Finite injective dimension 
Let $A$ be a commutative noetherian ring. Is it true that if $A$ is regular then any module over it has a finite injective dimension? What if $A$ is Gorenstein?

Any reference who discuss this?
 A: If $A$ is a regular non-local ring of infinite Krull dimension, then there exists an $A$-module whose injective dimension is not finite. (See my answer to this question and change everywhere "projective" by "injective".)
If $A$ is a regular (local or not) ring of finite Krull dimension, then every $A$-module has finite injective dimension. Let $M$ be an $A$-module. Then $\operatorname{inj\ dim}_AM=\sup_{m\in\operatorname{Max}(A)}\operatorname{inj\ dim}_{A_m}M_m$. Since $A_m$ is a regular local ring, its global dimension equals $\dim A_m$, and thus $\operatorname{inj\ dim}_{A_m}M_m\le\dim A_m$. Since $\sup_{m\in\operatorname{Max}(A)}\dim A_m=\dim A$ we get $\operatorname{inj\ dim}_AM\le\dim A$.
If $(A,m)$ is a Gorenstein non-regular local ring, then the $A$-module $A/m$ has infinite injective dimension as @zcn has pointed out. (I have included this part in order to have a complete picture.) 
A: The global dimension of a ring is both the sup of projective dimensions and the sup of injective dimensions (both are measured by vanishing of $\text{Ext}$), so if $A$ is regular (and local), then every module has finite injective dimension. In fact, if $A$ is local with residue field $k$, then $\text{gldim } A = \text{injdim } k$, so if $A$ is (Gorenstein but) not regular, then $k$ has infinite injective dimension.
