Cancelling the canonical module in tensor products

Question

Let $R$ be Cohen-Macaulay local ring with the canonical module $\omega_R$ and let $M$ and $N$ be two finitely generated $R$-modules. Assume that
$$ \omega_R \otimes_R M= \omega_R \otimes_R N $$
Can we conclude that $M \cong N$ ?

Answer 1

Let $R=k[x,y]/(x^2,xy,y^2)$. For any non-zero linear form $l$, we have, $$R\stackrel{l}{\to} R\to M_l\to 0$$ modules and $M_l\cong M_m$ if and only if $l=am$ for some non-zero $a\in k$. (For definiteness, take $l=x, m=y$ and then $M_x\not\cong M_y$.)

Tensoring with $\omega_R$, we get an exact sequence, $$\omega_R\stackrel{l}{\to}\omega_R\to \omega_R\otimes M_l\to 0.$$ Multiplication by $l$ is non-zero, since $\omega_R$ is faithful, and the image is contained in the socle, since $lx=ly=0$. But $\omega_R$ has one dimensional socle, so the image is precisely the socle. Thus, $\omega_R\otimes M_l=\omega_R/\operatorname{soc}\omega_R$ and thus they are isomorphic independent of $l$. This gives an example where your question has a negative answer.