Colon ideals in $k[x,y]\mathrm{/}(x^2,xy)$ recover its associated primes

Let $\bar{x}=x+I$. Then $f\bar{x}=0$ exactly when $fx\in (x^2,xy)=x(x,y)$, that is, exactly when $f\in (x,y)$. Hence ${\mathrm{Ann}}_R(\bar{x})=(I:x)=(x,y)\mathrm{.}$

Let $\bar{y}=y+I$. Then $f\bar{y}=0$ exactly when $fy\in (x^2,xy)=x(x,y)$. Because $x$ and $y$ are relatively prime in $k[x,y]$, this happens exactly when $x$ divides $f$. Thus ${\mathrm{Ann}}_R(\bar{y})=(I:y)=(x)\mathrm{.}$ Both ideals are prime.

Every class in $R\mathrm{/}I$ has a unique representative $g(y)+cx$ with $g(y)\in k[y]$ and $c\in k$, because $x^2=xy=0$ modulo $I$. Let $r=g(y)+cx\mathrm{\notin }I$. If $g=0$, then $c\mathrm{\ne }0$ and $(I:r)=(x,y)$ by step 1.1. Assume $g\mathrm{\ne }0$. For $a(y)+dx\in R\mathrm{/}I$, multiplication gives $(a(y)+dx)(g(y)+cx)=a(y)g(y)+(a(0)c+dg(0))x\mathrm{.}$ If this vanishes, then $a(y)g(y)=0$ in the domain $k[y]$, so $a=0$. Thus $(I:r)=(x)$ when $g(0)=0$, while $(I:r)=I$ when $g(0)\mathrm{\ne }0$. The ideal $I$ is not prime because $x^2\in I$ but $x\mathrm{\notin }I$. Hence the only prime colon ideals are $(x)$ and $(x,y)$.

By [L1] and step 3.1, the associated primes of $R\mathrm{/}I$ are exactly $(x)$ and $(x,y)$.