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Let $\mathbb{Z}_p$ be the ring of $p$-adic numbers, $\Phi _p(x)$ the cyclotomic polynomial and $\zeta$ be a $p$-th root of unity. So $\zeta \not \in \mathbb{Z}_p$ and we get $$ \mathbb{Z}_p[X]/\langle\Phi_p(x)\rangle \cong \mathbb{Z}_p[\zeta] = \mathbb{Z} \oplus \zeta \mathbb{Z} \oplus \dots \oplus \zeta^{p-2} \mathbb{Z} $$ a free $\mathbb{Z}_p$-module of rank $p-1$.

Let $\pi := \zeta-1$ and I want to know how to see that $\mathbb{Z}_p[\zeta]$ is a local ring with maximal ideal $\pi \mathbb{Z}_p[\zeta]$.

My idea is to use the obviously fact that $\Phi _p(\pi +1) =0$ and the extension formula $ \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} $.

From this I can conclude that we have following inclusions: $p \mathbb{Z}_p[\zeta] \subset \pi \mathbb{Z}_p[\zeta]$ and $\pi^{p-1} \mathbb{Z}_p[\zeta] \subset p \mathbb{Z}_p[\zeta]$

It's obviosly that it would be enough to show that $\mathbb{Z}_p[\zeta] / \pi\mathbb{Z}_p[\zeta] \cong \mathbb{F}_p$ and that every maximal ideal $M \subset \mathbb{Z}_p[\zeta]$ contains $\pi$ but unfortunately I haven't a idea how to realize this two steps using the information above...

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$\Phi_p(\pi + 1) = 0$ tells you that

$$ \Phi_p(x + 1) = p + \binom{p}{2}x + \binom{p}{3}x^2 + \dots + \binom{p}{p - 1} x^{p - 2} + x^{p - 1} $$

is the minimal polynomial of $\pi$. What does that imply about $(\pi\mathbf{Z}_p[\zeta]) \cap\mathbf{Z}_p$? Which ideal is that?

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    $\begingroup$ If $\Phi_{p}(x) = \frac{x^p-1}{x-1}$ then $\Phi_{p}(x+1) =\frac{(x+1)^p-1}{x} = \sum_{k=0}^{p-1} x^k {p \choose k+1}$ $\endgroup$ – reuns Jul 16 '17 at 19:57
  • $\begingroup$ D you mean that from $\Phi_p(\pi + 1) = 0$ and $ \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} $ I can conclude $(\pi\mathbf{Z}_p[\zeta]) \cap\mathbf{Z}_p = 0$? But why does it hold? Futhermore I don't see the next step... $\endgroup$ – KarlPeter Jul 17 '17 at 14:38
  • $\begingroup$ @KarlPeter No $(\pi\mathbf{Z}_p[\zeta]) \cap \mathbf{Z}_p = p\mathbf Z_p$. $\endgroup$ – Trevor Gunn Jul 17 '17 at 14:39
  • $\begingroup$ ups ok yes that's understandable ... but what I have to do next? $\endgroup$ – KarlPeter Jul 17 '17 at 14:44
  • $\begingroup$ @KarlPeter Think. You now have to think. $\endgroup$ – Trevor Gunn Jul 17 '17 at 14:47

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