Show that the ramification of the extension index is $ \ (p-1) \ $

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Consider the p-adic field $ \ \mathbb{Q}_p \ $ and let us extend $ \ \mathbb{Q}_p \ $ by adjoining the $ \ p^{th} \ $ roots of unity.

$(i) \ $ Show that the ramification of the extension index is $ \ (p-1) \ $.

$ (ii) \ $ Show that $ \ v(\zeta_p-1)=\frac{1}{p-1} \ $ , where $ \ \zeta_p \ $ is a primitive $ \ p \ $ -root of unity.

Answer:

Let the extension of $ \ \mathbb{Q}_p \ $ is denoted by $ \ K=\mathbb{Q}_p(\zeta_p) \ $.

Clearly , $ \ K=\mathbb{Q}_p(\zeta_p) \ $ formes a vector space of dimension $ \ p-1 \ $ over $ \ \mathbb{Q}_p \ $

Does this mean that the ramification index or degree is $ \ p-1 \ $ ?

But I don't know the proof.

Can someone give me the proof?

Next,

I got a formula as follows:

$ N_{\mathbb{Q}_p(\zeta_p)/\mathbb{Q}_p} \ =p $ , where $ \ N \ $ denote norm.

Thus for the valuation $ \ v \ $ , we \ have

$ v (N_{\mathbb{Q}_p(\zeta_p)/\mathbb{Q}_p})=1 \ $

Now, $ \ [\mathbb{Q}_p(\zeta_p): \mathbb{Q}_p]=p-1 \ $

Therefore,

$ \ v(\zeta_p-1)=\frac{1}{[\mathbb{Q}_p(\zeta_p): \mathbb{Q}_p]} v (N_{\mathbb{Q}_p(\zeta_p)/\mathbb{Q}_p}) \ =\frac{1}{p-1} \cdot 1=\frac{1}{p-1} $

But I am not sure with any of the above proofs.

Can I get help with the correct proof of my questions?

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I’ll drop the subscript from $\zeta_p$, ’cause it’s the only root of unity under discussion. First, the minimal polynomial of $\zeta$ is $$ f(X)=\frac{X^p-1}{X-1}=X^{p-1}+X^{p-2}+\cdots+X+1\,. $$ Form $\,f(X+1)=X^{p-1}+pX^{p-1}+\cdots+\frac{p(p-1)}2X+p$, which you see is the minimal polynomial of $\zeta-1$, and an Eisenstein polynomial. Ask about what $v(\zeta-1)$ must be, and you see, using the fact that if $v(\alpha)<v(\beta)$, then $v(\alpha+\beta)=v(\alpha)$, that $\zeta-1$, the root of $f(X+1)$, must have $v$-value equal to $\frac1{p-1}$.

That explains why both the degree and the ramification index of $\Bbb Q_p(\zeta)$ over $\Bbb Q_p$ are equal to $p-1$.

This quite independent of the general formula for an extension $K\supset k$ that $v_K(\alpha)=\frac1{[K:k]}N^K_k\alpha$ which you quoted.