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Let $A,B$ be two K-algebras over a field K.

  1. $A$ and $B$ are said to be $Morita $ $equivalent$ if the category $Mod A$ and $Mod B$ are equivalent.
  2. $A$ and $B$ are said to be $derived$ $equivalent$ if $\mathcal{D}^b(Mod A)$ and $\mathcal{D}^b(Mod B)$ are equivalent as triangulated categories.
  3. Given a minimal injective resolution of $A$ as an $A$-module$$0 \rightarrow A \rightarrow I_0 \rightarrow I_1 \rightarrow \dots$$ If n is maximal with the property that all modules $I_j$ are projective for $j<n$, then n is called the $dominant$ $dimension$ of $A$.

The following are my questions：

1) Is there any relation between $Morita$ $equivalent$ and $derived$ $equivalent$? (I think $A$ and $B$ are $Morita$ $equivalent$ can induces $A$ and $B$ are $derived$ $equivalent$. Is it right? Conversely, if $A$ and $B$ are $derived$ $equivalent$, when $A$ and $B$ are $Morita$ $equivalent$?)

2) If $A$ and $B$ are $Morita$ $equivalent$, someone tells me that the dominant dimensions of $A$ and $B$ are also equal. But I don't know the reason. Who can tell me? Thank you for any help.

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1) Yes, Morita equivalence trivially implies derived equivalence. Note that two algebras over an algebraically closed field are Morita equivalent iff their quiver algebras are isomorphic. So compared to derived equivalence, being Morita equivalent is rather easy to check and can be reduced to calculating quivers and checking isomorphism in the algebraically closed case. For which classes of algebras a derived equivalence implies a Morita equivalence seems to be unknown. It is true for examples when one restrics to local algebras, since they are derived equivalent iff they are morita equivalent. Of course it is not true in general, since hereditary algebras have many algebras (tilted algebras) of global dimension two as derived equivalent algebras.

2) This is because a Morita equivalence maps projectives to projectives and projective-injective modules to projective-injective modules and of course is exact. So apply the equivalence to a minimal injective resolution of the regular module to get the result.

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  • $\begingroup$ How easy is it really to compute a quiver presentation from say structure constants for the algebra? There are very few monoid algebras for which I know how to do it. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 22, 2016 at 13:12
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    $\begingroup$ It is not really easy, I should have said "easy, compared to check derived equivalence". $\endgroup$
    – Mare
    Commented Oct 22, 2016 at 13:13
  • $\begingroup$ @Mare For question 2), I know that Morita equivalence holds these properties. But given a minimal injective resolution of $A$, can you make sure the equivalence maps $A$ to $B$? $\endgroup$
    – Xiaosong Peng
    Commented Oct 22, 2016 at 13:16
  • $\begingroup$ no, but to get the necessary information you just need a resolution of a module which contains every indecomposable projective module (and just such modules) as a direct summand and this is the case. $\endgroup$
    – Mare
    Commented Oct 22, 2016 at 13:18
  • $\begingroup$ @Mare Hello, sir. I don't know this result " two algebras over an algebraically closed field are Morita equivalent iff their quiver algebras are isomorphic", could you tell me where to find it? $\endgroup$
    – Xiaosong Peng
    Commented Oct 22, 2016 at 13:26

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