I am reading Chapter 3 of Yoshino's book. There is a proposition that the author didn't prove. It looks interesing. But I don't know how to prove. The proposition is,
Let $(R,\mathfrak{m})$ be a Henselian Cohen-Macaulay local ring with dimension $d$. Suppose $M$ is a finitely generated $R$-module and $M$ is locally free on the punctured spectrum $\rm{Spec}(R)\setminus\{\mathfrak{m}\}$ of $R$. Then $M$ is a maximal Cohen-Macaulay if and only if ${\rm Ext}_R^i({\rm tr}(M),R)=0~ (1\leq i\leq d)$.
Here ${\rm tr}(M)$ is the Auslander transpose of $M$. Any complete local ring is Henselian. The assumption that $R$ is Henselian seems superflous.
After taking a look of Auslander and Bridger's book Stable Module Theory, I found that the condition ${\rm Ext}_R^i({\rm tr}(M),R)=0~ (1\leq i\leq d)$ implies that $M$ is a $d$-th syzygy of an $R$-module. In particular, this condition yields that $M$ is maximal Cohen-Macaulay.
How to prove forward direction of the above proposition? Thank you in advance.
Best Answer
Henselian is superfluous. Here need $(R,\mathfrak m)$ is a Cohen-Macaulay local ring.
$(\Leftarrow)$ In order to prove this, we just need to prove that the second condition implies that $M$ is a $d-$th syzygy of some module for any $d-$dimensional commutative noetherian ring.
First, recall that there is a long exact sequence $$ 0\rightarrow \rm{Ext}^1_R(\rm{tr}(M),R)\rightarrow M\rightarrow M^{\ast\ast}\rightarrow \rm Ext_R^2(\rm tr(M),R)\rightarrow 0. $$ We may assume that $d\geq 3$. The condition that $\rm Ext_R^i(\rm tr(M),R)=0$ for $1\leq i\leq d$ is equivalent to that $M\cong M^{\ast\ast}$ and $\rm Ext_R^i(\rm tr(M),R)=0$ for $3\leq i\leq d$. Since $\Omega^2_R(\rm \rm tr(M))\cong M^\ast$. Thus $\rm Ext_R^i(\rm tr(M),R)=0$ for $3\leq i\leq d$ is equivalent to $\rm Ext^i_R(M^\ast,R)=0$ for $1\leq d-2$.
Choose a free resolution of $M^\ast$ $$ F_{d-1}\rightarrow \cdots\rightarrow F_1\rightarrow F_0\rightarrow M^\ast\rightarrow 0. $$ The argument above means that the dual of above exact sequence is exact. That is, the following exact sequence is exact $$ 0\rightarrow M\rightarrow F_0^\ast\rightarrow F_1^\ast\rightarrow \cdots\rightarrow F_{d-1}^\ast. $$ In particular, $M$ is a $d-$syzygy of an $R$-module.
$(\Rightarrow)$ $d=0$ is OK. For $d=1$, recall that there is an exact sequence $$ 0\rightarrow \rm Ext_R^1(\rm tr(M),R)\rightarrow M\rightarrow M^{\ast\ast}. $$ Denote $L=\rm Ext^1(\rm tr(M),R)$. Since $M$ is locally free on the punctured spectrum, $\rm Supp L=\{\mathfrak m\}$ if $L\neq 0$. So $L$ is of finite length. Denote the cokernel of the map $L\rightarrow M$ is $L^\prime$. We observe that $M^{\ast\ast}$ is MCM. It follows that $\rm depth(L^\prime)>0$. However, the depth lemma yields that $$ \rm depth(L)\geq \min\{\rm depth(M),\rm depth(L^\prime)+1\}>0. $$ This contradicts with that $L$ has finite length. Hence $L=0$.
Similarly, one can prove the case of $d=2$. Then $M$ is reflexive. This yields that $\rm Ext_R^i(\rm tr(M),R)=0$ for $i=1,2$.
Now assume that $d\geq 3$. As before, one can prove $M$ is reflexive ($\iff \rm Ext_R^i(\rm tr(M),R)=0$ for $i=1,2$). Since $\Omega^2_R(\rm tr(M))\cong M^\ast$, it remains to show that $\rm Ext^i_R(M^\ast,R)=0$ for $1\leq i\leq d-2$. Assume hat $\rm Ext^i_R(M^\ast,R)\neq 0$ for some $1\leq i\leq d-2$. Let $j$ be the smallest integer such that $\rm Ext^j_R(M^\ast,R)\neq 0$. Choose a free resolution $$ P_j\rightarrow P_{j-1}\rightarrow \cdots\rightarrow P_1\rightarrow P_0\rightarrow M^\ast\rightarrow 0. $$ Dualizing the above resolution, our choice of $j$ yields that we have an exact complex $$ 0\rightarrow M\rightarrow P_0^\ast\rightarrow P_1^\rightarrow \cdots\rightarrow P_{j-1}^\ast\rightarrow P_j^\ast\rightarrow C\rightarrow 0. $$ We observe that $\rm Ext^j_R(M^\ast,R)$ is a submodule of $C$. Note that $\rm Ext^j_R(M^\ast,R)$ has finite length as $M$ is locally free on the punctured spectrum. Hence $\rm depth(C)=0$. Then the depth lemma yields that $\rm depth(M)=j+1\leq d-1$. This contradicts with $M$ is MCM. Thus we have $\rm Ext_R^i(M^\ast,R)=0$ for $1\leq i\leq d-2$.