Let $A$ be a finite dimensional algebra over a field K. Let $M$ be an $A$-module and $0 \rightarrow I^0 \rightarrow I^1 \rightarrow \cdots$ be a minimal injective resolution of $M$. The dominant dimension of $M$, denoted by domdim$(M)$, is the largest number $t$ or $\infty$ such that $I^0, I^1, \ldots, I^{t-1}$ are projective. If $M=A$, we get the dominant dimension of $A$, domdim$(A)$.


1. I want to know if there are any relation between domdim$(A)$ and domdim$(M)$ for any $M \in mod A$. (For example, if $A$ is selfinjective, then domdim$(A)$=domdim$(M)=\infty$)
2. We know there are algebras such that domdim$(A) \geq 2$. I want to know if there are algebras such that dom.dim$(M) \geq 2$ for any $M \in modA$. (I know selfinjetive algebras satisfy the condition, is there other examples?)



Questions about dominant dimension
1.There is no such relation in general. For specific algebras, such as non-semisimple higher auslander algebras, the dominant dimension of noninjective modules is always bounded by domdim(A).


2. Let $A$ be not selfinjective. Then there is a injective module which is not projective. this module has always dominant dimension zero as have all of its submodules. Interesting subcategories C, have the proeprty that domdim(M) >=2 for every M in C in case domdim(A) >=2, for example the subcategory C of gorenstein projective modules.


