Why is $M$ torsion-free?

Question

I am studying the following article

https://www.math.nagoya-u.ac.jp/~takahashi/tc9.pdf

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

1. 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.

2. 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?

Answer 1

1. The authors reduce to the case of $R$ complete with infinite residue field and use them implicitly at a couple of places in the proof. This is a fairly standard practice. For instance, to assert that $I$ has a principal reduction $(r)$ we need the residue field to be infinite. As for completeness, I think it was because they used the normalization $\overline R$ and needs it to be finite.

2. Cohen-Macaulayness implies torsion-freeness. You can continue with your proof. Such $s$ must be in one of associated primes $P$ of $M$, and since $s$ is nzd $P$ must have height at least one. But as $P$ is associated depth of $M_P$ is only $0$, violating Cohen-Macaulayness.

Answer 2

2.Every maximal Cohen-Macaulay module is torsion-free. Here a proof attempt for $R$ a local CM ring with $K=R/m$, where $m$ is the maximal ideal of $R$.

Lemma: A noetherian module M is torsion-free if and only if $Hom_R(K,M)=0$.

(proof: The maps in $Hom_R(R/m,M)$ correspond to left multiplications $L_z$ with $z \in Hom_R(R,M)=M$ such that $z m=0$. )

Now assume $M$ is MCM, then $depth(M) \geq 1$ and thus $Hom_R(K,M)=0$, thus $M$ is torsionfree.

(1.I have not read the article, but assuming completeness often garantees the existence of a canonical module which is needed in some arguments.)