Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 5 down vote accepted

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

If $a=p^aq^bc$, where $c$ is any element of the ring coprime to the primes $p$ and $q$, $p\ne q$, 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 $n$.

share|cite|improve this answer
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? – Jacqueline Pauwels 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.

share|cite|improve this answer
I am not about what comes after "via", could you explain a bit further? – Pedro Tamaroff Apr 30 '14 at 18:19
@Pedro Continue peeling prime powers $\ j\mid x^k = q^i y,\ j\nmid q^i\Rightarrow\, j\mid y^{k'},\,\ldots$ – Bill Dubuque Apr 30 '14 at 18:40
You mean use a PID is noetherian? I am not sure about what the details of the proof would be. I would appreciate if you could include a full proof. I am interested in the direct approach. – Pedro Tamaroff Apr 30 '14 at 18:46
Yes, I know that. But I don't know what proof you have in mind. – Pedro Tamaroff Apr 30 '14 at 18:48
Details about the proof in your post. It'd be nice if you could give a usual worded proof. – Pedro Tamaroff Apr 30 '14 at 18:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.