# In $K[X,Y]$, is the power of any prime also primary?

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.

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?

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 conclude 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 proved 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 $m$.

• I doubt this holds for $n\geq 3$. Feb 24, 2012 at 20:27
• Thanks Andrea. Do you have a proof or reference to a proof of the second fact (b) stated above? Feb 25, 2012 at 1:16
• It is quite easy. Try to prove it on your own. Feb 25, 2012 at 13:46
• @Alex: you are right to doubt! See a non-primary square of a prime in Atiyah-Macdonald, page 51, Example 3). Sep 3, 2015 at 6:42
• @GeorgesElencwajg That doubt refers to the polynomial ring over a field in (at least) three variables. There is such an example in Northcott, Ideal Theory, page 29, Example 3. Oct 5, 2016 at 21:06