# Norm of ideals in quadratic number fields

I do not really understand how to factor ideals in a quadratic field $K = \mathbb{Q}(\sqrt{d})$, mainly because I have some trouble computing the norm of ideals. I think I understand what is going on when $I=(a)$ is principal, as it is then easy to calculate its norm (then using Legendre symbol we can determine if it's split, inert or ramified and deduce how it factorises).

I have tried several methods to find the norm of an ideal of the form $(a,b)$ but they all seem to fail. For example, is it true that if $I=(a,b)$, then $N(I) \mid N(a)$ ? Because if it is true, then for example the norm of $(22,2+\sqrt{-7})$ should divide $11$ hence can only be $11$ (as it is not $1$), which is the correct answer. This method seems to work sometimes, but intuitively I would say the statement I made above is not necessarily true.

Also, there is another approach which might work, but I do not know how to apply it. We know that $N(I) = | \mathcal{O}_K/I |$, which works nicely when $I$ is principal, but how can we apply this to $(22,2+\sqrt{-7})$ say ?

I am also interested in any other method by the way. Thank you very much for your help !

• If $I=(a,b)$, then $(a)\subset I$, so $I\mid(a)$, and yes, $N(I)\mid N(a)$. Similarly, $N(I)\mid N(b)$. When has this method not worked for you? Commented Apr 19, 2013 at 12:07
• The statement is true because the norm is multiplicative $NIJ) = N(I)N(J)$. To contain is to divide for ideas here so your statement follows. Commented Apr 19, 2013 at 12:09
• Also, when you say you're having trouble factoring ideals, I guess this is factoring ideals of $\mathcal{O}_K$ into prime ideals? Commented Apr 19, 2013 at 12:15
• @WarrenMoore Sorry I meant that this method is not always conclusive, as it might leave several divisors to be checked (for example if it gives $N(I) \mid 30$ we cannot deduce the norm from just this). In any case thank you for your reply ! EDIT: oops, and yes I mean factorising ideals of $\mathcal{O}_K$ into prime ideals. Commented Apr 19, 2013 at 12:15
• Ahh okay. Well I'll certainly write up an answer to try to help with the factoring of ideals! Commented Apr 19, 2013 at 12:17

Are you familiar with this theorem?

Theorem. (Kummer-Dedekind) Let $K$ be a number field with ring of integers $\mathcal{O}_K$, and let $Z[\alpha]$ be an order of $K$, defined by a integral generator of $K/\mathbb{Q}$. Let $f$ be the minimal polynomial of $\alpha$, $p$ a rational prime, and let $\overline{f}$ denote the reduction of $f$ modulo $p$. Let $\overline{f}=\overline{g}_1^{e_1}\cdots\overline{g}_k^{e_k}$ be the factorisation of $\overline{f}$ into powers of distinct irreducibles $\overline{g}_i\in(\mathbb{Z}/p\mathbb{Z})[x]$. Then:

1. The prime ideals of $\mathbb{Z}[\alpha]$ over $p$ are in bijection with the irreducible factors $\overline{g}_i$ of $\overline{f}$, where each factor $\overline{g}_i$ corresponds to a prime ideal $\mathfrak{p}_i=(p,g_i(\alpha))$, where $g_i\in\mathbb{Z}[x]$ is any lift of $\overline{g}_i$.

2. Writing $f=q_ig_i+r_i$, with $q_i,r_i\in\mathbb{Z}[x]$, the prime ideal $\mathfrak{p}_i$ is non-invertible if $e_i>1$ and $r_i\in p^2\mathbb{Z}[x]$.

3. $(p)\supseteq\mathfrak{p}_1^{e_1}\cdots\mathfrak{p}_k^{e_k}$, with equality if and only if all the $\mathfrak{p}_i$ are invertible. Then $p\nmid|\mathcal{O}_K:\mathbb{Z}[\alpha]|$, and $N(\mathfrak{p}_i)=p^{\deg(g_i)}$.

4. If $\mathfrak{p}_i$ is non-invertible, then $\frac{q_i(\alpha)}{p}\in\mathcal{O}_K\setminus\mathbb{Z}[\alpha]$.

Now this is probably considerably more than you need, but it's a useful theorem for computing a plethora of things, not just the prime ideals of $\mathcal{O}_K$, but also (along with some other useful results) class groups and class numbers.

So suppose that $I=(a,b)\lhd\mathcal{O}_K$. Then if a prime ideal $\mathfrak{p}\mid I$, then $a,b\in\mathfrak{p}$. Let $(p)=\mathfrak{p}\cap\mathbb{Z}$. Then by the multiplicity of the norm, $N(\mathfrak{p})$ (which is a power of $p$ by the above theorem) divides $N(a)$ and $N(b)$. This means that the primes dividing $I$ are above the rational primes dividing $\gcd(N(a),N(b))$, which gives you a finite list of primes to check.

Simplifying further, suppose that $\mathcal{O}_K=\mathbb{Z}[\alpha]$. Since you are primarily concerned with the quadratic case, for $K=\mathbb{Q}(\sqrt{d})$:

$$\mathcal{O}_K=\begin{cases} \mathbb{Z}[\sqrt{d}] & \text{ if }d\equiv 2,3\text{ mod }4 \\ \mathbb{Z}[\frac{1+\sqrt{d}}{2}] & \text{ if }d\equiv 1\text{ mod }4 \end{cases}$$

So we can always write $\mathcal{O}_K$ in terms of a single integral element. Then find the possible $\mathfrak{p}=(p,g(\alpha))$. Then $\mathfrak{p}\mid I$ if and only if $a,b\in(\overline{g})\lhd(\mathbb{Z}/p\mathbb{Z})[\alpha]$. This can be checked by evaluating $a$ and $b$ under the homomorphism $\mathcal{O}_K\rightarrow(\mathbb{Z}/p\mathbb{Z})[\alpha]$ corresponding to $\mathfrak{p}$.

The only real difficulty comes when a prime $\mathfrak{p}$ has multiplicity greater than 1. But this is usually resolvable by using a norm argument.

• Thank you very much ! I had seen some parts of this theorem but in a specific setting only. Commented Apr 19, 2013 at 13:15