Some questions I can't understand.

Question

Let $A$ be a finite-dimensional k-algebra, where k is a fixed field. For an arbitrary $A$-module $_AX$, we denote by proj.dim$_AX$(resp. inj.dim$_AX$) the projective dimension (resp. the injective dimension) of the module $_AX$. We denote by $_A\mathcal{P}$ (resp. $_A \mathcal{I}$) the full subcategory of mod $A$ formed by the projective (resp. injective) $A$-modules. $D=Hom_k(-,k)$.

There are some questions confused me:

1. I have seen in a lemma that"suppose proj.dim$_AD(A_A)=r < \infty$. If inj.dim$_AA < \infty$, then proj.dim$_AD(A_A)$=inj.dim$_AA$." So I want to know if there is an algebra $A$ such that proj.dim$_AD(A_A)=r < \infty$ while inj.dim$_AA = \infty$?

2. How to prove that "proj.dim$_AD(A_A) < \infty \Leftrightarrow K^b(_A\mathcal{I}) \subseteq K^b(_A \mathcal{P})$"? (Here $K^b(_A\mathcal{I})$ is the category of bounded homotopy complex consisting of injective modules and similarly $K^b(_A\mathcal{P})$).

3. I have seen that $K^b(_A\mathcal{I})$ and $K^b(_A\mathcal{P})$ can be natural embedded into $D^b(A)$, where $D^b(A)$ is the derived category of bounded complexes over mod$A$, when $A$ is a selfinjective algebra or Gorenstein algebra. I know $D^b(S) \cong K^{-,b}(_A \mathcal{P})$ or $D^b(S) \cong K^{+,b}(_A \mathcal{I})$, so how could $K^b(_A\mathcal{I})$ and $K^b(_A\mathcal{P})$ be natural embedded into $D^b(A)$?

Answer 1

1. There is no known example. This is known as the Gorenstein symmetry conjecture. Usually it is stated as "$\operatorname{inj.dim}A_A<\infty$ if and only if $\operatorname{inj.dim}{}_AA<\infty$". As $D$ is a duality, $\operatorname{inj.dim} {}_AA=\operatorname{proj.dim}D(A_A)$.
2. This is more or less obvious once you write it out. $\operatorname{proj.dim} D(A_A)<\infty$ means that each injective module has a projective resolution of finite length. Thus, you can write $K^b({}_A\mathcal{I})$ alternatively as extensions of these projective resolutions and their shifts. This naturally embedds into $K^b({}_A\mathcal{P})$. On the other hand, $\operatorname{proj.dim} D(A_A)=\infty$ implies that there is an injective module of infinite projective dimension, which is therefore not in $K^b({}_A\mathcal{P})$.
3. If you already know that $D^b(A)\cong K^{-,b}({}_A\mathcal{P})$, then it should be clear how to embed it, just compose it with the obvious inclusion $K^b({}_A\mathcal{P})\subseteq K^{-,b}({}_A\mathcal{P})$.