Factorization of ideals in number fields of the form $K=\mathbb{Q}(\alpha,\beta)$

Question

Given a number field $K=\mathbb{Q}(\alpha)$, I am able to factorize ideals of the form $p\mathcal{O}_K$ where $p$ is some unramified prime, via the Dedekind Kummer theorem.

Now I have two questions:

1. How can I find the factorization of ideals $p\mathcal{O}_K$ for some $K=\mathbb{Q}(\alpha,\beta)$?
2. In particular, let $K=\mathbb{Q}(\sqrt[4]{2},i)$ the splitting field of $X^4-2$. Then, how can I factorize $3\mathcal{O}_K$? I know that $$X^4-2 \equiv (X^2+3X+2)(X^2+2X+2) \pmod{3}$$

Answer 1

Concerning 1: One possibility is to find a primitive element over $\Bbb Q$ and apply the Dedekind-Kummer theorem that you mention. However, when you have a large degree algebraic number field, especially if it's the Galois closure of a smaller degree field, it's sometimes possible to 'bootstrap' the factorization from information about factorizations in certain subfields (without finding a primitive element and its minimal polynomial).

Concerning 2: Consider the subfield $\Bbb Q(i)$ with ring of integers $\Bbb Z[i]$. $3$ is inert in this subfield. The discriminant of $X^4-2$ is coprime to $3$. So we can apply the more general Dedekind-Kummer theorem for Dedekind domains (which is also in the linked Wikipedia page) for the extension $\mathcal O_{\Bbb Q(\sqrt[4]{2},i)}/ \Bbb Z[i]$.

So we consider the factorization of $X^4-2$ over $\Bbb Z[i]/3\Bbb Z[i] \cong \Bbb F_9=\Bbb F_3[i]$. Well, let's first factorize it over $\Bbb F_3$. We get $X^4-2=(X^2+X+2)(X^2+2X+2)$

Now we have to factor $X^2+X+2$ which can be done, for example, with the quadratic formula. We obtain $X^2+X+2=(X-i-1))(X-i+1))$. Similarly, we find that $(X^2+2X+2)=(X+i+1)(X+i-1)$. All in all, we obtain $$X^4-2=(X+i+1)(X+i-1)(X-i+1)(X-i-1)$$ Note that all these factorizations are in $\Bbb F_3[i]$.

Finally, applying the general Dedekind-Kummer theorem, we get a factorization:

$$3\mathcal O_{\Bbb Q(\sqrt[4]{2},i)}=(3,\sqrt[4]{2}+i+1)(3,\sqrt[4]{2}+i-1)(3,\sqrt[4]{2}-i+1)(3,\sqrt[4]{2}-i-1)$$ (All these factors are unramified and have inertial degree $2$)

Answer 2

Your assertion "I am able to factorize ideals of the form $p\mathcal O_K$ where $p$ is some unramified prime, via the Dedekind Kummer theorem" is not true as generally as you claim.

Example. When $K = \mathbf Q(\alpha)$ where $\alpha$ is a root of $x^3 - x^2 - 2x - 8$, you can't use Dedekind Kummer to show $2$ is unramified in $K$, or the more precise statement that $2\mathcal O_K = \mathfrak p\mathfrak p'\mathfrak p''$ where the prime ideal factors on the right are distinct, since $2 \mid [\mathcal O_K:\mathbf Z[\gamma]]$ for every $\gamma$ in $\mathcal O_K - \mathbf Z$.