Two minimal decompositions of $(x^2,xy)$ share radicals but not the embedded component

The inclusion $(x^2,xy)\subseteq (x)\cap (x,y{)}^2$ is immediate, and every element of the right side has the form $xf$ with $f\in (x,y)$, so $(x^2,xy)=(x)\cap (x,y{)}^2$. Likewise $(x^2,xy)\subseteq (x)\cap (x^2,y)$, and if $xf\in (x^2,y)$ then $xf=x^{2}a+yb$ for some $a,b$. The right-hand side lies in $(x)$, so $yb\in (x)$; thus $b=xc$ and $xf=x(xa+yc)\in (x^2,xy)$. Hence $(x^2,xy)=(x)\cap (x^2,y)$.

The field $k$ is Noetherian because its only ideals are $0$ and $k$, so [L3] makes $R=k[x,y]$ Noetherian. The ideal $(x)$ is prime because $R\mathrm{/}(x)\cong k[y]$. The quotients $R\mathrm{/}(x,y{)}^2$ and $R\mathrm{/}(x^2,y)\cong k[x]\mathrm{/}(x^2)$ are local rings with square-zero maximal ideals, so $(x,y{)}^2$ and $(x^2,y)$ are $(x,y)$-primary. Both decompositions are irredundant: $x\in (x)$ lies in neither $(x,y{)}^2$ nor $(x^2,y)$, while $y^2\in (x,y{)}^2\setminus (x)$ and $y\in (x^2,y)\setminus (x)$. Hence both displayed decompositions are minimal primary decompositions with prime radicals $(x)$ and $(x,y)$.

Fact [L1] now predicts exactly the common radical set from step 2.1. The second component differs: one decomposition uses $(x,y{)}^2$, the other uses $(x^2,y)$.

Localizing at the minimal prime $(x)$ kills the $(x,y)$-primary component in either decomposition, so [L2] recovers the same isolated component $(x)$ from both. The difference therefore lies only in the embedded component.

This is the standard warning that first uniqueness does not imply componentwise uniqueness of embedded pieces.