0

This is the problem 1.1 from the book, A Gentle Course in Local Class Field Theory.

Let $p$ be a prime, let $F$ be a field of characteristic $\neq p$, and let $a \in F^{\times} \backslash F^{\times p}$. We make no assumption on the presence of roots of unity in $F$, and we wish to prove that $[F(\alpha):F]=p$, where $\alpha$ is a root of $X^p-a$. For this, we assume $[F(\alpha):F] \neq p$, and work toward a contradiction:

  1. Consider the composite field $F(\mu_p, \alpha)$, and show that $F(\alpha) \subset F(\mu_p)$.

2. Deduce that $F(\alpha)/F$ is Galois, and then that $F(\alpha)=F(\mu_p)$

  1. From the fact that $[F(\alpha):F]$ does not depend on the choice of root $\alpha$, conclude.

Note that $\mu_p$ stands for a pth primitive root of unity in the algebraic closure of $F$, aka $F^{alg}$. Also, $F^{\times p} = \{f^n|f \in F\}$.

I've had a hard time proving that $F(\alpha) \subset F(\mu_p)$ and $F(\alpha)/F$ is Galois, and didn't get how the assumption for absurdness would fit in the proof. Could anyone please give some help on this?

I have been aware that the problem is kind of equivalent to the proposition that $x^p-a$ is irreducible iff $x$ has no root in $F$ (there is an answer regrading this). However, I am trying to understand the way from the book.

A. Gong
  • 13
  • The $F$-minimal polynomial of $a^{1/p}$ divides $x^p-a$ thus it is $f(x)=\prod_{j \in S} (x-\mu_p^j a^{1/p})$ for some $S \subset 1\ldots p$, we obtain $F \ni f(0) = C a^{\deg(f)/p}$ with $C \in F(\mu_p)$ and hence either $\deg(f) = p$ or $a^{1/p}\in F(\mu_p)$. – reuns Jul 02 '19 at 23:54
  • @reuns may I ask how you concluded the "hence" part from $f(0) \in F$? – A. Gong Jul 03 '19 at 08:29
  • $a^{\deg(f)/p} \in F(\mu_p) \implies ...$ – reuns Jul 03 '19 at 15:01
  • @reuns Could u explain how did you deduce $a^{deg(f)/p} \in F(\mu_p)$ from $f(0) \in F$ and $C \in F(\mu_p)$?? – A. Gong Jul 04 '19 at 03:06
  • ?? $f(0)=C a^{\deg(f)/p} \in F \subset F(\mu_p) \implies a^{\deg(f)/p} \in C^{-1} F(\mu_p) = F(\mu_p)$ – reuns Jul 04 '19 at 10:32
  • @reuns crystal clear. thanks – A. Gong Jul 04 '19 at 10:35

0 Answers0