${{{\mathcal{I}}_C}\text{-}{\operatorname{id}}}_R({\operatorname{Hom}_{R}(C,M)})<\infty$. It follows that $M$ has locally finite ${{{\mathcal{I}}_C}\text{-}{\operatorname{id}}}$ if and only if ${C\otimes_{R}M}$ has locally finite injective dimension, and $M$ has locally finite injective dimension if and only if ${\operatorname{Hom}_{R}(C,M)}$ has locally finite ${{{\mathcal{I}}_C}\text{-}{\operatorname{id}}}$.

\[lem110612b\] Assume that $R$ is Cohen-Macaulay with a pointwise dualizing module $\omega$. Let $C$ be a semidualizing $R$-module, and let $x=x_1,\ldots,x_i\in R$ be an $R$-regular sequence. Then ${\operatorname{Hom}_{R}(C,\omega/(\underline x)\omega)}\cong{C^{\dagger}}/(\underline x){C^{\dagger}}$ where ${C^{\dagger}}={\operatorname{Hom}_{R}(C,\omega)}$.

Let $K$ be the Koszul complex $K^R(\underline x)$, and let $K^+$ denote the augmented Koszul complex $$K^+=(0\to R\to\cdots\to R\to \underbrace{R/(\underline x)}_{\text{deg. $-1$}}\to 0).$$ Since $x$ is $R$-regular, $K^+$ is an exact sequence of $R$-modules of finite projective dimension. Thus, each module $K^+_j$ is in ${{\mathcal{A}}}_{\omega}(R)$, so the induced complex $${\omega\otimes_{R}K^+}=(0\to \omega\to\cdots\to \omega\to \underbrace{\omega/(\underline x)\omega}_{\text{deg. $-1$}}\to 0)$$ is an exact sequence of $R$-modules locally of finite injective dimension. In particular, the modules in this exact sequence are in ${{\mathcal{B}}_C}(R)$, so the next sequence is also exact: $$\begin{aligned}
{\operatorname{Hom}_{R}(C,{\omega\otimes_{R}K^+})}\hspace{-2cm}\\
&=(0\to {\operatorname{Hom}_{R}(C,\omega)}\to\cdots\to {\operatorname{Hom}_{R}(C,\omega)}\to \underbrace{{\operatorname{Hom}_{R}(C,\omega/(\underline x)\omega)}}_{\text{deg. $-1$}}\to 0).\end{aligned}$$ It is straightforward to show that the differential on this sequence in positive degrees is the same as the differential on ${{C^{\dagger}}\otimes_{R}K}$. It follows that $${\operatorname{Hom}_{R}(C,\omega/(\underline x)\omega)}
\cong{\operatorname{H}}_0({{C^{\dagger}}\otimes_{R}K})
\cong{C^{\dagger}}/(\underline x){C^{\dagger}}$$ as desired.

\[proof110609b\] Assume that $R$ is Cohen-Macaulay with 