is not split, so the object $Q_0$ is not an injective object of $\mathcal{B}$. Hence $\mathcal{B}$ has noninjective projectives.

  • $\begingroup$ Thanks. Do we have $Ext^i(I,P)=0$ for all $i>0$ for any injective object I and projective object P in $\mathcal{B}$? I think not, so the main question should not hold in a more general setting. $\endgroup$
    – Mare
    Oct 26, 2019 at 13:23
  • $\begingroup$ @Mare I agree, there are two indecomposable non-projective injectives and two indecomposable non-injective projectives, and two cases where $\text{Ext}^1(I,P)\neq0$. $\endgroup$ Oct 26, 2019 at 18:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .