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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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    @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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    @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