1
$\begingroup$

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?

$\endgroup$
3
  • 2
    $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Apr 3 '19 at 0:17
  • $\begingroup$ 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 $(y)$. $\endgroup$ – SSF Apr 3 '19 at 3:06
  • 1
    $\begingroup$ Yes, that’s right. $\endgroup$ – Qiaochu Yuan Apr 3 '19 at 3:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.