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Let $\zeta_p$ be the $p^{th}$ root of unity in $p$-adic field.

I know that $v(\zeta_p-1)=\frac{1}{p-1}$ but I couldn't prove it.

I have tried in the following way:

$f(x)=(x-\zeta_p)(x-\zeta_p^2) \cdots (x-\zeta_p^{p-1})=\sum_{j=0}^{p-1}x^j=1+x+\cdots+x^{p-2}+x^{p-1}=\prod_{i=1}^{p-1}(x-\zeta_p^{i}).$

Now putting $x=1$, we get $$f(1)=(1-\zeta_p)(1-\zeta_p^2) \cdots (1-\zeta_p^{p-1}),$$ and also $f(1)=p$. Thus $$1-\zeta_p=\frac{f(1)}{(1-\zeta_p^2) \cdots (1-\zeta_p^{p-1})}=\frac{p}{(1-\zeta_p^2) \cdots (1-\zeta_p^{p-1})}.$$ So $v(1-\zeta_p)=v(p)-v[(1-\zeta_p^2) \cdots (1-\zeta_p^{p-1})]=1-v(1-\zeta_p^2)- \cdots-v(1-\zeta_p^{p-1})$

How to finish the proof ?

Help me in the above proof ?

MAS
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    Find the minimal polynomial $g(x)$ of $\zeta_p-1$ as follows. Just write $f(x)=x^p-1=(x-1)(x^{p-1}+\dots+x+1)$ and $g(x)=f(x+1)=x\cdot(\cdots\cdots)$, and in the second factor here, you see that all lower coefficients are divisible by $p$, while the constant is exactly $p$. Eisenstein, of degree $p-1$. Voilà! – Lubin Aug 13 '20 at 19:30
  • @Lubin, I got the Eisenstein polynomial of degree $p-1$.But still I didn't see how $v(\zeta_p-1)=\frac{1}{p-1}$ ? Can you please explain little bit more ? – MAS Aug 13 '20 at 19:46
  • Related: https://math.stackexchange.com/q/1327782/96384, https://math.stackexchange.com/q/2977896/96384. – Torsten Schoeneberg Aug 13 '20 at 22:09
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    Well, as @reuns says in his answer, if $g(x)$ is Eisenstein of degree $n$ with constant term $c$ and $v(c)=1$, then a root $\rho$ will have $v(\rho)=1/n$. Use Newton Polygon, or this: if a sum of $p$-adic terms is zero, the minimal $v$-value will be taken at least twice. In our situation, if that minimum value is $v(c)$, then you see that $v(\rho)=1/n$. If that minimum value is less than $v(c)$, you see that you get a contradiction. – Lubin Aug 14 '20 at 15:25
  • @Lubin, Thank you very much for your nice explanation. – MAS Aug 15 '20 at 06:38

1 Answers1

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For $p\nmid n$, $v(\zeta_p-1)=v(\zeta_p^n-1)$ because $$\frac{\zeta_p^n-1}{\zeta_p-1}= \sum_{m=0}^{n-1} \zeta_p^m \equiv \sum_{m=0}^{n-1} 1 \bmod\mathfrak{m}$$ Next $$\Phi_p(x)=\prod_{n=1}^{p-1}(x-\zeta_p^n)=\sum_{m=0}^{p-1} x^m, \qquad \Phi_p(x+1) = \prod_{n=1}^{p-1}(x-(\zeta_p^n-1))$$ $$v(p) = v(\Phi_p(1)) = \sum_{n=1}^{p-1} v(\zeta_p^n-1)=(p-1) v(\zeta_p-1)$$

This implies the irreducibility of $\Phi_p(x)\in \Bbb{Q}_p[x]$ and that $\Bbb{Q}_p(\zeta_p)/\Bbb{Q}_p$ is totally ramified.

A similar idea applies to any Eisenstein polynomial, an element and its $\Bbb{Q}_p$-conjugates have the same valuation.

reuns
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  • As $v(\zeta_p-1)=\frac{1}{p-1}$, $\zeta_p-1$ is a uniformizer and root of $\Phi_p(x)$. So the extension $\mathbb{Q}_p(\zeta_p-1)/\mathbb{Q}_p$ is totally ramified. Am I right ? – MAS Aug 14 '20 at 02:18
  • It is nice answer.I got it. The first line was unknown to me. I didn't realise before that $v(\zeta_p-1)=v(\zeta_p^n-1), \ p \nmid n$. Thanks – MAS Aug 14 '20 at 02:21
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    I think your first congruence should not be modulo $p$, but something else. After all, for $n=2$, $\zeta_p+1\not\equiv2\pmod p$. (I really do prefer an argument using or motivated by Newton Polygon.) – Lubin Aug 14 '20 at 15:33