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?

$\endgroup$
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 '20 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 '20 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.