3
$\begingroup$

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ May 2, 2020 at 20:31
  • $\begingroup$ 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. $\endgroup$
    – Fortox
    May 2, 2020 at 21:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .