Primary ideals in $\mathbb{Z}[\sqrt{-5}]$?

Let $$R$$ be a ring. An ideal $$I$$ of $$R$$ is primary if it is a proper ideal and if $$fg\in I$$ then $$f\in I$$ or $$g^n\in I$$ for some positive integer $$n$$.

How are the primary ideals of $$\mathbb{Z}[\sqrt{-5}]$$ categorised?

For example neither 2 or 3 are prime in this ring but $$\langle 3 \rangle$$ is not primary whereas $$\langle 2 \rangle$$ is. This is to be expected as primary is a weaker condition than prime, but which ideals exactly are primary?

In any Dedekind domain, the nonzero primary ideals are precisely the powers of prime ideals. Indeed, if $$I=P^n$$ is a power of a prime ideal and $$fg\in P^n$$, then we have $$P^n\mid (f)(g)$$ (divisibility of ideals), so either $$P^n\mid(f)$$ and $$f\in P^n=I$$, or $$P\mid(g)$$ and so $$g\in P,g^n\in P^n$$. Hence $$P^n$$ is primary.

Conversely, suppose $$I$$ is not a power of a prime ideal. By uniqueness of factorization, there are some relatively prime proper ideals $$A,B$$ such that $$I=AB$$. Relatively prime means that $$A+B=R$$, so $$a+b=1$$ for some $$a\in A,b\in B$$. Then we have $$ab\in AB=I$$, but no power of $$a$$ nor of $$b$$ is in $$I$$ (since no power of $$a$$ is in $$B$$ and vice versa).

In the particular case of $$R=\mathbb Z[\sqrt{-5}]$$, we note that $$(2)=(2,1+\sqrt{-5})^2$$ is a power of a prime ideal, while $$(3)=(3,1+\sqrt{-5})(3,1-\sqrt{5})$$ is a product of two distinct primes, so the former is primary while the latter is not.

We will need to find all the prime ideals of $$\mathbb Z[\sqrt{-5}]$$ and then find each $$P$$-primary ideal for each prime ideal $$P$$.

Let $$p$$ be a prime in $$\mathbb Z$$. The non-zero prime ideals of $$\mathbb Z[\sqrt{-5}]\cong \mathbb {Z}[X]/(X^2+5)$$ are $$\{(p,g(\sqrt{-5})):g(X)\text{ is an irreducible factor of }X^2+5\text{ in }\mathbb F_p[X]\}.$$ See this post for a justification of why this is true.

Finding the primary ideals is a little more tricky. Read through this MathOverflow question and the comments.

If $$X^2+5$$ is irreducible in $$\mathbb F_p[X]$$ then $$(p,g(\sqrt{-5}))$$ is the same as $$(p)$$, and in this case all the $$(p)$$-primary ideals are given by a power of $$(p)$$.

When $$X^2+5$$ is not irreducible mod $$p$$ but $$p^2$$ does not divide $$\text{disc}(X^2+5)=-20$$ then again all the $$(p,g(\sqrt{-5}))$$-primary ideals are powers of $$(p,g(\sqrt{-5}))$$.

This leaves only one more case. Since $$X^2+5$$ is reducible modulo $$2$$ with irreducible factor $$X+1$$ and $$2^2 \mid \text{disc}(X^2+5)$$ then all the powers of $$(2,\sqrt{-5}+1)$$ are still $$(2,\sqrt{-5}+1)$$-primary ideals, but there may be more $$(2,\sqrt{-5}+1)$$-primary ideals that are not of this form.

Unfortunately I am not sure how you would go about finding all the remaining $$(2,\sqrt{-5}+1)$$-primary ideals.

• This is a good answer, but I was hoping for something more constructive on how to show when individual ideals are or are not primary in this ring. For example the ideals generated by 2 and 3 in this question. As I know they are and are not but don't know why. Presumably something to do with the non-uniqueness of factorisation in this ring. May 2 '20 at 20:31
• It is hard to classify primary ideals even in a UFD. For example, if $P$ is a prime ideal of a UFD then it is not even necessarily true that $P^2$ is primary. Generally you would have to use an ad hoc approach to show ideals are primary. In your examples we have ideals of the form $(p)$ where $p$ is a prime of $\mathbb Z$. When $p\neq 2$ you have to find out if $X^2+5$ is irreducible mod $p$, which is easy to do. If it is irreducible then $(p)$ is primary, otherwise it is not. In the case of $(2)$ you can show it is equal to $(2,\sqrt{-5}+1)^2$ which we know to be primary. May 2 '20 at 21:08