A redundant four-term decomposition cleans up to $(x)\cap (x,y{)}^2$

The ideal $(x)$ is prime because $R\mathrm{/}(x)\cong k[y]$. The quotients $R\mathrm{/}(x^2,y)\cong k[x]\mathrm{/}(x^2)$ and $R\mathrm{/}(x,y{)}^2$ are local rings whose maximal ideals are square-zero, so every zero divisor is nilpotent. Hence $(x^2,y)$ and $(x,y{)}^2$ are $(x,y)$-primary, while $(x)$ is $(x)$-primary. Thus the displayed intersection is a primary decomposition.

Since $(x,y{)}^2=(x^2,xy,y^2)\subseteq (x^2,y)$, the factor $(x^2,y)$ is redundant, and the repeated copy of $(x,y{)}^2$ is redundant as well. Fact [L1] therefore cleans the four-term intersection down to $I=(x)\cap (x,y{)}^2\mathrm{.}$

The two surviving radicals are $(x)$ and $(x,y)$, already distinct. If one first combines the two equal-radical components $(x,y{)}^2$ and $(x,y{)}^2$, [L2] replaces them by their intersection, which is again $(x,y{)}^2$. Hence both cleanup orders lead to the same two-term presentation.

Finally, $(x)\cap (x,y{)}^2=(x)\cap (x^2,xy,y^2)=(x^2,xy),$ because an element in the intersection has the form $xf$ with $f\in (x,y)$. The decomposition is irredundant: $x\in (x)\setminus (x,y{)}^2$ and $y^2\in (x,y{)}^2\setminus (x)$. Together with the distinct radicals, this proves minimality.

This example isolates the two routine cleanup moves in a primary decomposition: deletion of redundant components and combination of equal radicals.