# When are powers of prime ideals primary?

This is a follow up to:

Normal domains and powers of height one primes

In the comments to the linked question, user26857 noted that the prime ideal $P = (x,z)$ in the Noetherian normal domain $k[x,y,z]/(z^2-xy)$ is such that even though $z^2=xy\in P^2 = (x^2,xz,z^2)$, $P^2$ doesn't contain $x$ or any power $y^n$, and hence $P^2$ isn't primary.

However, if we consider $k[x,y,z]/(z^2-xy)$ as a scheme over $k[y]$, then it is a flat family of affine genus 0 curves which is smooth except above $y = 0$, where the fiber is $k[x,z]/(z^2)$. From this perspective, the prime $P = (x,z)$ corresponds to a section of this family of curves.

Now, if one restricts this family to the locus $y \ne 0$, then we get a smooth family, and it's easy to see that argument which shows that $P^2$ is not primary fails when we work in the ring $k[x,y,z,y^{-1}]/(z^2-xy)$.

My question is then:

1. Is every power of the prime $(x,z)$ in $k[x,y,z,y^{-1}]/(z^2-xy)$ primary?

2. Let $A$ be a Noetherian normal domain, and $B$ a smooth $A$-algebra with dimension 1 fibers, and $P\subset B$ a locally principal prime ideal such that $B/P$ is finite etale over $A$. Then must powers of $P$ be primary?

Basically, my intuition is that the failure of $P^n$ to be primary stems from either the failure of $P$ to be locally principal and/or the failure of the fiber $y = 0$ to be smooth.

• Someone correct me: What about the localization at $\mathfrak{p}$? $\mathfrak{p}^n \subseteq A_{\mathfrak{p}}$ is a power of a maximal ideal, and hence is primary. Finally, primary ideals correspond through localization... Jul 4 '16 at 5:52
• @basket The problem is that the contraction of $p^nA_p$ might not be $p^n$ (it might be strictly larger than $p^n$) Jul 4 '16 at 5:57
• Thank you. very simple that one. Jul 4 '16 at 5:58

1. The ring in question is just $k[z,y,y^{-1}]$. Any prime ideal in this ring is either zero, principal or maximal. All these have the property that that powers are primary.
• If $P$ is a prime ideal, the $P$-primary component of $P^n$ is denoted by $P^{(n)}$, the symbolic power and $P^n\subset P^{(n)}$. Checking equality is local and locally, if $P$ is a principal prime, it is trivial. UFD does not play any role for principal prime ideals. Jul 4 '16 at 17:33
• I really don't see why if $P$ is a principal prime, then $P^n = P^{(n)}$. If it's so trivial perhaps you could edit it into your answer? Jul 4 '16 at 17:59
• Assume $P=(f)$. If $xy\in P^n$, then $f^n$ divides $xy$ and then $f$ divides $x$ or $y$, $f$ being a prime. I hope the rest is clear. Jul 4 '16 at 18:13