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Let $(R,\frak m)$ be a Cohen-Macaulay local ring. If there exists a maximal Cohen-Macaulay $R$-module $M$ with finite injective dimension, then can we conclude that $R$ has a canonical module?

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In short, the answer to your question is no.

Let us recall the following definition due to R. Sharp:

Definition Let $(R,\mathfrak{m},k)$ be a commutative noetherian local ring with $\text{dim}\,R=d$. A finitely generated $R$ module $G$ is said to be Gorenstein of rank (or type) $t$ if $$\mu_{i}(\mathfrak{m},G)= \begin{cases} 0 \mbox{ if }i \neq d \\ t \mbox{ if }i = d \end{cases}$$

Clearly any Gorenstein module is a maximal Cohen-Macaulay $R$-module, as its depth is $d$, and has finite injective dimension (this is clear from Proposition 3.1.14 in Bruns-Herzog). From its definition, the canonical module is just a Gorenstein module of rank 1.

Sharp showed that any local ring admitting a Gorenstein module is a Cohen-Macaulay ring (this is Corollary 3.9 in the paper Gorenstein Modules).

Consequently any local ring admitting a Gorenstein module of rank at least 2 provides a counterexample to your question. I believe the first example of such a ring was described by D. Weston in the paper On Descent in Dimension Two and Non-split Gorenstein Modules, and then examples of Gorenstein modules of arbitrary odd rank were given here.

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