Depth and Cohen Macaulay Modules — Examples

Since $\mathrm{Spec}(A)$ has one point, $\dim A=0$. The element $\epsilon$ is a zero divisor, so no positive-length regular sequence lies in the maximal ideal and $\mathrm{depth}A=0$.

Thus depth equals dimension. Equivalently, apply the general zero-dimensional corollary to the finite $A$-module $A$.

Iterating thm-polynomial-extension-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 $n$.

Hence $\dim A=\dim k\mathrm{⟦}y\mathrm{⟧}=1$. The nonzero class of $x$ is annihilated by both $x$ and $y$, so ${\mathrm{ann}}_A(x)=\mathfrak{m}$ and $\mathfrak{m}\in {\mathrm{Ass}}_A(A)$.

The depth-zero criterion gives $\mathrm{depth}A=0$. Since $0<1=\dim A$, the ring is not Cohen--Macaulay.

Put $Q=k\mathrm{⟦}x,y\mathrm{⟧}$. The sequence $x,y$ is a regular system of parameters of $Q$, so cor-one-regular-system-of-parameters-implies-cohen-macaulay makes $Q$ Cohen--Macaulay. The nonzero element $xy$ is $Q$-regular, and $(xy,x+y)$ is a system of parameters because $Q\mathrm{/}(xy,x+y)\cong k\mathrm{⟦}x\mathrm{⟧}\mathrm{/}(x^2)$ has finite length. Therefore thm-regular-quotients-and-cohen-macaulayness makes $A=Q\mathrm{/}(xy)$ Cohen--Macaulay of dimension one.

The classes of $x$ and $y$ are nonzero but their product is zero. Hence $A$ is Cohen--Macaulay although it is not a domain.

Multiplication by $y$ is injective on $M\cong k\mathrm{⟦}y\mathrm{⟧}$, and $M\mathrm{/}yM\cong k$ is nonzero. Thus ${\mathrm{depth}}_A(M)\ge 1$, while the dimension bound gives equality.

Hence ${\mathrm{depth}}_A(M)=\dim A=1$, so $M$ is maximal Cohen--Macaulay. It is not free: the nonzero element $x\in A$ annihilates all of $M$, whereas a nonzero free $A$-module has zero annihilator.

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 standard fibre-product sequence is $0\to A\to S\mathrm{/}I\oplus S\mathrm{/}J\to S\mathrm{/}(I+J)\to 0.$ The middle term has depth $2$, while the last term is $k$ and has depth $0$.

Since the middle depth is strictly greater than the quotient depth, the unequal-depth consequence of the Depth Lemma gives $\mathrm{depth}A=0+1=1$. 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 first sequence is split exact. The second and third sequences are exact because $t$ is a nonzerodivisor and its cokernel is $k$.

Substitution of the depth triples $(0,0,1)$ and $(1,1,0)$ gives the three equalities displayed above, so no lower bound can be increased in general.

The quotient by $x+y$ and $z$ is $A\mathrm{/}(x+y,z)\cong k\mathrm{⟦}x\mathrm{⟧}\mathrm{/}(x^2),$ which is nonzero and zero-dimensional. Thus $x+y,z$ is a system of parameters.

Every system of parameters on a Cohen--Macaulay module is regular, so $x+y$ is a nonzerodivisor on $A$ and $z$ is a nonzerodivisor on $A\mathrm{/}(x+y)$.

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 $0<1$.

In $A$, the annihilator of $x$ is $(y)$ and the annihilator of $y$ is $(x)$, so both primes are associated. The hypersurface is reduced with these two minimal primes; the Cohen--Macaulay associated-prime theorem rules out embedded primes.

Finally $A\mathrm{/}(x)\cong k\mathrm{⟦}y\mathrm{⟧}$ and $A\mathrm{/}(y)\cong k\mathrm{⟦}x\mathrm{⟧}$, each of dimension $1$. This verifies unmixedness directly.

If $(x+y)g=0$ modulo $(xy)$, reduction modulo $(x)$ and modulo $(y)$ shows that $g$ lies in both $(x)$ and $(y)$, hence in $(xy)$; thus $x+y$ is regular in $A$. Flat base change preserves its regularity in $\widehat{A}$.

The regular element gives depth at least $1$, and the dimension bound gives depth at most $1$, on both sides. This also illustrates the general completion depth equality.