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.



Whether Morita equivalence holds the following properties?
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.


