ramification index of $p$ in $\mathbb{Z}\left[e^{\frac{2\pi i}{p}}\right]$

Question

I am attempting to show that $p$ has ramification index $p-1$ in $\mathbb{Z}[\omega]$ where $\omega=e^{2\pi i/p}$. The issue is I want to do so avoiding actually factoring $p$. I was hoping to use the fundamental identity somehow. This is the hope.

$$\prod_{i=1}^{p-1}(1-\omega)=p,$$ Thus $(1-\omega)|p\mathbb{Z}[\omega]$. Then I wanted to show that

$$\mathbb{Z}[\omega]/(1-\omega)\cong \mathbb{Z}/p\mathbb{Z}$$

which would imply that the inertial degree is $1$. Then I would hopefully show that $(1-w^{j})=(1-\omega^{i})$ for all $i$ and $j$. Thus by the fundamental identity the ramification index must be $p-1$. I have tried the obious maps $\alpha\mapsto \alpha^{p-1}\pmod{p}$ and $\alpha\mapsto\text{Norm}(\alpha)\pmod{p}$, but am having trouble showing that the kernal is $(1-\omega)$.

Is this a reasonable method? Or should I just stick to the classic way to doing this? I was hoping to present this as an example before we showed the all ramified primes must divide the discriminant. Thanks for any ideas.

Answer 1

Hints: (assuming I understood what the question was)

• The minimal polynomial of $\omega$ is $m(x)=1+x+x^2+\cdots+x^{p-1}$. So $\Bbb{Z}[\omega]\cong \Bbb{Z}[x]/\langle m(x)\rangle$. If you simply consider the map $\Bbb{Z}[x]\to \Bbb{Z}_p$, $x\mapsto 1$, then why wouldn't that work? All of $m(x),p, 1-x$ are in the kernel, and $m(x)\equiv (1-x)^{p-1}\pmod p$.
• Because $\gcd(i,p)=1$ there exists an integer $i'$ such that $ii'\equiv1\pmod p$. Then $$\frac{1-\omega^i}{1-\omega}=1+\omega+\omega^2+\cdots+\omega^{i-1}$$ and $$\frac{1-\omega}{1-\omega^i}=\frac{1-\omega^{ii'}}{1-\omega^i}=1+\omega^i+\omega^{2i}+\cdots+\omega^{(i'-1)i}$$ are both in $\Bbb{Z}[\omega]$, so the ratio $(1-\omega)/(1-\omega^i)$ is a unit.