Localizing $(x^2,xy)=(x)\cap (x,y{)}^2$ keeps only the matching component

The field $k$ is Noetherian because its only ideals are $0$ and $k$, so [L3] makes $R=k[x,y]$ Noetherian. As an $R$-module, $R$ is finitely generated by $1$. The inclusion $(x^2,xy)\subseteq (x)\cap (x,y{)}^2$ is immediate. Conversely, every element of $(x)\cap (x,y{)}^2$ has the form $xf$ with $f\in (x,y)$, hence lies in $(x^2,xy)$. So $(x^2,xy)=(x)\cap (x,y{)}^2$.

Since a field is an integral domain, [L4] makes $k[y]\cong R\mathrm{/}(x)$ an integral domain, so [L5] makes $(x)$ prime. If multiplication by a class $\bar{a}$ on the domain $R\mathrm{/}(x)$ has nontrivial kernel, then $\bar{a}=0$; hence that multiplication is zero and therefore nilpotent. Also ${\mathrm{Ann}}_R(R\mathrm{/}(x))=(x)$, whose radical is $(x)$. Thus [L6] shows that $(x)$ is $(x)$-primary.

Put $Q=(x,y{)}^2$. Every class in $R\mathrm{/}Q$ has the form $c+\alpha \bar{x}+\beta \bar{y}$. If $c\mathrm{\ne }0$, this class is a unit, with inverse $c^{-1}-c^{-2}(\alpha \bar{x}+\beta \bar{y})$ because $(\bar{x},\bar{y}{)}^2=0$. Therefore every zero divisor lies in $(\bar{x},\bar{y})$ and is square-zero, so every zero divisor on $R\mathrm{/}Q$ acts nilpotently. Moreover ${\mathrm{Ann}}_R(R\mathrm{/}Q)=Q$ and $\sqrt{Q}=(x,y)$: every element of $(x,y)$ has square in $Q$, while $f^n\in Q$ forces the constant term of $f$ to vanish. Finally, $R\mathrm{/}(x,y)\cong k$ is a domain, so [L5] makes $(x,y)$ prime. Thus [L6] shows that $(x,y{)}^2$ is $(x,y)$-primary.

The two prime radicals $(x)$ and $(x,y)$ are distinct, with $(x)⊊(x,y)$. The decomposition from step 1.1 is irredundant: $y^2\in (x,y{)}^2\setminus (x)$, so the component $(x)$ is not redundant, while $x\in (x)\setminus (x,y{)}^2$, so the component $(x,y{)}^2$ is not redundant. By [L7], the displayed primary decomposition is minimal, and its $(x)$-primary component is isolated because $(x)$ is the smaller of the two radicals.

At the prime $(x)$, the element $y$ becomes a unit. Since $y^2\in (x,y{)}^2$, fact [L1] and steps 1.1–2.1 give $((x,y{)}^2{)}_{(x)}=R_{(x)}$, while $(x{)}_{(x)}$ survives. Hence $(x^2,xy{)}_{(x)}=(x{)}_{(x)}\mathrm{.}$ Since step 2.1 proves that $(x)$ is the isolated component of a minimal primary decomposition, fact [L2] says it contracts back to $(x)$.

At the maximal ideal $(x,y)$, neither prime radical meets the denominator set, so [L1] and steps 1.1–2.1 preserve both primary components. Thus the whole decomposition survives in $R_{(x,y)}$.

This computation shows concretely how localization removes exactly the components whose radicals meet the denominator set.