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This set of questions pertains to the proof of Proposition 3.3.18(b) in Bruns and Herzog, Cohen-Macaulay Rings:

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Question 1: It seems to me that under the hypothesis (a) of the theorem, the torsion-freeness of $\omega_R$ forces the underlying ring $R$ to be an integral domain. Is that correct or am i missing something? Here is my argument: We have that $\omega_R$ is torsion-free, so let $\xi \in \omega_R$ be a non-zero element. Now assume that $R$ is not an integral domain. Then there exist nonzero elements $a,b \in R$ such that $ab=0$. Since $b \neq 0$ and $\xi$ is not torsion, we must have that $b \xi \neq 0$. Since $0 \neq b \xi \in \omega_R$, $\omega_R$ is torsion-free and $a \neq 0$ we must also have that $a (b \xi) \neq 0$, which is a contradiction.

Question 2: I am puzzled by the inequality $\operatorname{depth} (R_p/\omega_R R_p) \ge \operatorname{depth}(R_p) -1$. How did the authors arrive to this? The way i understood this part of the proof is by noticing that $\omega_R R_p$ is a principal ideal generated by an $R_p$-regular element and so $R_p/\omega_R R_p$ is Cohen-Macaulay, which also shows that $R/\omega_R$ is CM.

Manos
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  • There must be $=$ since $\omega_R$ becomes principal after localization. However, this doesn't show that $R/\omega_R$ is CM. You need to consider all primes $p$ containing $\omega_R$, not only one.
    • – user26857 Nov 24 '14 at 20:56
  • Torsion element $x$ in a module means that there is a regular element $r$ such that $rx=0$. – user26857 Nov 24 '14 at 20:59
  • @user26857: I am aware of what you are pointing out for 2. I just used that this is true for every $p$ containing $\omega_R$. Hm...i was missing this detail in the definition of torsion element. Thanks! – Manos Nov 24 '14 at 21:02
  • @user26857: I think i understand now your comment. Do you mean that in the inequality chain in the proof we should have $depth(R_p/\omega_R R_p) = depth R_p -1$ because $R_p / \omega_R R_p$ is CM? And then we need to use this inequality chain for all such $p$ to actually conclude that $R/\omega_R$ is CM? – Manos Nov 24 '14 at 21:10
  • Not really: the equality of depths holds for $\omega R_p$ is a principal ideal (in a local ring) generated by a regular element, so depth must decrease with exactly one. (To the second question the answer is yes.) – user26857 Nov 24 '14 at 21:13
  • @user26857: Ok i understand. Thanks :) – Manos Nov 24 '14 at 21:20