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:
- $(1-\omega)$ is a prime ideal in the ring of integers of $K=\mathbb{Q}(\omega)$, and that $(3)=(1-\omega)^2$
- 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 $(1-\omega)$.
- 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:
- 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 $(1-\omega)$.
- 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
Best Answer
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 $(3)=(3,\frac{\sqrt{-3}}{2^{1/3}+1})^6$)