Depth and Cohen Macaulay Modules

The first assertion is exactly the exceptional convention in the definition of $I$-depth; it includes $M=0$.

In the local case, $IM=M$ with $I\subseteq \mathfrak{m}$ would give $\mathfrak{m}M=M$. Under the stated AC hypothesis, thm-nakayama-lemma then forces $M=0$, contrary to the hypothesis.

An element of $R$ is a zero divisor on $M$ exactly when it belongs to an associated prime: one direction follows from an associated element, and the other from the zero-divisor lemma. Thus an $M$-regular element of $I$ exists exactly when $I$ is not contained in the union of the associated primes.

The associated-prime set is finite for a finite module over a Noetherian ring. Finite prime avoidance therefore says that $I$ is contained in that union exactly when it is contained in one member. Combining this with step 1.1 produces a nonzerodivisor $x\in I$ exactly under the displayed condition; $M\mathrm{/}xM\mathrm{\ne }0$ because $xM=M$ would imply $IM=M$. Thus $x$ is regular in the adopted sense, proving both directions.

Because $IM\mathrm{\ne }M$, depth zero means precisely that the empty regular sequence cannot be extended by an $M$-regular element of $I$.

The regular-element lemma identifies failure of such an extension with $I$ being contained in an associated prime of $M$. This proves the biconditional.

Nakayama gives $\mathfrak{m}M\mathrm{\ne }M$, so the ideal-relative criterion applies with $I=\mathfrak{m}$. It says depth is zero exactly when $\mathfrak{m}\subseteq \mathfrak{p}$ for some associated prime $\mathfrak{p}$.

Every associated prime is proper and every proper ideal of a local ring is contained in $\mathfrak{m}$. Hence $\mathfrak{m}\subseteq \mathfrak{p}$ forces $\mathfrak{p}=\mathfrak{m}$, proving both directions.

The sequence can be extended in $I$ exactly when $I$ contains a nonzerodivisor on $Q$ whose quotient is nonzero. If such an element $y$ had $Q\mathrm{/}yQ=0$, then $Q=yQ\subseteq IQ$, contrary to $IQ\mathrm{\ne }Q$.

Thus maximality is equivalent to ${\mathrm{depth}}_I(Q)=0$. The depth-zero associated-prime criterion converts this exactly into the displayed containment.

An $R$-linear map $f:R\mathrm{/}I\to M$ is determined by $m=f(1+I)$, and $Im=0$. Conversely, every $m\in (0{:}_{M}I)$ defines $f_m(r+I)=rm$.

These constructions are inverse and natural. Since degree-zero Ext is Hom, they give the asserted isomorphism.

Apply ${\mathrm{Ext}}_R(R\mathrm{/}I,-)$ to $0\to M\stackrel{x}{\to }M\to M\mathrm{/}xM\to 0$. Multiplication by $x$ on every ${\mathrm{Ext}}_R^i(R\mathrm{/}I,M)$ is zero because $x$ annihilates $R\mathrm{/}I$. The long exact sequence therefore gives $0\to {\mathrm{Ext}}_R^i(R\mathrm{/}I,M)\to {\mathrm{Ext}}_R^i(R\mathrm{/}I,M\mathrm{/}xM)\to {\mathrm{Ext}}_R^{i+1}(R\mathrm{/}I,M)\to 0.$

For $i<d-1$ both outer groups vanish, while at $i=d-1$ the right-hand group is nonzero and the left-hand group is zero. This proves the shift.

Put $Q=M\mathrm{/}(\mathbf{x})M$. Regularity makes $Q\mathrm{\ne }0$, and Nakayama gives $IQ\mathrm{\ne }Q$. Maximality and the stopping criterion yield $(0{:}_{Q}I)\mathrm{\ne }0$, hence ${\mathrm{Ext}}_R^0(R\mathrm{/}I,Q)\mathrm{\ne }0$.

Iterating the Ext shift backwards along the $r$ quotients shows that ${\mathrm{Ext}}_R^i(R\mathrm{/}I,M)=0$ for $i<r$ and is nonzero for $i=r$. Therefore the displayed minimum is $r$, independently of the sequence.

If $IM=M$, depth is infinite by definition. In this case $R\mathrm{/}I$ and $M$ have disjoint support, so all the displayed Ext groups vanish; this includes $M=0$.

If $IM\mathrm{\ne }M$, maximal regular sequences exist and the preceding lemma identifies their common length with the first nonzero Ext degree. The definition of depth gives the same length.

By lem-maximal-regular-sequences-have-common-length-ext, the length of every maximal regular sequence is the first nonzero Ext degree.

The Ext characterization identifies that same degree with ${\mathrm{depth}}_I(M)$, proving both assertions.

Noetherianity gives $a,b>0$ with $I^a\subseteq J$ and $J^b\subseteq I$. The standard finite-filtration Ext devissage says that vanishing of ${\mathrm{Ext}}_R^i(R\mathrm{/}I,M)$ for $i<d$ is equivalent to the same vanishing for every finite module annihilated by a power of $I$.

Apply this first to $R\mathrm{/}J$, annihilated by $I^a$, and then symmetrically to $R\mathrm{/}I$. The two Ext families vanish through the same initial range, including the all-vanishing case.

Radical invariance gives equality of the two first nonzero Ext degrees.

The Ext characterization of depth identifies those degrees with the two depths, with the same infinity convention.

Regularity includes $M\mathrm{/}xM\mathrm{\ne }0$. Since $I$ lies in the Jacobson radical, Nakayama excludes $IM=M$ and $I(M\mathrm{/}xM)=M\mathrm{/}xM$; both depths are finite.

The Ext shift lowers the first nonzero degree by one. Applying the Ext characterization on both sides yields the formula.

Embed $N=R\mathrm{/}\mathfrak{p}$ in $M$. Artin--Rees gives $n\ge 1$ such that $N\cap x^{n}M\subseteq xN$. Hence $L=N\mathrm{/}(N\cap x^{n}M)\subseteq M\mathrm{/}{x}^{n}M$ has quotient $N\mathrm{/}xN\cong R\mathrm{/}(\mathfrak{p}+(x))$.

We have $x^{n}N\subseteq N\cap x^{n}M\subseteq xN$. Viewing $N$ as $R\mathrm{/}\mathfrak{p}$, the annihilator of $L$ therefore lies between $\mathfrak{p}+(x^n)$ and $\mathfrak{p}+(x)$, so its radical is $\sqrt{\mathfrak{p}+(x)}$. Thus $\mathfrak{q}$ is minimal in $\mathrm{Supp}(L)$ and hence associated to $L$. An associated element of the submodule $L$ has the same annihilator in $M\mathrm{/}{x}^{n}M$.

Induct on $d=\mathrm{depth}(M)$. The case $d=0$ is immediate. If $d>0$, choose an $M$-regular $x\in \mathfrak{m}$. Then $x\mathrm{\notin }\mathfrak{p}$. Choose $\mathfrak{q}$ minimal over $\mathfrak{p}+(x)$. The principal ideal theorem and $\mathfrak{p}⊊\mathfrak{q}$ give $\dim (R\mathrm{/}\mathfrak{q})\le \dim (R\mathrm{/}\mathfrak{p})-1$.

The power-quotient lemma puts $\mathfrak{q}$ in $\mathrm{Ass}(M\mathrm{/}{x}^{n}M)$ for some $n\ge 1$. Since $x^n$ is regular, the quotient has depth $d-1$. Induction gives $d-1\le \dim (R\mathrm{/}\mathfrak{q})\le \dim (R\mathrm{/}\mathfrak{p})-1$, as required.

The claim is immediate if $M_{\mathfrak{p}}=0$ or if $\mathrm{depth}M\le \dim (R\mathrm{/}\mathfrak{p})$. Otherwise the associated-prime dimension bound shows that $\mathfrak{p}$ is contained in no member of $\mathrm{Ass}(M)$. Finite prime avoidance supplies an $M$-regular $x\in \mathfrak{p}$.

Localizing preserves regularity of $x$, and quotienting by it lowers both $\mathrm{depth}M$ and $\mathrm{depth}{M}_{\mathfrak{p}}$ by one. Induction on $\mathrm{depth}M$ applied to $M\mathrm{/}xM$ proves the inequality after adding one.

Part 1 is radical invariance, and part 2 is the localization inequality with its zero-localization convention.

Part 3 is the regular-element quotient formula. The three cited results prove the package.

By cor-depth-as-first-nonzero-ext, ${\mathrm{Ext}}_R^i(R\mathrm{/}\mathfrak{m},A)$ and ${\mathrm{Ext}}_R^i(R\mathrm{/}\mathfrak{m},C)$ vanish for every $i<\min \{a,c\}$.

Fix $i<\min \{a,c\}$. A finite partial free resolution of $R\mathrm{/}\mathfrak{m}$ through degree $i+1$ exists because $R$ is Noetherian. Applying its finite free terms to the short exact sequence gives the exact cohomology fragment ${\mathrm{Ext}}_R^i(R\mathrm{/}\mathfrak{m},A)⟶{\mathrm{Ext}}_R^i(R\mathrm{/}\mathfrak{m},B)⟶{\mathrm{Ext}}_R^i(R\mathrm{/}\mathfrak{m},C)\mathrm{.}$ Its outer terms vanish by step 1.1, so the middle term vanishes. This argument is finite for each $i$ and uses no choice principle. The Ext characterization now gives $b\ge \min \{a,c\}$.

Put $s=\min \{b,c+1\}$. At $i=0<s$, left exactness gives an injection ${\mathrm{Hom}}_R(R\mathrm{/}\mathfrak{m},A)\hookrightarrow {\mathrm{Hom}}_R(R\mathrm{/}\mathfrak{m},B),$ whose target is zero. For $0<i<s$, the Ext characterization gives ${\mathrm{Ext}}_R^i(R\mathrm{/}\mathfrak{m},B)=0$ and ${\mathrm{Ext}}_R^{i-1}(R\mathrm{/}\mathfrak{m},C)=0$.

For each fixed $0<i<s$, choose a finite partial free resolution of $R\mathrm{/}\mathfrak{m}$ through degree $i+1$. Applying its finite free terms to the short exact sequence gives the exact fragment ${\mathrm{Ext}}_R^{i-1}(R\mathrm{/}\mathfrak{m},C)⟶{\mathrm{Ext}}_R^i(R\mathrm{/}\mathfrak{m},A)⟶{\mathrm{Ext}}_R^i(R\mathrm{/}\mathfrak{m},B)\mathrm{.}$ The outer terms vanish by step 1.1. Together with the separately proved $i=0$ case, this gives vanishing for every $i<s$, so $a\ge s$.

If $a=0$ there is nothing to prove. Otherwise put $s=\min \{a-1,b\}$. For every $i<s$, the Ext characterization gives ${\mathrm{Ext}}_R^i(R\mathrm{/}\mathfrak{m},B)=0$ and ${\mathrm{Ext}}_R^{i+1}(R\mathrm{/}\mathfrak{m},A)=0$.

For each fixed $i<s$, a finite partial free resolution of $R\mathrm{/}\mathfrak{m}$ through degree $i+2$ gives the exact fragment ${\mathrm{Ext}}_R^i(R\mathrm{/}\mathfrak{m},B)⟶{\mathrm{Ext}}_R^i(R\mathrm{/}\mathfrak{m},C)⟶{\mathrm{Ext}}_R^{i+1}(R\mathrm{/}\mathfrak{m},A)\mathrm{.}$ The outer terms vanish by step 1.1. Thus the middle term vanishes for every $i<s$, and $c\ge s$. The construction is finite for each $i$ and needs no choice principle.

The middle, left, and right lower bounds are respectively lem-depth-lemma-lower-bound-middle, lem-depth-lemma-lower-bound-left, and lem-depth-lemma-lower-bound-right.

Substituting $a,b,c$ into those three bounds gives the three displayed inequalities, including the one-degree shifts and the stated $a=0$ boundary.

The first Depth Lemma inequality and either of the other two give $a<c\Rightarrow b=a$ and $a>c+1\Rightarrow b=c$. Cyclically applying the same comparison to the remaining inequalities gives the other four displayed implications.

For example, if $a<c$, then $b\ge a$ while $a\ge \min \{b,c+1\}$ forces $a\ge b$; hence $b=a$. If $a>c+1$, then $c\ge \min \{a-1,b\}$ and $a-1>c$ force $c\ge b$, while $b\ge c$; hence $b=c$. If $a<b$, then $c\ge a-1$, while $a\ge \min \{b,c+1\}$ forces $a\ge c+1$; hence $c=a-1$. The other three cases are the same comparisons.

Induct on ${\mathrm{depth}}_I(M)$. At depth $0$, prime avoidance gives an associated prime $\mathfrak{p}=\mathrm{ann}(m)$ containing $I$; hence $0\mathrm{\ne }m\in 0{:}_{M}I=H^0(K^{\bullet }(\mathbf{x};M))$. If the depth is positive, choose an $M$-regular $y\in I$. Adjoining $y$, an $R$-linear combination of the $x_i$, does not change the first nonzero Koszul cohomology degree: multiplication by $y$ on $K^{\bullet }(\mathbf{x};M)$ is null-homotopic, so its mapping cone has cohomology $H^i(K)\oplus H^{i-1}(K)$.

Reorder the enlarged sequence with $y$ first. Since $y$ is regular, its two-term Koszul cochain complex has only $M\mathrm{/}yM$ in degree $1$. Thus the first nonzero degree for the enlarged complex is one plus that for the induced Koszul complex on $M\mathrm{/}yM$; explicitly, lem-depth-quotient-by-regular-element gives ${\mathrm{depth}}_I(M\mathrm{/}yM)={\mathrm{depth}}_I(M)-1$. Induction proves the first equality. The finite free Koszul complex is self-dual: $K^i(\mathbf{x};M)\cong K_{n-i}(\mathbf{x};M)$, so $H^i\cong H_{n-i}$ and the first nonzero cohomology degree is $n$ minus the last nonzero homology degree.

Nakayama gives $M\mathrm{/}IM\mathrm{\ne }0$, so the Koszul homology in degree $0$ is nonzero and the depth is finite.

By lem-koszul-depth-first-nonzero-cohomology, the depth is the first nonzero degree of a cochain complex concentrated in degrees $0,\dots ,n$. Therefore it is at most $n$.

lem-koszul-depth-first-nonzero-cohomology proves both displayed equalities.

Their common value is ${\mathrm{depth}}_I(M)$, which is defined from $I$ and $M$ rather than from the chosen generators. Hence the numerical Koszul expression is generator-independent.

Put $A=R_{\mathfrak{p}}$. Let $N$ be a nonzero finite $A$-module and let $x$ in the maximal ideal be $N$-regular. Then $\mathrm{Supp}(N\mathrm{/}xN)=\mathrm{Supp}(N)\cap V(x)$ by localization and Nakayama. A minimal prime $\mathfrak{q}$ of $\mathrm{Supp}(N)$ cannot contain $x$: otherwise $N_{\mathfrak{q}}$ has support only at the maximal ideal of $A_{\mathfrak{q}}$, so a power of $x$ annihilates it, contradicting injectivity on that nonzero localization. Any prime chain in $\mathrm{Supp}(N\mathrm{/}xN)$ can therefore be extended downward by a strict inclusion from a minimal prime of $\mathrm{Supp}(N)$. Hence $\dim \mathrm{Supp}(N\mathrm{/}xN)\le \dim \mathrm{Supp}(N)-1$. Conversely, lift a parameter tuple for $N\mathrm{/}xN$ and prepend $x$. Its quotient has finite length, so thm-dimension-and-parameters-for-modules gives $\dim \mathrm{Supp}(N)\le 1+\dim \mathrm{Supp}(N\mathrm{/}xN)$. Thus the dimension drops exactly one, without any catenarity assumption. All dimensions are finite by that module-dimension theorem (ultimately the local height bound).

Starting with $M_{\mathfrak{p}}$, every regular sequence of length $r$ has nonzero successive quotients by def-regular-sequence-on-a-module. Applying step 1.1 successively gives $r\le \dim {\mathrm{Supp}}_A(M_{\mathfrak{p}})$. Extending a sequence while possible must therefore stop, so a maximal sequence exists. The maximal ideal is the Jacobson radical, and Nakayama gives $\mathfrak{p}A{M}_{\mathfrak{p}}\mathrm{\ne }{M}_{\mathfrak{p}}$. Thus thm-depth-equals-maximal-regular-sequence-length identifies its length with depth, proving the asserted inequality. This also handles dimension zero, when the maximal sequence is empty.

Apply lem-depth-at-a-prime-bounded-by-local-dimension at $\mathfrak{p}=\mathfrak{m}$.

Localization at the maximal ideal changes neither $R$, $M$, nor the support dimension, giving the displayed inequality.

Depth is the length of a regular sequence and is therefore nonnegative. The upper bound is cor-depth-of-a-finite-local-module-at-most-its-dimension.

These give the displayed double inequality. The last sentence follows directly from the separately declared zero-module depth convention.

Depth is nonnegative, while cor-depth-of-a-finite-local-module-at-most-its-dimension gives ${\mathrm{depth}}_R(M)\le 0$.

Thus ${\mathrm{depth}}_R(M)=0={\dim }_R(M)$, which is precisely the definition of Cohen--Macaulayness.

By lem-depth-quotient-by-regular-element, $\mathrm{depth}(M\mathrm{/}xM)=\mathrm{depth}(M)-1$.

Put $A=R\mathrm{/}{\mathrm{Ann}}_R(M)$. The finite-length terminal quotient shows that the images of $x,x_2,\dots ,x_d$ form a system of parameters of the local ring $A$, as in thm-dimension-and-parameters-for-modules. Moreover ${\mathrm{Supp}}_R(M\mathrm{/}xM)={\mathrm{Supp}}_R(M)\cap V(x)\cong \mathrm{Spec}(A\mathrm{/}(x))\mathrm{.}$ Applying lem-parameter-dimension-drop-is-exact to $A$ gives ${\dim }_R(M\mathrm{/}xM)=d-1$.

Subtracting the equalities in steps 1.1 and 1.2 cancels the common decrement and gives the displayed identity.

The gap lemma says the two modules have the same $\dim -\mathrm{depth}$ value.

Each module is Cohen--Macaulay exactly when that value is zero. Hence one is Cohen--Macaulay exactly when the other is.

Apply cor-regular-quotient-cohen-macaulay-equivalence first to $M$ and $x_1$, then to each successive quotient and $x_i$.

Chaining the resulting equivalences proves the assertion. The case $r=0$ is the identity statement.

Cohen--Macaulayness and the associated-prime depth bound give $d={\mathrm{depth}}_R(M)\le \dim (R\mathrm{/}\mathfrak{p})$.

Since $V(\mathfrak{p})\subseteq \mathrm{Supp}(M)$, the reverse inequality $\dim (R\mathrm{/}\mathfrak{p})\le d$ holds. The two inequalities give equality.

Suppose $\mathfrak{q}\in \mathrm{Supp}(M)$ and $\mathfrak{q}⊊\mathfrak{p}$. A chain of length $d$ above $\mathfrak{p}$, preceded by the strict inclusion $\mathfrak{q}⊊\mathfrak{p}$, gives $\dim (R\mathrm{/}\mathfrak{q})>d$.

But $V(\mathfrak{q})\subseteq \mathrm{Supp}(M)$ implies $\dim (R\mathrm{/}\mathfrak{q})\le \dim M=d$, a contradiction. Hence no such $\mathfrak{q}$ exists.