# Characterization of primary ideals in a principal ideal domain?

On the commutative algebra wiki, a table of properties lists that "for a PID, the primary ideals coincide with the powers of prime ideals."

I played around with it, couldn't produce a proof, and have been searching around for a proof, since I'm sure this is a standard fact. I couldn't find a reference online. Can someone please provide a proof, or reference where I can read such a proof?

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You can identify an ideal with its generator. If $x \in (a)$ if and only if $a|x$. Suppose $a=p^n$. If $x \notin (a)$ but $xy \in (a)$, since $p^n | xy$, $p|y$, hence $p^n|y^n$, and $y^n \in (a)$

If $a=p^aq^bc$, where $c$ is any element of the ring coprime to the primes $p$ and $q$, then let $x=p^a$ and $y=q^bc$. Then $xy\in (a)$ but $x^n$ and $y^n$ are not in $(a)$ for any $n$.

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You wrote $x \in (a)$ if and only if $x \, | \, a$, but you did not seem to follow this afterwards, so I guess it's a typo (you meant $a \, | \, x$). –  Patrick Da Silva Feb 25 '12 at 22:25
Thanks. The post has now been corrected. –  Brett Frankel Feb 25 '12 at 22:28
It would be nice if you also mentioned that you implicitly used that a PID is also a UFD. But nice proof anyway. +1 –  Patrick Da Silva Feb 25 '12 at 22:30
Thanks Brett. Also, when you write $a=p^aq^bc$, I assume the $a$s are different things, yes? –  Jakucha Feb 28 '12 at 21:08
Yeah, bad choice of notation there. –  Brett Frankel Feb 29 '12 at 3:25

Hint $\$ Peel off prime factors of an element $\ne 0$ in $\rm\:\! J = (j)\:$ till only one prime $\rm\:q\:$ remains, via

$$\rm\ j\ |\ p^n\: x,\ \ j\nmid p^n\ \Rightarrow\ \ j\ |\ x^k \ \Rightarrow\ \cdots\ \Rightarrow\ \ j\ |\ q^m,\quad p,\:q\ \ prime$$

More generally, a similar proof shows that the radical of a primary ideal is prime.

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