Why Does (3) Completely Ramify in $\mathbb{Q}(\omega, \sqrt[3]{2})$?

Question

I'm working through this notes on Algebraic Number Theory. In section 2.6 they claim (3) is completely ramified over the larger field:

If we know that the ring of integers of $\mathbb{Q}(\omega, \sqrt[3]{2})$ is in fact, $\mathbb{Z}[\omega, \sqrt[3]{2}]$, then we can just apply Kummer-Dedekind (proven in the previous section). But it looks like they don't want to use that fact here (and I'm not sure if that's even true).

From reading the notes, I was able to see that:

1. $(1-\omega)$ is a prime ideal in the ring of integers of $K=\mathbb{Q}(\omega)$, and that $(3)=(1-\omega)^2$
2. Because $N_{L/K}(1+\sqrt[3]{2})=3$ (where $L=\mathbb{Q}(\omega, \sqrt[3]{2})$), there exists a prime $P$ in the ring of integers of $L$ such over $(1-\omega)$.
3. Because the extension $L/K$ is Galois, and $Q=(1-\omega, 1+\sqrt[3]{2})$ is invariant under Gal($L/K$), every prime ideal over $1-\omega$ must in fact, contain $Q$ (this follows from transitivity of the action of Gal($L/K$) on the set of primes over $(1-\omega)$.

My questions are:

4. How do we know that $Q$ is prime in the ring of integers of $L$? (remember, the notes seems to avoid wanting to characterize the ring of integers of $L$ in a concrete way, so how do we know this?) I know if we do know $Q$ is prime, then $Q$ is the only prime over $(1-\omega)$.
5. How do we know that $Q$ must have ramification index $3$ over $(1-\omega)$? If we know the explicit characterization of the ring of integers of $L$, then I can see how we can show the residue degree is $1$, therefore, ramification index $3$?

Thank you all

Answer 1

The subtlety is that $\frac{\sqrt{-3}}{2^{1/3}+1}$ is an algebraic integer whose minimal polynomial is of degree $6$ and Eisenstein at $3$.

Showing that every prime numbers factorize in products of prime ideals we get that $$O_K = \Bbb{Z}[\omega,2^{1/3},\frac{\sqrt{-3}}{2^{1/3}+1}]$$

(if $p\ge 5$ then $p$ is unramified in $\Bbb{Z}[\omega,2^{1/3}]$ so it factorizes in prime ideals, then $(2)=(2^{1/3})^3$ and $(3)=(3,\frac{\sqrt{-3}}{2^{1/3}+1})^6$)

Answer 2

You do not need to know the ring of integers of a number field $K$ in order to show a prime $p$ is totally ramified in $K$: $p$ is totally ramified in $K$ if and only if there is some $\alpha \in \mathcal O_K$ such that (i) $K = \mathbf Q(\alpha)$ and (ii) the minimal polynomial of $\alpha$ over $\mathbf Q$ is Eisenstein at $p$. This is proved as Theorems 3.1 and 3.2 here.

Example. Let $K = \mathbf Q(\sqrt[3]{2})$. In $K$, $\sqrt[3]{2}+1$ is a root of $(x-1)^3 - 2 = x^3 - 3x^2 + 3x - 3$, which is Eisenstein at $3$, so $3$ is totally ramified in $K$: $3\mathcal O_K = \mathfrak p^3$ for some prime ideal $\mathfrak p$.

Let $F = \mathbf Q(\omega)$, where $\omega$ is a nontrivial cube root of unity. Since $F = \mathbf Q(\sqrt{-3})$, $3$ is totally ramified in $F$ since $(3) = (\sqrt{-3})^2$ (or $\sqrt{-3}$ is a root of $x^2 + 3$, which is Eisenstein at $3$, but such reasoning is more technical than it has to be).

Let $L = \mathbf Q(\sqrt[3]{2},\omega)$. Since $3\mathcal O_K$ is a cube and $3\mathcal O_F$ is a square, $3\mathcal O_L$ is a $6$th power, so each prime ideal dividing $3\mathcal O_L$ has multiplicity divisible by $6$. Then the formula $\sum e_if_i = n = 6$ implies $3$ is totally ramified in $\mathbf Q(\sqrt[3]{2},\omega)$.

Remark. The ring $\mathcal O_L$ is bigger than $\mathbf Z[\sqrt[3]{2},\omega]$. See Theorem 9 here.