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In a commutative ring, if one takes a primary ideal $I$, then $\sqrt I$ is prime. It is not true in general that an ideal with such property is primary. For example, given a prime ideal $\mathfrak p$, one has that the radical of $\mathfrak p^n $ is $\mathfrak p$, clearly, but a $\mathfrak p^n $ is not always primary. The notes from which I'm studying prove that, in $\mathbb Z$, every ideal $I$ such that $\mathfrak q:=\sqrt I$ is prime is a power of $\mathfrak q$. Then they conclude noticing that every power of a prime ideal is primary; however, it seems to me that they only proved that in $\mathbb Z $, for an ideal, being a power of a prime is equivalent to have the property that its radical is prime. This proves that every primary is a power of a prime (in $\mathbb Z$) but not the converse, which is what we need. What am I missing? Thanks

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    $\begingroup$ You can note that all non-zero prime ideals in $\mathbb{Z}$ are maximal - and powers of maximal ideals are primary (see Atiyah-Macdonald) $\endgroup$ Oct 31 '20 at 16:44
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Take $n=p_1^{a_1}p_2^{e_2}\dotsm p_k^{e_k}$. Then it's not difficult to show that $\sqrt{\smash[b]{(n)}}=(p_1p_2\dots p_n)$, so the radical is prime if and only if $n$ is a prime power.

Conversely, $(p)^e=(p^e)$ being a $(p)$-primary ideal is easy to show.

It is generally false that the power of a prime ideal is primary, but it can hold for particular prime ideals. Indeed, if $p$ is a prime element of a domain, then $(p^e)=(p)^e$ is $(p)$-primary.

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  • $\begingroup$ Thanks, there is only a thing not too clear to me; as you said, it's easy to verify that every power of a prime ideal $\mathfrak p\subset \mathbb Z$ is $\mathfrak p$-primary. However from my notes it seems that this thing can be deduced from the fact that, in $\mathbb Z$, $\sqrt I$ $\mathfrak q$-prime implies $I=\mathfrak q^n$. To me, it seems that this fact is not needed to prove that all powers of prime ideals in $\mathbb Z$ are primary (but one can actually use this fact to prove that in $\mathbb Z$ being primary and being the power of a prime is equivalent). Do you confirm? $\endgroup$
    – Dorian
    Oct 31 '20 at 18:15
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    $\begingroup$ @Dorian Why not checking by the definition? Consider $xy\in(p^e)$, for $e\ge1$. Then $p\mid x$ or $p\mid y$ and so either $x^e\in(p^e)$ or $y^e\in(p^e)$. $\endgroup$
    – egreg
    Oct 31 '20 at 18:41
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We can prove that any power of a prime in $\mathbb{Z}$ is primary by hand. Let $I=(p^n)$ for a prime $p\in\mathbb{Z}$. Suppose $xy\in I$, but $x\not\in I$. Since $xy\in I$, $p^n m = xy$ for some $m$. $p^n$ divides the LHS, so $p^n$ must divide the RHS. Since $x\not\in I$, we have $p^j\mid x$ and $p^k\mid y$ for some $j$ strictly less than $n$ and some $k$ strictly greater than 0. Thus $y^n\in (p^{kn})\subset (p^n)$.

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