For a local Noetherian Cohen-Macaulay ring $(R,m,k)$ the canonical module is defined to be any maximal Cohen-Macaulay module of finite injective dimension and of type $1$. The canonical module is unique up to isomorphism.

So let $\omega_R$ be the canonical module of $R$ and let $\hat{}$ denote $m$-adic completion. I am interested in seeing an elegant proof of the fact that $\omega_{\hat{R}} = \hat{\omega_R}$, i.e. the canonical module of $\hat{R}$ is just the completion of the canonical module of $R$.

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    $\begingroup$ I wonder if you tried reducing to the Artinian case. But I have no ideal what you meant by "elegant". $\endgroup$ – Youngsu Mar 7 '14 at 20:09
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    $\begingroup$ I must say I'm also not too sure what you mean by elegant. Youngsu's suggestion to reduce to the Artinian case is a good one. The proof that Bruns-Herzog give, in Theorem 3.3.5(c), is also (in my opinion) quite elegant $\endgroup$ – zcn Mar 7 '14 at 20:50
  • $\begingroup$ @user115654: Actually the reason i ask this question is because i don't really understand the 2-line proof that Bruns-Herzog give. So i suppose by elegant i mean including a little bit more details. For example, what does it mean that the fibre of $R \rightarrow \hat{R}$ is $k$? Also, even though i have studied completion i still don't feel comfortable with it, so elegant would also mean a little bit instructive. Finally, i find the placement of the $\hat{}$ operator in the text of Bruns&Herzog unelegant and confusing. $\endgroup$ – Manos Mar 7 '14 at 22:52
  • $\begingroup$ But both you and Youngsu are totally right in asking me what "elegant means". I hope it is clear now. $\endgroup$ – Manos Mar 7 '14 at 22:54

Here is a fleshing out of the $2$-line (!) proof given in Bruns-Herzog: $\newcommand{\Ext}{\text{Ext}}$

Recall that $C$ is a canonical module of $R$ iff $\dim_k \Ext^i_R(k, C) = \delta_{id}$, where $d = \dim R$, i.e. $\Ext^i_R(k,C) = 0$ if $i \ne d$, and $\Ext^d_R(k,C) \cong k$. The remark that the fibre of $R \to \hat{R}$ is $k$ seems to just say that the special fiber ring of $\hat{R}$ is isomorphic to $k$, i.e. $\hat{k} \cong \widehat{(R/m)} \cong \hat{R}/m\hat{R} \cong R/m \cong k$. The main result that is being used is the following:

Proposition: Let $R \to S$ be a ring map making $S$ flat over $R$, and $M, N$ $R$-modules. If $R$ is Noetherian and $M$ is finite, then $S \otimes_R \Ext^i_R(M,N) \cong \Ext^i_S(S \otimes_R M, S \otimes_R N)$ for every $i$.

Now as to the proof: $R \to \hat{R}$ is flat, $R$ is Noetherian, and $k$ is finite over $R$. By the proposition, $\hat{R} \otimes_R \Ext^i_R(k,\omega_R) \cong \Ext^i_{\hat{R}}(\hat{k}, \widehat{\omega_R}) \cong \Ext^i_{\hat{R}}(k, \widehat{\omega_R})$ for every $i$. Now $\dim R = \dim \hat{R} = d$, and if $i \ne d$, then $\Ext^i_{\hat{R}}(k, \widehat{\omega_R}) = 0$, and if $i = d$, then $\Ext^d_{\hat{R}}(k, \widehat{\omega_R}) \cong k \otimes_R \hat{R} \cong k$. Thus $\widehat{\omega_R}$ is a canonical module for $\hat{R}$.

  • $\begingroup$ Excellent answer! What is the definition of the special fiber ring? $\endgroup$ – Manos Mar 10 '14 at 19:29
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    $\begingroup$ If $\varphi : R \to S$ is a ring homomorphism and $p \in \text{Spec}(R)$, the fiber ring over $p$ is $S \otimes_R k(p)$, where $k(p)$ is the residue field at $p$ (note $\text{Spec}$ of the fiber ring is the (scheme-theoretic) fiber). The special fiber is just the fiber over the maximal ideal when $R$ is local (and similarly the generic fiber is the fiber over $0$ when $R$ is a domain) $\endgroup$ – zcn Mar 10 '14 at 20:19

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