Compute $\text{Ass}(R/p^2)$

Problem: Consider the ring $R=\mathbb{Q}[X,Y,Z]/\langle{XY-Z^2}\rangle$, and the ideal $p=\langle{x,z}\rangle$, where $x,y,z$ are the residues of $X,Y,Z$. Show that
($1$) $p$ is prime;
($2$) $p^2$ is not primary;
($3$) $p^2=\langle{x}\rangle\cap\langle{x,y,z}\rangle^2$;
($4$) and calculate $\text{Ass}(R/p^2)$.

Attempt: To show that $p$ is prime I take the quotient ring $R/p$ and prove it is a domain. But I don't really know how to work this out. How do I simplify $R/p$?

That $p^2$ is not primary, I don't see a way to find this.

I note that $xy=z^2$, therefore $p^2=\langle{x^2,xz,z^2}\rangle=\langle{x^2,xz,xy}\rangle=x\langle{x,z,y}\rangle$. So, $p\subset\langle{x}\rangle$ and $p\subset\langle{x,y,z}\rangle^2$, since $\langle{x,y,z}\rangle^2=\langle{x^2,xy,xz,y^2,yz,z^2}\rangle$. Hence, $p\subset \langle{x}\rangle\cap\langle{x,y,z}\rangle^2$. For the opposite inclusion, I am also without ideas.

Lastly, what I think I need to do here is prove that $(3)$ is a primary decomposition of $p^2$. Then $\text{Ass}(R/p^2)\subset\{p_1,p_2\}$, where $p_1=\langle{X}\rangle$, since $\langle{X}\rangle$ is prime, and $p_2=\text{Ass}(R/\langle{x,y,z}\rangle^2)$.

I appreciate any help. Thanks in advance!

Solution 1:

To prove $p$ is prime as you said we can take the quotient $R/p$. Another way to think of taking $\Bbb{Q}[x,y,z]/\langle x,z\rangle$ which is isomorphic to $\Bbb{Q}[y]$ (since taking the quotient is the same as saying that all element of the form $xf+zy$ are now $0$ (of the quotient), which is a domain.

$p^2=\langle x^2,z^2,xz\rangle=\langle x^2,xy,xz\rangle$. To prove this is not a primary ideal, notice that $xy\in\langle x^2,xy,xz \rangle$ but neither $x$ nor $y^n$ are in $p^2$.

To prove the other inclusion, notice that it's often easier to think of ideals as span (especially when generators are monomials), meaning $p^2=\langle x^2,xy,xz\rangle$=$$\operatorname{Span}_\Bbb{C}\{x^iy^jz^l:i\geq2,\text{or }i=1\text{ and }l=1,\text{or } i=1 \text{ and } z=1\}.$$ Use this method to write the other ideal which is easy and figure out the intersection to prove the equality.

Lastly, by definition $\mathrm{Ass}(R/p^2)$ is the radical of each ideal in the primary decomposition which is not too hard to compute - the radical of $\langle x\rangle$ is himself since it's prime. $\langle x,y,z\rangle^2=\langle x^2,y^2,z^2,xy,xz,yz\rangle$, and computing the radical gives us $\langle x,y,z\rangle$. So those give us the associated primes as needed.

Edit: To calculate the radical of $\langle x^2,y^2,z^2,xy,xz,yz\rangle$ one usually first gets rid of all powers of generators (since they are monomials) and gets $\langle x,y,z,xy,xz,yz\rangle=\langle x,y,z\rangle$. If you want to show it's maximal, you can simply look at $\Bbb{Q}[x,y,z]/\langle x,y,z\rangle$ which is isomorphic to $\Bbb{Q}$, a field. One should be careful and notice that this only works since $\Bbb{Q}[X,Y,Z]$ is a finitely generated algebra (by $X,Y,Z$) then $\Bbb{Q}[X,Y,Z]/\langle XY-Z^2\rangle$ is a finitely generated algebra by $x,y,z$ - following your notation. So, $(\Bbb{Q}[X,Y,Z]/\langle XY-Z^2\rangle)/\langle x,y,z\rangle$ is isomorphic to $\Bbb{Q}$ as I mentioned and therefore $\langle x,y,z\rangle$ is maximal.

I didn't give all the details in my proof the ofcourse leave some of it for you but I hope this does clarify things a little.