Given a prime ideal $\mathfrak p\subset k[x,y,z]$, is $\mathfrak{p}^2$ a primary ideal?

Question

As mentioned, we have $R = k[x,y,z]$, $\mathfrak{p} = (xz - y^2, x^3 - yz, x^2y - z^2)$, and we want to determine if $\mathfrak{p}^2$ is $\mathfrak{p}$-primary.

So my first thought was no, mostly because I think the professor is pointing out that $\mathfrak p$ being prime (which I have already shown) does not imply powers of $\mathfrak p$ are $\mathfrak p$-primary. However, I cannot seem to develop a solid argument. I tried finding an element whose multiplication map from $R/\mathfrak p^2$ to itself is neither injective nor isomorphic, but I don't know how to pick a clever enough element. I also tried thinking about elements $ab\in \mathfrak p^2$ such that $a\notin \mathfrak p$ and $b^n\notin \mathfrak p$ for any $n$, but again I don't see how to come up with these. I think I would maybe want $b = z$, and come up with a crazy $a$ somehow. I even tried looking at some associated prime stuff, but that did not seem too fruitful.

Now I am second guessing my intuition like I am trying to fit a square peg in a round hole.

Any hints/suggestions are appreciated. Thanks.

Answer 1

In M2:

R=QQ[x,y,z] 
I=ideal(x*z - y^2, x^3 - y*z, x^2*y - z^2) 
I^2 -- ideal(y^4-2*x*y^2*z+x^2*z^2,-x^3*y^2+x^4*z+y^3*z-x*y*z^2,-x^2*y^3+x^3*y*z+y^2*z^2-x*z^3,x^6-2*x^3*y*z+y^2*z^2,x^5*y-x^2*y^2*z-x^3*z^2+y*z^3,x^4*y^2-2*x^2*y*z^2+z^4) 
primaryDecomposition I^2 -- {ideal(y^4-2*x*y^2*z+x^2*z^2,x^2*y^3-x^3*y*z-y^2*z^2+x*z^3,x^3*y^2-x^4*z-y^3*z+x*y*z^2,x^5+x*y^3-3*x^2*y*z+z^3), ideal(z,y^4,x^2*y^3,x^3*y^2,x^5*y,x^6)}

So $V(I^2)$ has an embedded fat point at the origin in ${\Bbb A}^3$.