Factorization of $(2)$ in $\mathbb{Z}[\zeta_p]$

Question

Let $p$ be a prime number such that $p\equiv 1(\text{mod }3)$ and $x^3\equiv 2(\text{mod }p)$ for some $x\in\mathbb{Z}$. Prove that:

1) There is a unique intermediate field $\mathbb{Q}\subset K\subset \mathbb{Q}(\zeta_p)$ such that $[K:\mathbb{Q}]=3$.

2) Show that $2$ is not ramified in $\mathbb{Q}(\zeta_p)$ and that the inertia degree divides $\frac{p-1}{3}$. Conclude that $2$ splits completely in $K$.

3) Show that $\mathcal{O}_K\neq\mathbb{Z}[a]$ for every $a\in\mathcal{O}_K$.

Now here's where I'm at:

For 1), I used that $G:=\text{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})\simeq(\mathbb{Z}/p\mathbb{Z})^{\times}\simeq\mathbb{Z}/(p-1)$, so that it is enough to prove that $\mathbb{Z}/(p-1)$ has only one subgroup with order $\frac{p-1}{3}$ so that $K=\mathbb{Q}(\zeta_p)^{H}$.

In $3)$, I argue by contraditcion supposing that $\mathcal{O}_K=\mathbb{Z}[a]$ for some $a$. Since $\Phi_p(x)=x^{p-1}+...+1$ is the minimal polynomial of $\zeta_p$, then both $\Phi_p(0)$ and $\Phi_p(1)$ are not zero modulo $2$, so $(2)$ is a prime ideal (contradiction, since by $2)$, we have that $2$ splits completely in $K$).

For $2)$, I used the fact that $2\nmid \Delta_{\mathbb{Q}(\zeta_p)}=(-1)^{\frac{p-1}{2}}p^{p-2}$, so $2$ is not ramified. But now comes my problem: I know that $\mathcal{O}_{\mathbb{Q}(\zeta_p)}=\mathbb{Z}[\zeta_p]$, so I just need to analyse $\Phi_p(x)$ modulo $2$, which is indecomposable in $\mathbb{F}_2[x]$, so $(2)=\frak{p}_1^{e_1}...\frak{p}_s^{e_s}$ is such that $s=1$, $e:=e_1=1$, so $fe=\varphi(p)=p-1\Rightarrow f=p-1$, which doesn't divide $\frac{p-1}{3}$. What am I missing here? Besides, how do I conclude from this that $2$ splits completely in $K$?

Answer 1

$ 2 $ splits completely because it is unramified, and the Frobenius element at a prime lying over $ 2 $ in the Galois group is $ \zeta_p \to \zeta_p^2 $, and this automorphism fixes the field $ K $ since $ 2 $ is a cube modulo $ p $. The order of the Frobenius map is the inertia degree, and this holds even when it is restricted to $ K $; which means there is no inertia at any prime of $ K $ lying over $ 2 $.

For (3), the result follows since the fact that $ 2 $ splits completely in $ \mathbf Z[a] $ would imply that the minimal polynomial of $ a $ factors into three distinct monic linear factors over $ \mathbb F_2 $, which is impossible: there are only two monic linear polynomials in $ \mathbb F_2[X] $.