${{{\mathcal{P}}_C}\text{-}{\operatorname{pd}}}_R(M)={\operatorname{pd}}_R({\operatorname{Hom}_{R}(C,M)})$ and ${\operatorname{pd}}_R(M)={{{\mathcal{P}}_C}\text{-}{\operatorname{pd}}}_R({C\otimes_{R}M})$ by [@takahashi:hasm Theorem 2.11]. In particular, one has ${{{\mathcal{P}}_C}\text{-}{\operatorname{pd}}}_R(M)<\infty$ if and only if ${\operatorname{pd}}_R({\operatorname{Hom}_{R}(C,M)})<\infty$, and one has ${\operatorname{pd}}_R(M)<\infty$ if and only if ${{{\mathcal{P}}_C}\text{-}{\operatorname{pd}}}_R({C\otimes_{R}M})<\infty$.

\[proof110530a\] Assume that $M$ is a non-zero finitely generated $R$-module such that ${{{\mathcal{P}}_C}\text{-}{\operatorname{pd}}}_R(M) = r < \infty$. Then Fact \[disc01xyz\] implies that ${\operatorname{pd}}_R({\operatorname{Hom}_{R}(C,M)})=r<\infty$. Since $M\neq 0$, we have ${\operatorname{Hom}_{R}(C,M)}\neq 0$, by Fact \[fact110527a\](b), so Theorem \[thm110527a\] provides an $R$-regular sequence $\underline z = z_1, \ldots, z_r$, integers $n_0, n_1, \ldots , n_r  {\geqslant}0$ with $n_r{\geqslant}1$, and an exact sequence $$\label{eq110530a}
0 \to {\operatorname{Hom}_{R}(C,M)} \to Z' \to N' \to 0$$ with $Z' = \oplus_{i=0}^r(R/(\underline z_{[i]}))^{n_i}$ and ${\operatorname{pd}}_R(N') {\leqslant}r$. In particular, we have $N'\in{{\mathcal{A}}_C}(R)$ which implies that ${\operatorname{Tor}^{R}_{1}(C,N')}=0$. Thus, an application of ${C\otimes_{R}-}$ to the sequence  yields the next exact sequence: $$\label{eq110530b}
0 \to {C\otimes_{R}{\operatorname{Hom}_{R}(C,M)}} \to {C\otimes_{R}Z'} \to {C\otimes_{R}N'} \to 0.$$ Fact \[disc01xyz\] implies that $M\in{{\mathcal{B}}_C}(R)$, so we have ${C\otimes_{R}{\operatorname{Hom}_{R}(C,M)}}\cong M$. The equality $Z' = \oplus_{i=0}^r(R/(\underline z_{[i]}))^{n_i}$ implies that $Z:={C\otimes_{R}Z'} = \oplus_{i=0}^r(C/(\underline z_{[i]})C)^{n_i}$. Because of Fact \[disc01xyz\], the condition ${\operatorname{pd}}_R(N') {\leqslant}r$ implies that ${{{\mathcal{P}}_C}\text{-}{\operatorname{pd}}}_R({C\otimes_{R}N'}){\leqslant}r$. Thus, with $N:={C\otimes_{R}N'}$, the sequence  satisfies the conclusion of Theorem \[thm110527b\].

The remainder of the proof of Theorem \[thm110527b\] is similar to the above proof, but it requires a bit more technology.

