The natural inclusion $R\varphi\to\mathrm{Hom}_R(C,C')$ induces the above isomorphism modulo $x$.

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This is a part in proof of Theorem 3.3.4 in Cohen-Macaulay Rings, Bruns and Herzog. I don't understand it. Here is the Theorem:

Theorem 3.3.4. Let $(R,m,k)$ be a Cohen-Macaulay local ring and let $C$ and $C'$ be canonical modules of $R$ Then

b. the canonical modules $C$ and $C'$ are isomorphic,

c. $\mathrm{Hom}_R(C,C')\cong R$ and any generator $\varphi$ of $\mathrm{Hom}_R(C,C')$ is an isomorphism.

In proof (b) and (c) we have

$$\mathrm{Hom}_R(C,C')\otimes_RR/(x)\cong \mathrm{Hom}_{R/(x)}(C/xC,C'/xC')\cong R/(x)\qquad(*)$$

and so $\mathrm{Hom}_R(C,C')$ is cyclic by Nakayama's lemma. Let $\varphi$ be a generator of this module. Then the natural inclusion $R\varphi\to \mathrm{Hom}_R(C,C')$ induces the above isomorphism modulo $x$.

I want to use the isomorphism $(*)$ to prove $R\varphi\to \mathrm{Hom}_R(C,C')$ is isomorphism. Then, I want to prove $R\varphi\otimes R/(x)\cong R/(x)$.

I have not proved it yet. Can you help me please?