Compute the inertial degrees of two prime ideals e.g the inertial degree of $P/(2)$ where P is prime in $Q[e^{\frac{2\pi i}{23}}]$ lying over (2))

Question

I was reading through Marcus Number Field chapter 3 and I got stuck on exercise 17

Let $K=\mathbb{Q}[\sqrt{-23}]$, $L=\mathbb{Q}[\omega]$ where $\omega=e^{\frac{2\pi i}{23}}$. We know that K $\subset$ L.

Let Q be one of the primes of $R=O_K$ lying over 2; specifically, take Q = (2, $\theta$) where $\theta=(\left(1+\frac{\sqrt{-23}}{2}\right))$. Let P be a prime of $\mathbb{Z}[\omega]$ lying over Q.

1. Show that f (P|Q) = 11. . Conclude that in fact $P = (2, \theta)$ in $\mathbb{Z}[\omega]$.

2. Show that $Q^3 = (\theta − 2)$, but that Q is not principal in R. (use the fact that for $\alpha\neq 0\in R$ then we have for the principal ideal generated by $(\alpha)$ ||( $\alpha$ )||=|$N_{\mathbb{Q}^K(\alpha)$| )

3. Show that P is not principal.

4. Show that if 2 = $\alpha\beta$, with $\alpha, \beta \in \mathbb{Z}[\omega]$, then $\alpha$ or $\beta$ is a unit in $\mathbb{Z}[\omega]$.

I having a problem with the first point (and without that I don't seem to be able to solve the others). I think I get the theory, id est that I should look after the degrees of f(Q/(2)) and f(P/(2)) but I actually don't know how to compute them.

Is there a general method to find the degree of such an extension? Is maybe P easy to define explicitly?

UPDATE: I've done everything but two small points:

1. I don't know how to conclide that P=(2,$\theta$)
2. I'don't get the 4th point, the hint is to use the proof of the following theorem but I don't get how

Theorem 18. A Dedekind domain is a UFD iff it is a PID.

Proof. As we have noted, PID always implies UFD; for Dedekind domains we can also get this result by using Theorem 16. Conversely, assuming that the Dedekind domain R is a UFD, let I be any ideal in R. By Theorem 15, I divides some principal ideal (a). The element a is a product of prime elements in R, and it is easily shown that each prime element p generates a principal prime ideal (p): If ab $\in$ (p), then p | ab, and then p | a or p | b, implying that a or b is in (p). Thus I divides a product of principal prime ideals. By unique factorization of ideals in R, it follows that I is itself a product of principal primes and therefore a principal ideal.

Answer 1

We'll use the formula $$f(Q|(2))=f(Q|P)f(P|(2)) $$ and notice that in $\mathbb{F}_2$ 2 has order 11 and that in $R $ we have that $$(2,\theta)(2,\overline{\theta})=(4,\frac{1+\sqrt{-23}}{2},\frac{1-\sqrt{-23}}{2},3)=(2)$$ and so f(Q|(2))=11 and f(P|(2))=1 and so f(Q|P)=11.

Regarding the second and third point we have the following

2. $$ P^3=(8,4\theta,2\theta^2,\theta^3)=$$$$\left(8,\frac{4+4\sqrt{-23}}{2},\frac{-22+2\sqrt{-23}}{2},\frac{-17-5\sqrt{-23}}{2}\right)$$ and $$ (\theta-2)=(\frac{-3+\sqrt{-23}}{2})$$ and now we notice that $$4\theta+2\theta^2+\theta^3+2*8=$$$$\frac{4-22-17+32+(4+2-5)\sqrt{-23}}{2}=\frac{-3+\sqrt{-23}}{2}$$ so $P^3\supset (\theta-2)$.\ On the other hand $$\frac{-3+\sqrt{-23}}{2}\frac{-3-\sqrt{-23}}{2}=\frac{9+23}{4}=8\Rightarrow 8\in (\theta-2)$$ $$ (\theta-2)(\theta^2+\theta+1)=\theta^3-8\Rightarrow \theta^3\in (\theta-2)$$ $$ 2(\theta-2)(\theta+2)=2\theta^2-8\Rightarrow 2\theta^2\in (\theta-2)$$ $$ 4(\theta-2)=4\theta-8\Rightarrow 4\theta\in (\theta-2)$$ and so $$ P^3\subset (\theta-2)$$.\ We can now prove that P is not principal by noticing that for a principal ideal ($\alpha$) it holds $$||(\alpha)||=|N(\alpha)| $$ now any element of $P$ has norm $a^2+ab+6b^2$ for $a,b\in \mathbb{Z}$ and $||(\alpha)||=||(2,\theta)||=|(2)^{f(P|(2))}|=2$ and since the equation $$a^2+ab+6b^2=2$$ has no integer solutions we proved that P can't be principal;

3. From the previus exercise we have that, if we call $d_Q$ the order of the class containing Q in G(S) and simlarly $d_P$ the order of the class containing P in G(R), $$ d_Q|d_Pf(P|Q)$$ but in our case $1<d_Q\le3$ and $f(P|Q)=11$ so we have that $d_P\not|11\Rightarrow d_Q|d_P\Rightarrow d_P\neq 1 $ so P is not prime (the prime class has order 1)