\[lem0701’iii\] ${\operatorname{Gid}}_R(C\otimes_RM)<\infty$ and $M\in{{\mathcal{A}}_C}(R)$.

When these conditions are satisfied, we have $$\begin{aligned}
 {{\mathcal{G}}}({{\mathcal{I}}_C})\text{-}{\operatorname{id}}_R(M)
  &
  ={{\mathcal{G}}}{{\mathcal{I}}_C}\text{-}{\operatorname{id}}_R(M)
  ={\operatorname{Gid}}_R(C\otimes_RM).  \end{aligned}$$

We actually need the following corollary of the previous result.

\[lem110613a\] Let $C$ be a semidualizing $R$-module. For an $R$-module $M$, the following conditions are equivalent:

1.  \[lem110613ai\] $M$ has locally finite ${{\mathcal{G}}}({{\mathcal{I}}_C})\text{-}{\operatorname{id}}$;

2.  \[lem110613aii\] $M$ has locally finite ${{\mathcal{G}}}{{\mathcal{I}}_C}\text{-}{\operatorname{id}}$ and $M\in{{\mathcal{A}}_C}(R)$; and

3.  \[lem110613aiii\] ${C\otimes_{R}M}$ has locally finite ${\operatorname{Gid}}$ and $M\in{{\mathcal{A}}_C}(R)$.

This follows from the definitions of the locally finite Gorenstein injective dimensions and the local-global principal for Auslander classes from Fact \[disc01xy\](b).

\[proof110609c\] Assume that $R$ is Cohen-Macaulay with a pointwise dualizing module $\omega$ and that $M$ is a non-zero finitely generated $R$-module in ${{\mathcal{A}}_C}(R)$ with locally finite ${{\mathcal{G}}}{{{\mathcal{I}}_C}\text{-}{\operatorname{id}}}$. Lemma \[lem110613a\] implies that ${C\otimes_{R}M}$ has locally finite G-injective dimension. Since $M\neq 0$, we have ${C\otimes_{R}M}\neq 0$, by Fact \[fact110527a\](b), so Theorem \[thm110608a\] provides an integer $r = {\mathrm{G}\text{-}\!\dim}_R({\operatorname{Hom}_{R}(\omega,{C\otimes_{R}M})})$, a proper $R$-regular sequence $\underline z = z_1, \ldots , z_r$, non-negative integers $n_0, n_1, \ldots , n_r$ with $n_r{\geqslant}1$, and an exact sequence $$\label{proof110609c1}
0 \to {C\otimes_{R}M} \to Z' \to N' \to 0$$ such that $Z' = \oplus_{i=0}^r(\omega/(\underline z_{[i]})\omega)^{n_i}$, $N'\in{{\mathcal{B}}}_{\omega}(R)$, and ${\mathrm{G}\text{-}\!\dim}_R({\operatorname{Hom}_{R}(\omega,N')}){\leqslant}r$.

Since $M$ is in ${{\mathcal{A}}_C}(R)$, we have ${\operatorname{Ext}_{R}^{1}(C,{C\otimes_{R}M})}=0$ and ${\operatorname{Hom}_{R}(C,{C\otimes_{R}M})}\cong 