A Cohen--Macaulay ring with zero divisors

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.