Let consider the ring $\mathbb{Z}_p$ and $\zeta$ be a $p$-th root of unity. Especially $\zeta \not \in \mathbb{Z}_p$. Denote with $\Phi _p(x)$ the cyclotomical polynomial in $p$. Since $p$ is a prime we know that it has the shape $\Phi _p(x)= 1 + x +x^2 +... +x^{p-1}$. This gives rise for the quotient ring

$$ \mathbb{Z}_p[X]/\langle\Phi_p(x)\rangle \cong \mathbb{Z}_p[\zeta] = \mathbb{Z}_p \oplus \zeta \mathbb{Z}_p \oplus \dots \oplus \zeta^{p-2} \mathbb{Z}_p $$

which is obviously a free $\mathbb{Z}_p$-module of rank $p-1$. Denote $g=\zeta -1$.

My question is how to see that $\mathbb{Z}_p[\zeta]$ isa local ring with maximal ideal $g \mathbb{Z}_p[\zeta]$?

I tried to argue in following way: Obviously observation provides $\Phi _p(g +1) =0$ and the formula above gives $$ \Phi_p(x + 1) = p + \binom{p}{2}x + \binom{p}{3}x^2 + \dots + \binom{p}{p - 1} x^{p - 2} + x^{p - 1} $$.

In light of this I can conclude following inclusions:

$p \mathbb{Z}_p[\zeta] \subset g \mathbb{Z}_p[\zeta]$ and $\pi^{p-1} \mathbb{Z}_p[\zeta] \subset p \mathbb{Z}_p[\zeta]$ which imply $(g\mathbb{Z}_p[\zeta]) \cap\mathbb{Z}_p = p\mathbb{Z}_p$.

From here I'm stuck.

Algèbre8, §2, exercise 10. $\endgroup$ – abx Apr 15 '18 at 14:08