Depth and Cohen Macaulay Modules — Examples

Every element is uniquely $a+b\epsilon$ with $a,b\in k$. It is a unit when $a\mathrm{\ne }0$, with inverse $a^{-1}-b{a}^{-2}\epsilon$; when $a=0$ it lies in $(\epsilon )$. Thus $(\epsilon )$ is the unique maximal ideal. Any nonzero ideal contained in $(\epsilon )$ contains $b\epsilon$ for some $b\mathrm{\ne }0$ and hence contains $\epsilon$, so the ideals are exactly $0$, $(\epsilon )$ and $A$. Consequently $A$ is Artinian and Noetherian, and it is nonreduced because $\epsilon \mathrm{\ne }0$ but ${\epsilon }^2=0$.

Every prime ideal contains the nilpotent element $\epsilon$, so the unique prime is $(\epsilon )$ and $\dim A=0$. Every element of $(\epsilon )$ annihilates the nonzero element $\epsilon$, hence no element of the maximal ideal is $A$-regular and $\mathrm{depth}A=0$. Therefore $A$ is Cohen--Macaulay by [F1].

Under the stated Choice premise, iterating Polynomial extensions of Cohen--Macaulay rings shows that $k[X_1,\dots ,X_n]$ is globally Cohen--Macaulay.

In the localization at $\mathfrak{m}$, multiplication by $X_i$ remains injective after quotienting by the preceding variables, and the terminal quotient is $k$. Thus the displayed variables are a regular parameter system of length $n$.

Every class in $A$ has a unique expression $f(y)+ax$, where $f\in k\mathrm{⟦}y\mathrm{⟧}$ and $a\in k$: all monomials involving $x$ except $x$ itself belong to $(x^2,xy)$. Thus $x\mathrm{\ne }0$, $x^2=yx=0$, and the projection $A\to k\mathrm{⟦}y\mathrm{⟧}$ has kernel $kx$. An element is a unit precisely when $f(0)\mathrm{\ne }0$: the series $f$ is then invertible by recursive coefficient multiplication, and the square-zero correction is inverted by $1-z$. If $f(0)=0$, its image in the quotient $A\mathrm{/}(x,y)=k$ is zero, so it is not a unit. Hence $A$ is local with maximal ideal $(x,y)$.

Every nonzero series in $k\mathrm{⟦}y\mathrm{⟧}$ is $y^m$ times a unit, by its least nonzero coefficient. Consequently every nonzero ideal in that ring is $(y^m)$, taking the least order among its nonzero elements. For an ideal $J$ of $A$, its projection is therefore principal. If nonzero, lift one generator to $z\in J$; subtracting suitable $k\mathrm{⟦}y\mathrm{⟧}$-multiples of $z$ from elements of $J$ leaves elements of $J\cap kx$. This intersection is either zero or $kx$, since it is a subspace of a one-dimensional $k$-space. Thus $J$ is generated by $z$ and, if needed, $x$; when its projection is zero it is generated by $x$ or is zero. Every ideal is finitely generated, so $A$ is Noetherian by Noetherian commutative rings and modules.

Every prime of $A$ contains the nilpotent $x$, so prime ideals correspond to those of $A\mathrm{/}(x)=k\mathrm{⟦}y\mathrm{⟧}$. The latter has exactly the primes $(0)$ and $(y)$ by the ideal description in step 2.1, and they are distinct. Hence $\dim A=1$.

Every element of $\mathfrak{m}$ annihilates the nonzero $x$. None is therefore an $A$-regular element. Since $A\mathrm{/}\mathfrak{m}=k\mathrm{\ne }0$, the empty sequence is regular and no positive-length regular sequence in $\mathfrak{m}$ exists. Thus $\mathrm{depth}A=0$ by [F1]. The support of the regular module is the whole spectrum, of dimension one, so [F2] shows that $A$ is not Cohen--Macaulay.

Put $A=k\mathrm{⟦}x,y\mathrm{⟧}\mathrm{/}(xy)$ and $\mathfrak{m}=(x,y)A$. A series is a unit exactly when its constant term is nonzero, so $A$ is local with maximal ideal $\mathfrak{m}$. The maps setting $y=0$ and $x=0$ identify $A\mathrm{/}(y)$ with $k\mathrm{⟦}x\mathrm{⟧}$ and $A\mathrm{/}(x)$ with $k\mathrm{⟦}y\mathrm{⟧}$. Every prime of $A$ contains $x$ or $y$ since $xy=0$. In either one-variable quotient the only primes are $(0)$ and the variable ideal: factor a nonzero member by its least-order term, whose remaining factor is a unit. Thus the prime chains of $A$ have length at most one, while $(x)A⊊\mathfrak{m}$ is such a chain. Hence $\dim A=1$.

The map $A\to k\mathrm{⟦}x\mathrm{⟧}\times k\mathrm{⟦}y\mathrm{⟧}$ given by the two branch restrictions is injective: a series divisible by both $x$ and $y$ is divisible by $xy$. The element $x+y$ maps to $(x,y)$, so multiplication by it is injective on both domains and therefore on $A$. The quotient $A\mathrm{/}(x+y)\cong k\mathrm{⟦}x\mathrm{⟧}\mathrm{/}(x^2)$ is nonzero. Thus $(x+y)$ is an $A$-regular sequence and $\mathrm{depth}A\ge 1$.

To see that no $A$-regular sequence has length two, let $u\in \mathfrak{m}$ be any nonzerodivisor. Its two branch restrictions are nonzero; write them $x^{a}h(x)$ and $y^{b}g(y)$, with $a,b\ge 1$ and $h,g$ units. Since $xu=x^{a+1}h(x)$ and $yu=y^{b+1}g(y)$ in $A$, both $x^{a+1}$ and $y^{b+1}$ lie in $(u)$. Hence a power of $\mathfrak{m}$ vanishes in the nonzero local ring $A\mathrm{/}uA$. Every element of its maximal ideal is nilpotent and therefore cannot be a nonzerodivisor there. By the depth and regular-sequence definitions, $\mathrm{depth}A=1=\dim A$. The classes of $x$ and $y$ are nonzero and have zero product, as claimed.

For $r=1,2$, the coordinate localization $P_r=k[x_1,\dots ,x_r{]}_{(x_1,\dots ,x_r)}$ is Noetherian local. Its dimension is $r$: the coordinate-prime chain has length $r$, while localization cannot increase the dimension $r$ of the polynomial ring. The quotients $P_r\mathrm{/}(x_1,\dots ,x_r{)}^q$ are the degree-$<q$ truncated polynomial rings. Their inverse limit is the formal power series ring $k\mathrm{⟦}{x}_1,\dots ,x_r\mathrm{⟧}$, because compatible truncations determine exactly one coefficient for every monomial and multiplication agrees coefficientwise. The completion results in the Given paragraph make this series ring Noetherian local of dimension $r$, with maximal ideal generated by its variables and residue field $k$.

Put $Q=k\mathrm{⟦}x,y\mathrm{⟧}$. Its quotient $A=Q\mathrm{/}(xy)$ is Noetherian local. Every prime of $Q$ containing $xy$ contains $x$ or $y$. Each quotient $Q\mathrm{/}(x)\cong k\mathrm{⟦}y\mathrm{⟧}$ and $Q\mathrm{/}(y)\cong k\mathrm{⟦}x\mathrm{⟧}$ has only the primes $(0)$ and its maximal ideal: every nonzero one-variable series is a unit times a power of its variable. Hence the primes of $A$ are $(x)A$, $(y)A$ and $(x,y)A$, and $\dim A=1$.

The cyclic $A$-module $M=A\mathrm{/}(x)\cong k\mathrm{⟦}y\mathrm{⟧}$ is nonzero and finite. Multiplication by $y$ is injective on it, and $M\mathrm{/}yM\cong k$ is nonzero, so ${\mathrm{depth}}_A(M)\ge 1$. If $a\in (x,y)A$ is any $M$-regular element, its image in $k\mathrm{⟦}y\mathrm{⟧}$ is $y^{t}u$ for some $t\ge 1$ and unit $u$. The nonzero class of $y^{t-1}$ in $M\mathrm{/}aM\cong k\mathrm{⟦}y\mathrm{⟧}\mathrm{/}(y^t)$ is annihilated by both $x$ and $y$, hence by the maximal ideal. Thus no second element of that ideal is regular on $M\mathrm{/}aM$, and ${\mathrm{depth}}_A(M)=1$.

By step 2.1 and the definition of maximal Cohen--Macaulay, ${\mathrm{depth}}_A(M)=\dim A=1$. The class of $x$ in $A$ is nonzero (reduce modulo $y$) but annihilates all of $M$; a nonzero free $A$-module has zero annihilator. Therefore $M$ is nonfree.

The depth quotient formula gives $\mathrm{depth}(Q\mathrm{/}(f))=\mathrm{depth}(Q)-1=d-1$.

The principal ideal theorem and regularity of $f$ give $\dim (Q\mathrm{/}(f))=d-1$. Thus the hypersurface quotient is Cohen--Macaulay.

The ring $A$ embeds in $S\mathrm{/}I\oplus S\mathrm{/}J$ by reduction modulo $I$ and $J$, because $I\cap J=IJ$. Put $\mathrm{\ell }=x+z$. Its image is $(z,x)$ in the two polynomial-domain quotients, so multiplication by $\mathrm{\ell }$ is injective on their direct sum and hence on $A$. Thus $\mathrm{depth}A\ge 1$.

More generally the embedding identifies $A$ with the fibre product of the two local polynomial rings $B=S\mathrm{/}I$ and $C=S\mathrm{/}J$ over their common residue field $k$: its elements are pairs $(f,g)$ with equal constant term. Let $a=(f,g)$ be any regular element of the maximal ideal of $A$. Since $B,C$ are domains, regularity forces $f,g\mathrm{\ne }0$, and both have zero constant term. The pair $(f,0)$ belongs to $A$ but is not in $aA$: an equation $a(h,j)=(f,0)$ would force $h=1,j=0$, while $(1,0)\mathrm{\notin }A$. For every maximal-ideal element $(u,v)$, however, $(u,v)(f,0)=(fu,0)=a(u,0)$ because $(u,0)\in A$. Thus the nonzero class of $(f,0)$ in $A\mathrm{/}aA$ is annihilated by the entire maximal ideal. No regular sequence can extend $a$, so $\mathrm{depth}A\le 1$. Step 1.1 gives the reverse bound. Both irreducible components have dimension $2$, so $\dim A=2$.

The exceptional clause in the definition applies to the zero module because $I0=0$. In the product-ring example, $I^2=I$, so $IM=I^2=M$.

lem-depth-infinity-when-ideal-acts-surjectively therefore assigns $+\mathrm{\infty }$ in both cases. This does not conflict with Nakayama: the ideal in the nonzero example is not contained in the Jacobson radical.

The maximal ideal of $R$ is $(t)$. Multiplication by $t$ is injective on $R$, so $(t)$ is an $R$-regular sequence of length one. For any nonzero $r\in (t)$, write $r=t^{m}u$ with $m\ge 1$ and $u$ a unit, using the first nonzero coefficient of the power series. In $R\mathrm{/}(r)$, the class of $t^{m-1}$ is nonzero and is annihilated by every element of $(t)$. Thus no regular sequence in $(t)$ has length two, and $\mathrm{depth}R=1$. The same annihilation gives ${\mathrm{depth}}_{R}k=0$.

The first sequence is split exact. The second and third sequences are exact because $t$ is a nonzerodivisor and its cokernel is $k$. Every element of $(t)$ annihilates the nonzero element $(1,0)$ of $k\oplus R$, so this middle module has depth zero.

Substitution of the depth triples $(0,0,1)$ and $(1,1,0)$ into the three bounds of The three Depth Lemma inequalities gives the displayed equalities, so no lower bound can be increased in general.

For $r=1,2,3$, the coordinate localization $P_r=k[x_1,\dots ,x_r{]}_{(x_1,\dots ,x_r)}$ is Noetherian local. Its dimension is $r$: the coordinate-prime chain has length $r$, while localization cannot increase the dimension $r$ of the polynomial ring. The quotients $P_r\mathrm{/}(x_1,\dots ,x_r{)}^q$ are the degree-$<q$ truncated polynomial rings. Their inverse limit is $k\mathrm{⟦}{x}_1,\dots ,x_r\mathrm{⟧}$, since compatible truncations give exactly one coefficient per monomial and multiplication agrees coefficientwise. By the completion results in the Given paragraph, this series ring is Noetherian local of dimension $r$, with maximal ideal generated by the variables and residue field $k$. It is a domain: the lowest nonzero homogeneous parts of two nonzero series multiply to a nonzero polynomial over the field $k$.

Put $Q=k\mathrm{⟦}x,y,z\mathrm{⟧}$. Its quotient $A=Q\mathrm{/}(xy)$ is Noetherian local. Every prime of $Q$ containing $xy$ contains $x$ or $y$; any chain of such primes lies above one of these two ideals and so has length at most $2$, because $Q\mathrm{/}(x)\cong k\mathrm{⟦}y,z\mathrm{⟧}$ and $Q\mathrm{/}(y)\cong k\mathrm{⟦}x,z\mathrm{⟧}$ both have dimension $2$. Either branch also supplies a chain of length $2$, hence $\dim A=2$.

Suppose $(x+y)g\in (xy)$ in $Q$. Reducing modulo $(x)$ gives $yg\equiv 0$ in the domain $k\mathrm{⟦}y,z\mathrm{⟧}$, so $g\in (x)$; reducing modulo $(y)$ similarly gives $g\in (y)$. Coefficientwise, $(x)\cap (y)=(xy)$ in $Q$, so $g\in (xy)$. Hence $x+y$ is a nonzerodivisor on $A$.

The quotient $A\mathrm{/}(x+y,z)\cong k\mathrm{⟦}x\mathrm{⟧}\mathrm{/}(x^2)$ is nonzero and has only one prime, $(x)$, since $x$ is nilpotent and the quotient by $(x)$ is $k$. Thus $(x+y,z)A$ has radical the maximal ideal $(x,y,z)A$. Together with step 2.1, this makes $x+y,z$ a system of parameters.

After quotienting by $x+y$ one has $A\mathrm{/}(x+y)\cong k\mathrm{⟦}x,z\mathrm{⟧}\mathrm{/}(x^2)\cong k\mathrm{⟦}z\mathrm{⟧}\oplus xk\mathrm{⟦}z\mathrm{⟧}$ as a $k\mathrm{⟦}z\mathrm{⟧}$-module. Multiplication by $z$ is injective on both summands, and the final quotient is nonzero by step 3.1. Therefore $x+y,z$ is a regular sequence on $A$, as claimed.

The quotient $A\mathrm{/}yA\cong k\mathrm{⟦}x\mathrm{⟧}\mathrm{/}(x^2)$ has dimension $0$, so $y$ is a parameter.

But the nonzero class of $x$ satisfies $yx=0$. Thus $y$ is a zero divisor and the parameter sequence is not regular, matching the depth gap $0<1$.

Put $Q=k\mathrm{⟦}x,y\mathrm{⟧}$. Its quotients by $(x)$ and $(y)$ are the domains $k\mathrm{⟦}y\mathrm{⟧}$ and $k\mathrm{⟦}x\mathrm{⟧}$. Also $(x)\cap (y)=(xy)$: a series divisible by both $x$ and $y$ has every monomial divisible by $xy$. Thus $(x)A$ and $(y)A$ are prime. In $A$, the annihilator of the nonzero class of $x$ is $(y)A$, and that of $y$ is $(x)A$, so both primes are associated.

Let $0\mathrm{\ne }a\in A$ be represented by $f\in Q$. If $f\in (x)\setminus (y)$, then $ga=0$ implies $gf\in (xy)\subseteq (y)$; reduction in the domain $Q\mathrm{/}(y)$ forces $g\in (y)$, and the reverse inclusion follows from $f\in (x)$. Hence ${\mathrm{Ann}}_A(a)=(y)A$. Symmetrically, if $f\in (y)\setminus (x)$, its annihilator is $(x)A$. If $f$ lies in neither ideal, reduction modulo each of $(x)$ and $(y)$ forces any annihilator $g$ into their intersection $(xy)$, so ${\mathrm{Ann}}_A(a)=0$. The remaining case $f\in (x)\cap (y)$ represents zero. Since $xy=0$ in $A$ with $x,y\mathrm{\ne }0$, the zero ideal is not prime. Therefore $(x)A$ and $(y)A$ are the only associated primes.

Finally $A\mathrm{/}(x)\cong k\mathrm{⟦}y\mathrm{⟧}$ and $A\mathrm{/}(y)\cong k\mathrm{⟦}x\mathrm{⟧}$, each of dimension $1=\dim A$ by the cited hypersurface example. This verifies unmixedness directly.

In either the localized polynomial ring or the formal power-series ring, if $(x+y)g=0$ modulo $(xy)$, reduction modulo $(x)$ and modulo $(y)$ gives $yg=0$ in the domain $k[y{]}_{(y)}$ or $k\mathrm{⟦}y\mathrm{⟧}$ and $xg=0$ in $k[x{]}_{(x)}$ or $k\mathrm{⟦}x\mathrm{⟧}$. Thus $g$ lies in both $(x)$ and $(y)$, whose intersection is $(xy)$ in each ambient ring. Hence $x+y$ is regular on both $A$ and $\widehat{A}$.

The regular element gives depth at least $1$, and the given dimension bound gives depth at most $1$, on both sides. Thus both depths equal $1$ directly.