If $I = (xy,(x-y)z)$ is an ideal in $k[x,y,z]$ where $k$ is a field, then $\operatorname{rad} I = (xy,xz,yz)$.

Question

This is taken from Dummit & Foote, chapter 15.2, exercise 5. I am only quoting the part I am currently having trouble with:

If $I = (xy,(x-y)z)$ is an ideal in $k[x,y,z]$ where $k$ is a field, then $\operatorname{rad} I = (xy,xz,yz)$.

I have been able to show that $(xy,xz,yz) \subseteq \operatorname{rad}I$, but I am having trouble showing that $\operatorname{rad}I \subseteq (xy,xz,yz)$.

I want to show that if $r \in k[x,y,z]$ is such that $r^n \in (xy,(x-y)z)$ for some $n \in \mathbb{Z}^+$, then $r \in (xy,xz,yz)$. Clearly we have $r^n = r_1(xy)+r_2((x-y)z) = r_1(xy)+r_2(xz)-r_2(yz)$ for $r_1,r_2 \in k[x,y,z]$.

But then I get stuck and don't exactly know how to proceed to show that there exists $r'_1,r'_2,r'_3 \in k[x,y,z]$ such that $r = r'_1(xy)+r'_2(xz)+r'_3(yz)$.

Edit: I mean, we clearly have that $r^n \in (xy,xz,yz)$. Now, if I can show that $(xy,zx,yz)$ is a prime (which I am unsure if it is true), then I believe we are done, we can then conclude (in a finite number of steps) that $r \in (xy,xz,yz)$. The ideal is not a prime, since we have that $f = x$ and $g = y+z$ is such that $fg \in I$ but $f,g \not \in I$, if I am not mistaken (courtesy GPT-4).

Answer 1

For any monomial $a\in k[x,y,z]$, if $a$ involves more than one of $x,y,$ and $z$, then $a\in (xy,zx,yz)$ since $a$ must contain distinct neighboring formal variables. So it suffices to check univariate monomials.

If $r=x^k$ for some $k\in\Bbb N$, then $r^n\notin (xy,zx,yz) $ for any $n$, and similarly for $y^k$ and $z^k$. Then since each term of $r$ is either in the ideal $(xy,zx,yz)$ or is a univariate monomial, the binomial (or multinomial) theorem implies that $r$ cannot have a term of univariate monomial. In other words, $r\in(xy,zx,yz)$.

Answer 2

K is a field so $-k= + (-k) $ and in your terms $ r_{1}' = r_{1} $; $ r_{2}' = r_{2} $;$ r_{3}' = -r_{2} $;