Associated primes are unmixed in a Cohen--Macaulay example

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.