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Let $p \ne 2$ be prime number and denote by $\zeta_p$ the p-th root of unity. It's well known that $K = \mathbb{Q}_p(\zeta_p)$ has $t=1 - \zeta$ as prime element (generator of the Ideal $P_K = \{ x\in \mathbb{Q} | |x|<1 \}$ in the ring of integers $O_K= \{x\in \mathbb{Q} | |x| \leq 1\}$.

Let $\sigma:\zeta_p \to \zeta_p^g$ be an automorphism of $K/\mathbb{Q}_p$ ($1\leq g\leq p-1$).

Show $\sigma(t) \equiv gt \;(t^2)$ (it means $\sigma(t)-gt \in (t^2)$, the ideal generated by $t^2$)

and $t^{-p+1} p \equiv -1 \; (t)$.

Any hints ?

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Hint: $\sigma(t)$ has the form $1 - \eta_{p}^{g}.$ This is all you really need to know. – Geoff Robinson Oct 29 '12 at 20:03
@Geoff I have tried it. We have $\sigma(t)-gt = 1-\zeta_p^g -g +g \zeta_p$ and I need to show $|\sigma(t)-gt| < |t| = |1-\zeta_p|$. I am stuck here. – colge Oct 29 '12 at 20:27
Take out a factor of $1- \eta_{p}$ from $1 - \eta_{p}^{g}$ and see what you have left. – Geoff Robinson Oct 29 '12 at 22:40
@Geoff: Thank you. Any hints for the second congruence ? – colge Oct 30 '12 at 10:58
Subtract the $1$ from each side, that is irrelevant. The $i$-th term of the product on the right side is congruent to $i$ (mod $t$), so what is the congruence of the product (mod $t$)? – Geoff Robinson Oct 31 '12 at 11:21

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