If the radical of an ideal is prime, must it be a prime power?

Easy fact: If an ideal is a prime power, then its radical is prime.

I'd like to give a counterexample to the converse. A good candidate is $$I=(y^2,xy) \subseteq K[x,y]$$, since this post shows that it has prime radical: Is it true that an ideal is primary iff its radical is prime?

How do I show that $$I$$ is not a prime power? I am not sure where to begin; I've done the following: suppose $$I=p^n$$ for some prime ideal $$p$$. I try localising at $$p$$: $$P_p^n=(y^2,xy)_p=(y)_p^2+(x)_p(y)_p$$. What next?

Aside: is there an easier counterexample?

• Try $I = (x^2, y^3)$. If it's a prime power it must be a power of its radical, namely $(x, y)$. So just show that it isn't. – Qiaochu Yuan Apr 3 '19 at 0:17
• I see. Just to see if I've got this right: since $I=p^n$, take radicals to get that $rad(I)=rad(p^n)=p$, so $I$ is a power of its radical. And this also works using my choice of $I=(y^2,xy)$, whose radical is $(y)$, right? In this case it's immediately clear that $I=(y^2,xy)$ is not a power of its radical $(y)$. – SSF Apr 3 '19 at 3:06
• Yes, that’s right. – Qiaochu Yuan Apr 3 '19 at 3:17