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Why $M$ is Torsion$M$ torsion-free?

I am studying the following article

The main theorem is the Theorem 3.3. Howewer, I have the following questions about yourthe proof:

Why How does it help to assume that $R$ is complete ring? I thought it is to conclude that R$R$ is isomorphic to a quotient of a formal series ring (Cohen's theorem) but I don't know how that would help the proof.
Why is $M$ is torsion-free? (A definition of torsion free is: $M$ is torsion-free if the natural map $M\to M\otimes_R S^{-1}R$ is injetiveinjective (or equivanlentlyequivalently the natural map $M\to S^{-1}M$ is injetiveinjective) where $S=\{\mbox{nondivizores of zero}\}$$S=\{\mbox{nondivizors of zero}\}$.

If $s \in S,m \in M$ and $s.m=0$, I have to show that $m=0$. Howewer I don't see how to conclude that $m=0$. With the assumptions of the proof and hipoteseshypotheses of theorem, I know that $M$ is Maximal Cohen Macaulay of rank one and $R$ Cohen-Macaulay complete local ring. Any suggestions?

I am studying the following article

The main theorem is the Theorem 3.3. Howewer, I have the following questions about your proof:

Why How does it help to assume that $R$ is complete ring? I thought it is to conclude that R is isomorphic to a quotient of a formal series ring (Cohen's theorem) but I don't know how that would help the proof.
Why $M$ is torsion-free? (A definition of torsion free is: $M$ is torsion-free if the natural map $M\to M\otimes_R S^{-1}R$ is injetive (or equivanlently the natural map $M\to S^{-1}M$ is injetive) where $S=\{\mbox{nondivizores of zero}\}$.

If $s \in S,m \in M$ and $s.m=0$, I have to show that $m=0$. Howewer I don't see how to conclude that $m=0$. With the assumptions of the proof and hipoteses of theorem, I know that $M$ is Maximal Cohen Macaulay of rank one and $R$ Cohen-Macaulay complete local ring. Any suggestions?

I am studying the following article

The main theorem is the Theorem 3.3. Howewer, I have the following questions about the proof:

How does it help to assume that $R$ is complete ring? I thought it is to conclude that $R$ is isomorphic to a quotient of a formal series ring (Cohen's theorem) but I don't know how that would help the proof.
Why is $M$ torsion-free? (A definition of torsion free is: $M$ is torsion-free if the natural map $M\to M\otimes_R S^{-1}R$ is injective (or equivalently the natural map $M\to S^{-1}M$ is injective) where $S=\{\mbox{nondivizors of zero}\}$.

If $s \in S,m \in M$ and $s.m=0$, I have to show that $m=0$. Howewer I don't see how to conclude that $m=0$. With the assumptions of the proof and hypotheses of theorem, I know that $M$ is Maximal Cohen Macaulay of rank one and $R$ Cohen-Macaulay complete local ring. Any suggestions?

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asked Sep 16, 2021 at 15:27

I am studying the following article

The main theorem is the Theorem 3.3. Howewer, I have the following questions about your proof:

Why How does it help to assume that $R$ is complete ring? I thought it is to conclude that R is isomorphic to a quotient of a formal series ring (Cohen's theorem) but I don't know how that would help the proof.
Why $M$ is torsion-free? (A definition of torsion free is: $M$ is torsion-free if the natural map $M\to M\otimes_R S^{-1}R$ is injetive (or equivanlently the natural map $M\to S^{-1}M$ is injetive) where $S=\{\mbox{nondivizores of zero}\}$.

If $s \in S,m \in M$ and $s.m=0$, I have to show that $m=0$. Howewer I don't see how to conclude that $m=0$. With the assumptions of the proof and hipoteses of theorem, I know that $M$ is Maximal Cohen Macaulay of rank one and $R$ Cohen-Macaulay complete local ring. Any suggestions?