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 !
Best Answer
Are you familiar with this theorem?
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.