A nonfree maximal Cohen--Macaulay module

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.