# Clarify about primary ideals in a commutative ring

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

• You can note that all non-zero prime ideals in $\mathbb{Z}$ are maximal - and powers of maximal ideals are primary (see Atiyah-Macdonald) Oct 31 '20 at 16:44

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.
• 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? Oct 31 '20 at 18:15
• @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 $y^e\in(p^e)$. Oct 31 '20 at 18:41
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 $$y^n\in (p^{kn})\subset (p^n)$$.