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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?

2 Answers 2

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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 $n$.

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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 $a \, | \, x$).2012-02-25
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    It would be nice if you also mentioned that you implicitly used that a PID is also a UFD. But nice proof anyway. +12012-02-25
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    Thanks Brett. Also, when you write $a=p^aq^bc$, I assume the $a$s are different things, yes?2012-02-28
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    Yeah, bad choice of notation there.2012-02-29
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    @Patrick Da Silva: what property of UFD is being used here? Thanks2017-09-09
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    @Quand Dao : He used only the definition of a PID to have a complete characterization of the ideals of A (i.e. those with a single generator) and the definition of a UFD to factor those generators. He then shows that powers of prime ideals are primary and that those which are not powers of prime ideals are not primary.2017-09-10
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    @PatrickDaSilva: I do not understand why p divides y? Is there some property of UFD being used here?2017-09-10
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    Just write x and y as a power of $p$ times an element coprime to $p$ and use the assumptions2017-09-10
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    @PatrickDaSilva: So $xy = \lambda p^n$, and $x$ will be the power of $p$ times an unit, and since $x$ is not in $(a)$, the power of $p$ in $x$ will be less than $n$, therefore $p$ divides $y$?2017-09-10
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    @Quang Dao : You are not working in a local PID, only a PID. The element $x$ will be divisible by a certain power of $p$ but the exponent will be less than $n$. You should only be careful with the definitions and assumptions in the question!2017-09-10
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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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    I am not about what comes after "via", could you explain a bit further?2014-04-30
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    @Pedro Continue peeling prime powers $\ j\mid x^k = q^i y,\ j\nmid q^i\Rightarrow\, j\mid y^{k'},\,\ldots$2014-04-30
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    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.2014-04-30
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    Yes, I know that. But I don't know what proof you have in mind.2014-04-30
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    Details about the proof in your post. It'd be nice if you could give a usual worded proof.2014-04-30
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    @Pedro The hinted proof shows, by induction on $\,n =$ #prime factors of $\ a = p_1^{e_1}\cdots p_n^{e_n},\ p_i$ prime, that primary $\,j\mid a\,\Rightarrow\, j\mid p_i^k,\,$ for some prime $\,p_i\mid a.\,$ Choosing $\,a = j\,$ implies that $\,j\mid p_i^k.\ \ $2014-04-30