# Is $\mathbb{Q}_p(\zeta_p)$ the same as $\mathbb{Q}_p(p^{\frac{1}{p-1}})$?

It seems so. $\mathbb{Q}_p(\zeta_p)$ is a $p-1^{th}$ extension of $\mathbb{Q}_p$ which doesn't extend the residue field; and so is $\mathbb{Q}_p(p^{\frac{1}{p-1}})$. However I can't see how to express $\zeta_p$ in $\mathbb{Q}_p(p^{\frac{1}{p-1}})$ or how to express $p^{\frac{1}{p-1}}$ in $\mathbb{Q}_p(\zeta_p)$.

Can you help?

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Note that while unramified extensions of $\mathbb Q_p$ are determined up to isomorphism by their degree, the same is not true of totally ramified extensions (even tame ones, as in this question). – Matt E Dec 11 '10 at 0:58
+1 It's a good question and the non-uniqueness of totally ramified extensions is a very common point of confusion. – Alex B. Dec 11 '10 at 3:30

The right result is that $\mathbb{Q}_p(\zeta_p) = \mathbb{Q}_p((-p)^{1/(p-1)})$.

I will show that $\zeta_p$ is in $\mathbb{Q}_p((-p)^{1/(p-1)})$. Since both fields are of dimension $p-1$ over $\mathbb{Q}_p$ (compute ramification degrees), this shows the fields are equal.

Let $K=\mathbb{Q}_p((-p)^{1/(p-1)})$ and let $\pi$ be a $(p-1)$st root of $-p$ in $K$. We want to show that the equation $x^p=1$ has $p$ roots in $K$. Make the change of variable $x=1+\pi y$. So $\pi^{p} y + p \pi^{p-1} y^{p-1} + \cdots + p \pi y + 1 =1$. Dividing out $\pi^{p} = -p \pi$, we get $$y^{p} - \pi^{p-1} y^{p-1} - \frac{1}{p} \binom{p}{2} \pi^{p-2} y^{p-2} - \frac{1}{p} \binom{p}{3} \pi^{p-3} y^{p-3} - \cdots - y=0 \quad (*).$$ Notice that $(1/p) \binom{p}{k}$ is an integer for $1 \leq k \leq p-1$. So, when we reduce $(*)$ modulo $\pi$, we get $$y^p - y =0.$$

The equation $y^p-y=0$ has $p$ distinct roots in $\mathbb{F}_p$. So, by Hensel's lemma, equation $(*)$ has $p$ roots in $K$.

We have shown that $K$ contains a nontrivial $p$-th root of unity, as desired.

Here is a direct argument for the other direction, when $p >2$. Let $1-\zeta = \lambda$ (where $\zeta$ is a primitive $p$th root of unity.) We know that $$\prod_{j=1}^{p-1} (1-\zeta^j) = p.$$ The left hand side is $$\prod_{j=1}^{p-1} (j \lambda - \binom{j}{2} \lambda^2 + \cdots) = (p-1)! \lambda^{p-1} \cdot \prod_{j=1}^{p-1} \cdot \left( 1 - \frac{j}{2} \lambda + \cdots \right) =$$ $$(p-1)! \cdot \lambda^{p-1} \cdot \left( 1 - \frac{p(p-1)}{4} \lambda + \cdots \right)$$

By Wilson's theorem, $(p-1)! \equiv -1 \mod p$. And, since $p$ is odd, $p(p-1)/4 \equiv 0 \mod p$. So we deduce that $$p = (-1) \lambda^{p-1} \left( 1+ \mbox{something divisible by} \lambda^2 \right).$$

Anything of the form $\left( 1+ \mbox{something divisible by} \lambda^2 \right)$ has a $(p-1)$st root in $\mathbb{Q}_p(\zeta)$, as the Taylor series for $(1+u)^{1/(p-1)}$ will converge for $u$ that small. So we see that $$(-p)^{1/(p-1)} = \lambda \left( 1+ \mbox{something divisible by} \lambda^2 \right)^{1/(p-1)}$$ is in $\mathbb{Q}_p(\zeta)$ as desired.

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