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I've recently been reading about primary decomposition, and was browsing the questions here. From this, I know that it is not true that every primary ideal is the power of a prime ideal.

I'm curious about a slight variation inspired by the comments. Consider $K[X,Y]$, for $K$ a field. For any prime $\mathfrak{p}\subset K[X,Y]$, and any $n\geq 1$, is the ideal $\mathfrak{p}^n$ primary, so that if $ab\in\mathfrak{p}^n$, then either $a\in\mathfrak{p}^n$ or $b^j\in\mathfrak{p}^n$ for some $j$?

I suspect it is indeed true, and it is obviously true when $n=1$. An inductive approach doesn't feel right, so is there another way to reach the conclusion?

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up vote 7 down vote accepted

The answer is yes. Recall the following important fact:

Theorem. (Reid, Undergraduate Commutative Algebra, Proposition page 22) The prime ideals of $k[x,y]$ are as follows:

  1. $0$;
  2. $(f)$, for irreducible $f \in k[x,y]$;
  3. maximal ideals $\mathfrak{m}$.

Now you can easily conlude with the following two facts: (a) a power of a maximal ideal is primary (Atiyah, Macdonald 4.2); (b) in a UFD the power of a principal prime ideal is primary.

I don't know what happens in general in $k[x_1, \dots, x_n]$, when $n \geq 3$. It can be prooved that if $\mathfrak{p}$ is a monomial prime in $k[x_1, \dots, x_n]$ then $\mathfrak{p}^m$ is $\mathfrak{p}$-primary for all $m$.

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I doubt this holds for $n\geq 3$. – Alex Becker Feb 24 '12 at 20:27
Thanks Andrea. Do you have a proof or reference to a proof of the second fact (b) stated above? – Kally Feb 25 '12 at 1:16
It is quite easy. Try to prove it on your own. – Andrea Feb 25 '12 at 13:46
@Alex: you are right to doubt! See a non-primary square of a prime in Atiyah-Macdonald, page 51, Example 3). – Georges Elencwajg Sep 3 '15 at 6:42

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