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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.2012-10-29
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    @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.2012-10-29
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    Take out a factor of $1- \eta_{p}$ from $1 - \eta_{p}^{g}$ and see what you have left.2012-10-29
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    @Geoff: Thank you. Any hints for the second congruence ?2012-10-30
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    Hint : $p = \prod_{i=1}^{p-1}(1-\eta_{p}^{i})$ then use last hint.2012-10-30
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    @Geoff: Thank you again. I got: $ 1+\frac{p}{t^{p-1}} = 1+ \frac{\prod_{i=1}^{p-1} (1-\zeta_p^i)}{(1-\zeta_p)^{p-1}}=1+\prod_{i=1}^{p-1} (1+\zeta_p + ... + \zeta_p^{i-1})$. How can I take out a factor of $1-\zeta_p$ ?2012-10-31
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    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 $t$)?2012-10-31
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    -1 by Wilson Theorem. Thank you !2012-10-31

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