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, Assuming the Axiom of Choice, Nakayama's 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 A zero divisor is contained in an associated prime under the stated Choice hypothesis. 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 by Finite modules over Noetherian rings have finitely many associated primes under the stated Choice hypothesis. It is nonempty because $0$ is a zero divisor on $M\mathrm{\ne }0$ and step 1.1 places every zero divisor in an associated prime. Thus An ideal contained in a finite union of prime ideals lies in one of them applies and 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.

Since $M\mathrm{\ne }0$, [F1] gives $\mathfrak{m}M\mathrm{\ne }M$, so [F2] applies with $I=\mathfrak{m}$. Its published proof uses finiteness of associated primes through an AC-qualified prime filtration; the assumed Choice supplies that premise. 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 ${\mathrm{depth}}_I(Q)=0$. By [L1], this is exactly 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 $M_j=M\mathrm{/}(x_1,\dots ,x_j)M$ for $0\le j\le r$, so $M_0=M$ and $Q=M_r$. By [F1], $Q$ is nonzero and finite, and $IQ\mathrm{\ne }Q$. Maximality and [L1] give a prime $\mathfrak{p}\in {\mathrm{Ass}}_R(Q)$ containing $I$. By the definition in [L1], some nonzero $q\in Q$ has ${\mathrm{Ann}}_R(q)=\mathfrak{p}$, hence $Iq=0$. Therefore $E^0(Q):={\mathrm{Ext}}_R^0(R\mathrm{/}I,Q)\cong (0{:}_{Q}I)$ is nonzero.

For $1\le j\le r$, regularity gives $0\to M_{j-1}\stackrel{x_j}{\to }{M}_{j-1}\to M_j\to 0$. Write $E^i(T)={\mathrm{Ext}}_R^i(R\mathrm{/}I,T)$. Since $x_j\in I$ annihilates $R\mathrm{/}I$, multiplication by $x_j$ on every $E^i(M_{j-1})$ is zero. The long exact sequence of [L2] therefore yields, for each $i\ge 0$, a short exact sequence $0\to E^i(M_{j-1})\to E^i(M_j)\to E^{i+1}(M_{j-1})\to 0$. Moreover $E^0(M_{j-1})=(0{:}_{M_{j-1}}I)=0$: an $I$-annihilated element is killed by the injective multiplication map $x_j$.

Suppose the first nonzero $E^i(M_j)$ occurs in degree $d_j$. Starting with $E^0(M_{j-1})=0$, the sequences in step 1.2 show inductively that $E^i(M_{j-1})=0$ for $0\le i\le d_j$, and at $i=d_j$ give $E^{d_j+1}(M_{j-1})\cong E^{d_j}(M_j)\mathrm{\ne }0$. Thus the first nonzero degree rises by one when moving backwards across $x_j$. Step 1.1 gives $d_r=0$; repeating this finite argument for $j=r,r-1,\dots ,1$ gives $d_0=r$. Hence the displayed Ext minimum equals the length of every maximal sequence, so all have the same length.

If $IM=M$, [F1] makes $M=0$. Its depth is infinite by definition and every displayed Ext group vanishes. Hence assume $M\mathrm{\ne }0$ and $IM\mathrm{\ne }M$. Write $E^i(T)={\mathrm{Ext}}_R^i(R\mathrm{/}I,T)$ and $d=\inf \{i:E^i(M)\mathrm{\ne }0\}$. If $d=0$, a nonzero element of $E^0(M)=(0{:}_{M}I)$ is killed by every $x\in I$, so no $M$-regular element exists in $I$ and depth is zero.

Suppose $d>0$, allowing $d=\mathrm{\infty }$. Then $E^0(M)=0$. If an associated prime $\mathfrak{p}={\mathrm{Ann}}_R(m)$ contained $I$, its nonzero associated element $m$ would belong to $(0{:}_{M}I)$, a contradiction. By [L1], choose an $M$-regular $x\in I$ and put $Q=M\mathrm{/}xM\mathrm{\ne }0$. Applying $E^{\ast }(-)$ to $0\to M\stackrel{x}{\to }M\to Q\to 0$, multiplication by $x$ on each $E^i(M)$ is zero because $x$ annihilates $R\mathrm{/}I$. Thus $0\to E^i(M)\to E^i(Q)\to E^{i+1}(M)\to 0$ is exact for every $i\ge 0$. If $d<\mathrm{\infty }$, the first nonzero degree for $Q$ is $d-1$; if $d=\mathrm{\infty }$, every $E^i(Q)$ vanishes.

Induct on finite $d$. The case $d=0$ is step 1.1. For $d>0$, step 1.2 and induction give an $M\mathrm{/}xM$-regular sequence of length $d-1$, so prepending $x$ gives an $M$-regular sequence of length $d$. Conversely, for any $M$-regular first element $y\in I$, the same argument gives first nonzero Ext degree $d-1$ on $M\mathrm{/}yM$; induction bounds every sequence beginning with $y$ by $d$. Therefore ${\mathrm{depth}}_I(M)=d$. If $d=\mathrm{\infty }$, step 1.2 applies successively to every nonzero quotient and keeps all its Ext groups zero; finite induction constructs regular sequences of every length. Their supremum, hence depth, is $\mathrm{\infty }$.

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.

Since $I$ and $J$ are finitely generated and have equal radicals, some positive powers satisfy $I^a\subseteq J$ and $J^b\subseteq I$: take powers of each generator lying in the other ideal, then a sufficiently high power of the whole ideal lies there. Fix $d\ge 0$ and suppose ${\mathrm{Ext}}_R^i(R\mathrm{/}I,M)=0$ for $0\le i<d$. The same vanishing holds for every finite $R\mathrm{/}I$-module $N$. Indeed, for each such $N$ choose a finite free $R\mathrm{/}I$-module $F$ surjecting onto it, with finite kernel $K$. For $i=0$, ${\mathrm{Hom}}_R(N,M)$ injects into ${\mathrm{Hom}}_R(F,M)=0$. Inductively, for $1\le i<d$, the long exact Ext sequence places ${\mathrm{Ext}}_R^i(N,M)$ between ${\mathrm{Ext}}_R^{i-1}(K,M)=0$ and ${\mathrm{Ext}}_R^i(F,M)=0$. If a finite module $N$ is killed by $I^r$, its finite filtration $N\supseteq IN\supseteq \cdots \supseteq I^{r}N=0$ has finite $R\mathrm{/}I$-module quotients, so the long exact sequences give the same vanishing for $N$. The converse follows by taking $N=R\mathrm{/}I$.

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.

Apply [L1] to $I$ and $J$: the least nonvanishing degrees of ${\mathrm{Ext}}_R^{\ast }(R\mathrm{/}I,M)$ and ${\mathrm{Ext}}_R^{\ast }(R\mathrm{/}J,M)$ agree, with both infinite if neither family has a nonzero group.

Apply [L2] separately to $I$ and $J$; both lie in the Jacobson radical by hypothesis. Their depths are the degrees equal in step 1.1, including the infinite case.

Regularity includes $M\mathrm{/}xM\mathrm{\ne }0$. Since $x\in I$ acts injectively on $M$, no nonzero element of $M$ is annihilated by $I$. Thus ${\mathrm{Ext}}_R^0(R\mathrm{/}I,M)=0$.

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 each Ext group is zero because $x$ annihilates $R\mathrm{/}I$, so the long exact sequence yields $0\to E^i(M)\to E^i(M\mathrm{/}xM)\to E^{i+1}(M)\to 0$, where $E^i(T)={\mathrm{Ext}}_R^i(R\mathrm{/}I,T)$. By step 1.1, the first nonzero Ext degree for $M$ is either a positive integer $d$ or $\mathrm{\infty }$. The sequence makes the first degree for $M\mathrm{/}xM$ equal $d-1$ when $d$ is finite and $\mathrm{\infty }$ when all groups vanish. The Ext characterization gives the displayed depth formula with its stated convention.

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, under the stated AC, associated to $L$ by [L1]. 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 follows from [L1], and part 2 follows from [L2] under the stated Choice assumption.

Part 3 follows from [L3]. 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 [L1], 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 $x$ lie in the maximal ideal and act injectively on $N$. By [L3], $N\mathrm{/}xN\mathrm{\ne }0$. Localization and [L3] give ${\mathrm{Supp}}_A(N\mathrm{/}xN)={\mathrm{Supp}}_A(N)\cap V_A(x)$: outside $V(x)$ the quotient is zero, while at a support prime containing $x$ its localized nonzero finite module cannot equal its $x$-multiple.

Let $\mathfrak{q}$ be minimal in ${\mathrm{Supp}}_A(N)$. By [L1], it is minimal over ${\mathrm{Ann}}_A(N)$. In $A_{\mathfrak{q}}$ the only support prime of $N_{\mathfrak{q}}$ is $\mathfrak{q}{A}_{\mathfrak{q}}$; [L1] therefore makes $\sqrt{\mathrm{Ann}(N_{\mathfrak{q}})}=\mathfrak{q}{A}_{\mathfrak{q}}$. This maximal ideal is finitely generated, so some power annihilates $N_{\mathfrak{q}}$. If $x\in \mathfrak{q}$, a power of multiplication by $x$ would be zero on the nonzero module $N_{\mathfrak{q}}$, contradicting injectivity after localization. Hence no minimal support prime contains $x$.

Take any strict chain ${\mathfrak{r}}_0⊊\cdots ⊊{\mathfrak{r}}_s$ in ${\mathrm{Supp}}_A(N\mathrm{/}xN)$. Since ${\mathfrak{r}}_0$ contains $\mathrm{Ann}N$, the finite minimal-prime decomposition [L2] gives a minimal support prime $\mathfrak{q}\subseteq {\mathfrak{r}}_0$. Because $x\in {\mathfrak{r}}_0$ but $x\mathrm{\notin }\mathfrak{q}$ by step 2.1, this inclusion is strict. Prepending $\mathfrak{q}$ gives a chain of length $s+1$ in ${\mathrm{Supp}}_A(N)$. Thus $\dim {\mathrm{Supp}}_A(N\mathrm{/}xN)\le \dim {\mathrm{Supp}}_A(N)-1$.