CM local codimension and Ext concentration over a regular local ring

Upper bound for the codimension formula. Since $R$ is Cohen--Macaulay, ${\mathrm{depth}}_R(R)=D$. Applying [F1] to $M=R$ gives ${\mathrm{depth}}_{R_{\mathfrak{p}}}(R_{\mathfrak{p}})+\dim R\mathrm{/}\mathfrak{p}\ge D$, and [F2] applied to the nonzero $R_{\mathfrak{p}}$-module $R_{\mathfrak{p}}$ gives ${\mathrm{depth}}_{R_{\mathfrak{p}}}(R_{\mathfrak{p}})\le \dim R_{\mathfrak{p}}$. Hence $D\le \dim R_{\mathfrak{p}}+\dim R\mathrm{/}\mathfrak{p}$.

Lower bound for the codimension formula. Take a saturated chain of primes ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_s=\mathfrak{p}$ with $s=\dim R_{\mathfrak{p}}$ and a saturated chain $\mathfrak{p}={\mathfrak{q}}_0⊊\cdots ⊊{\mathfrak{q}}_t$ with $t=\dim R\mathrm{/}\mathfrak{p}$. Concatenation gives a chain of primes of $R$ from ${\mathfrak{p}}_0$ to ${\mathfrak{q}}_t$ of length $s+t$, so $\dim R\ge \dim R_{\mathfrak{p}}+\dim R\mathrm{/}\mathfrak{p}$; both chains are finite because $R$ is Noetherian.

Projective dimension of $B$ and vanishing above $c$. Now $R$ is regular; by [F6] it is Cohen--Macaulay of depth $D$ and every finite $R$-module has finite projective dimension. Since $B$ is a nonzero finite $R$-module of depth $e$ (it is Cohen--Macaulay of dimension $e$), the Auslander--Buchsbaum formula [F6] gives ${\mathrm{pd}}_{R}B=D-e=c$. The forward vanishing implication in [F9] follows here directly: a projective resolution of length $c$ computes ${\mathrm{Ext}}_R^q(B,R)$ by its dual complex, which has no terms in degrees $q>c$; thus these groups vanish.

Codimension formula and heights. Steps 1.1 and 1.2 give $D=\dim R_{\mathfrak{p}}+\dim R\mathrm{/}\mathfrak{p}$, and $\mathrm{ht}\mathfrak{p}=\dim R_{\mathfrak{p}}$ by definition of height. For an ideal $I$ with $R\mathrm{/}I\mathrm{\ne }0$, the dimension $\dim R\mathrm{/}I$ is the maximum of $\dim R\mathrm{/}\mathfrak{q}$ over the minimal primes $\mathfrak{q}$ of $I$; applying the formula to those finitely many primes gives $\mathrm{ht}I={\min }_{\mathfrak{q}}\mathrm{ht}\mathfrak{q}=D-\dim R\mathrm{/}I$.

Inductive construction of a regular sequence in $I$. Maintain, for $j=0,\dots ,c$, a tuple $f_1,\dots ,f_j\in I$ that is an $R$-regular sequence with $Q_j:=R\mathrm{/}(f_1,\dots ,f_j)$ Cohen--Macaulay of dimension $D-j$. For $j=0$ this is [F6] and the empty tuple. Let $j<c$ and assume the tuple constructed. For $\mathfrak{q}\in {\mathrm{Ass}}_R(Q_j)$, [F3] gives $\dim R\mathrm{/}\mathfrak{q}=\dim Q_j=D-j$, hence $\mathrm{ht}\mathfrak{q}=j$ by step 2.1; since $\mathrm{ht}I=c>j$ and $I\subseteq \mathfrak{q}$ would force $\mathrm{ht}I\le \mathrm{ht}\mathfrak{q}=j$, we have $I⊈\mathfrak{q}$. Moreover $I{Q}_j\mathrm{\ne }{Q}_j$, because $Q_j\mathrm{/}I{Q}_j=R\mathrm{/}((f_1,\dots ,f_j)+I)=B\mathrm{\ne }0$. Thus [F4] with $M=Q_j$ and the ideal $I$ produces $f_{j+1}\in I$ that is $Q_j$-regular. Then $f_1,\dots ,f_{j+1}$ is an $R$-regular sequence with $Q_{j+1}=Q_j\mathrm{/}{f}_{j+1}{Q}_j$, and [F5] applied to the nonzero Cohen--Macaulay local ring $Q_j$ of dimension $D-j$ shows that $Q_{j+1}$ is nonzero and Cohen--Macaulay of dimension $D-j-1$, completing the induction. In particular the case $j=c$ gives a regular sequence $f_1,\dots ,f_c$ in $I$ with $Q_c=R\mathrm{/}(f_1,\dots ,f_c)$ Cohen--Macaulay of dimension $D-c=e$.

Derived adjunction along the regular sequence. For $0\le j<c$, put $Q_{j+1}=Q_j\mathrm{/}(f_{j+1})$, where $f_{j+1}$ is $Q_j$-regular by step 3.1; the quotient $Q_j\to Q_{j+1}$ is finite, and $B$ is a $Q_{j+1}$-module. The two-term resolution in [F7] gives $\mathcal{R}{\mathrm{Hom}}_{Q_j}(Q_{j+1},Q_j)\simeq Q_{j+1}[-1]$. Applying its finite-quotient derived adjunction to $M=B$ gives $\mathcal{R}{\mathrm{Hom}}_{Q_j}(B,Q_j)\simeq \mathcal{R}{\mathrm{Hom}}_{Q_{j+1}}(B,Q_{j+1})[-1]\mathrm{.}$ Iterating over $j=0,\dots ,c-1$ yields $\mathcal{R}{\mathrm{Hom}}_R(B,R)\simeq \mathcal{R}{\mathrm{Hom}}_{Q_c}(B,Q_c)[-c]\mathrm{.}$ The unshifted complex $\mathcal{R}{\mathrm{Hom}}_{Q_c}(B,Q_c)$ has no negative cohomology, so ${\mathrm{Ext}}_R^q(B,R)=0$ for $q<c$, and for $q\ge c$ it identifies with ${\mathrm{Ext}}_{Q_c}^{q-c}(B,Q_c)$. With step 1.3 this proves the full concentration statement; when $c=0$ the iteration is the identity.

The canonical module is nonzero. Apply [F8] with $A=R$, $L=R$ and $n=D$: the complex $D_B=\mathcal{R}{\mathrm{Hom}}_R(B,R[D])$ is a dualizing complex over $B$, with $H^q(D_B)={\mathrm{Ext}}_R^{q+D}(B,R)$, and biduality holds for $B\in D_{\mathrm{f}\mathrm{i}\mathrm{n}}^b(B)$. By steps 1.3 and 4.1, $D_B$ has cohomology only in degree $c-D=-e$, where it is $E$; hence $D_B\simeq E[e]$. Biduality for $M=B$ gives $B\simeq \mathcal{R}{\mathrm{Hom}}_B(E[e],E[e])\simeq \mathcal{R}{\mathrm{Hom}}_B(E,E)\mathrm{.}$ Since $B\mathrm{\ne }0$, this complex is nonzero, so $E\mathrm{\ne }0$.

A finite free resolution of $E$ and its dimension. Choose a finite free resolution $F_c\to \cdots \to F_0\to B\to 0$ of length $c$. Its dual complex $F_0^{\vee }\to \cdots \to F_c^{\vee }$ is a complex of finite free modules whose $q$-th cohomology is ${\mathrm{Ext}}_R^q(B,R)$; by step 4.1 the cohomology vanishes in degrees $q<c$, so the dual complex is exact except at its last term, and $0\to F_0^{\vee }\to \cdots \to F_c^{\vee }\to E\to 0$ is a finite free resolution of $E$; hence ${\mathrm{pd}}_{R}E\le c$. Since $IE=0$, the support of $E$ over $R$ lies in $V(I)$, so ${\dim }_{R}E\le \dim V(I)=\dim R\mathrm{/}I=e$, and $E$ is finite as the cokernel in this finite free resolution.

Full support. Suppose that $E_{\mathfrak{p}}=0$ for some $\mathfrak{p}\in \mathrm{Spec}B$. By [F6], $B$ has a finite free resolution over the regular local ring $R$; [F12] identifies its termwise localized dual with the dual of the localized resolution, so localizing computes the derived Hom over $R_{\mathfrak{p}\cap R}$. Therefore the localization of $D_B\simeq E[e]$ is zero. Put $\mathfrak{q}=\mathfrak{p}\cap R$ and $d=\dim R_{\mathfrak{q}}$. By [F11], $R_{\mathfrak{q}}$ is regular, and its dualizing complex for the quotient $B_{\mathfrak{p}}$ is $D_{B_{\mathfrak{p}}}:=\mathcal{R}{\mathrm{Hom}}_{R_{\mathfrak{q}}}(B_{\mathfrak{p}},R_{\mathfrak{q}}[d])\mathrm{.}$ The localized complex $(D_B{)}_{\mathfrak{p}}$ is $D_{B_{\mathfrak{p}}}[D-d]$, so $D_{B_{\mathfrak{p}}}=0$. Applying [F8] over $R_{\mathfrak{q}}$ to $M=B_{\mathfrak{p}}$, biduality would then give $B_{\mathfrak{p}}\simeq \mathcal{R}{\mathrm{Hom}}_{B_{\mathfrak{p}}}(0,0)=0$, contradicting $\mathfrak{p}\in \mathrm{Spec}B$. Thus $E_{\mathfrak{p}}\mathrm{\ne }0$ for every prime of $B$, and ${\mathrm{Supp}}_B(E)=\mathrm{Spec}B=V(I)={\mathrm{Supp}}_B(B)$.

$E$ is Cohen--Macaulay of dimension $e$. The module $E$ is nonzero and finite with ${\mathrm{pd}}_{R}E\le c$, so Auslander--Buchsbaum [F6] gives ${\mathrm{depth}}_{R}E=D-{\mathrm{pd}}_{R}E\ge D-c=e$, while [F2] gives ${\mathrm{depth}}_{R}E\le {\dim }_{R}E\le e$ by step 6.1. Hence ${\mathrm{depth}}_{R}E={\dim }_{R}E=e$, that is, $E$ is a Cohen--Macaulay $R$-module of dimension $e$; as $IE=0$ it is a $B$-module of dimension $e$ with the same depth.

Localization of the statement. For a prime $\mathfrak{p}$, the ring $R_{\mathfrak{p}}$ is local Noetherian, and ${\mathrm{depth}}_{R_{\mathfrak{p}}}(R_{\mathfrak{p}})=\dim R_{\mathfrak{p}}=D-\dim R\mathrm{/}\mathfrak{p}$ by [F10] applied to $M=R$ and step 2.1, so $R_{\mathfrak{p}}$ is Cohen--Macaulay. If $R$ is regular and $\mathfrak{p}\supseteq I$, then $R_{\mathfrak{p}}$ is regular by [F11], the module $B_{\mathfrak{p}}$ is nonzero and finite, and [F10] applied to the Cohen--Macaulay $R$-module $B$ gives ${\mathrm{depth}}_{R_{\mathfrak{p}}}(B_{\mathfrak{p}})={\dim }_{R_{\mathfrak{p}}}(B_{\mathfrak{p}})=e_{\mathfrak{p}}$; so $B_{\mathfrak{p}}$ is Cohen--Macaulay over the regular local ring $R_{\mathfrak{p}}$ of dimension $\dim R_{\mathfrak{p}}$. Apply the projective-dimension calculation of step 1.3 and the regular-sequence/change-of-rings argument of steps 3.1 and 4.1 with $(R,I,B,e)$ replaced by $(R_{\mathfrak{p}},I{R}_{\mathfrak{p}},B_{\mathfrak{p}},e_{\mathfrak{p}})$. They give ${\mathrm{Ext}}_{R_{\mathfrak{p}}}^q(B_{\mathfrak{p}},R_{\mathfrak{p}})=0$ for $q\mathrm{\ne }\dim R_{\mathfrak{p}}-e_{\mathfrak{p}}$.