Associated primes of the square of a monomial ideal

Question

Consider the ideal $J=(xy,yz,zx)$ in $R=\mathbb C[x,y,z]$. How to show that $(x,y,z) \in \mathrm{Ass}_R (R/J^2)$ ?

Answer 1

A down-to-earth method is: find an element whose annihilator is $(x, y, z)$.

An element that seems to work here is $xyz+J^2$. It is nonzero in $R/J^2$, since $J^2$ consists only of polynomials whose terms of degree $\leq 3$ are $0$.

We have \begin{align*} x \cdot xyz=x^2yz=(xy)(xz) \in J^2, \\ y \cdot xyz=xy^2z=(xy)(yz) \in J^2, \\ z \cdot xyz=xyz^2=(xz)(yz) \in J^2, \end{align*} showing that $(x, y, z) \subseteq \mathrm{Ann} (xyz + J^2)$. On the other hand, $(x, y, z)$ is a maximal ideal and $1 \notin \mathrm{Ann} (xyz + J^2)$, so we conclude equality.