Subfield of $\mathbb{Q}(\sqrt[n]{a})$

Question

Exercise 14.7.4 from Dummit and Foote

Let $K=\mathbb{Q}(\sqrt[n]{a})$, where $a\in \mathbb{Q}$, $a>0$ and suppose $[K:\mathbb{Q}]=n$ (i.e., $x^n-a$ is irreducible over $\mathbb Q$). Let $E$ be any subfield of $K$ and let $[E:\mathbb{Q}]=d$. Prove that $E=\mathbb{Q}(\sqrt[d]{a})$. [Consider $ N_{K/E}(\sqrt[n]a)\in E$]

I think I know how to solve it by considering $K=\mathbb{Q}(\sqrt[n]{a})$ as a subfield of $L=K(\sqrt[n]{a},\xi_n=e^{2\pi i/n})$, splitting field of $x^n-a$. $L$ over $\mathbb{Q}$ has Galois group $\left<\sigma,\tau\right>$ where $\sigma: \sqrt[n]{a}\rightarrow \sqrt[n]{a}\xi_n$, $\xi_n\rightarrow\xi_n$ and $\tau:\xi_n\rightarrow\xi_n^{m}$ ($\gcd(m,n)=1$, $m\neq 1$), $\sqrt[n]{a}\rightarrow \sqrt[n]{a}$. $K$ is the fixed field of $\left<\tau\right>$, therefore a subfield of $K$ must be fixed by a group containing $\left<\tau\right>$, and it's not hard to check that $\left<\tau,\sigma^{d}\right>$ fixes $K=\mathbb{Q}(\sqrt[d]{a})$.

However what I don't understand the given hint. Can anyone explain to me what that notation mean (I couldn't find it in the corresponding section in the book) and how does that (potentially) generate a easier solution of this problem?

Answer 1

Let $\alpha=\sqrt [n]a\in \mathbb R_+$ be the real positive $n$-th root of $a$, so that $K=\mathbb Q(\alpha)$.
Consider some intermediate field $\mathbb Q\subset E\subset K$ (with $d:=[E:\mathbb Q]$) and define $\beta=N_{K/E}(\alpha)\in E$.
We know that $\beta=\Pi _\sigma \sigma (\alpha) $ where $\sigma$ runs through the $E$-algebra morphisms $K\to \mathbb C$.
Now, $\sigma (\alpha)=w_\sigma \cdot\alpha$ for some suitable complex roots $w_\sigma$ of $1$ so that $$\beta=\Pi _\sigma \sigma (\alpha)=(\Pi _\sigma w_\sigma) \cdot\alpha^e\quad (\operatorname { where } e=[K:E]=\frac nd)$$ Remembering that $\beta\in E\subset K\subset \mathbb R$ is real and that the only real roots of unity are $\pm 1$ we obtain $\Pi _\sigma w_\sigma=\frac {\beta}{\alpha^e}=\pm 1$ and $\beta=\pm \alpha^e=\pm \sqrt [d]a$.
Thus we have $\mathbb Q(\beta)=\mathbb Q(\sqrt [d]a)\subset E$ with $ \sqrt [d]a$ of degree $d$ over $\mathbb Q$.
Since $[E:\mathbb Q]=d$ too we obtain $E=\mathbb Q(\sqrt [d]a)$, just as claimed in the exercise.

Answer 2

This answer is essentially same as Georges Elencwajg's. But $e = \frac{n}{d}$ is justified.

Let L be a Galois closure of K. It is easy to see that $L$ contains n-th primitive root of unit, which is denoted as $\xi$. Let $\beta := N_{K/E}(a^{1/n}) = \Pi_\sigma \sigma(a^{1/n})$, where the product is taken over all cosets of Gal$(L/K)$ in Gal$(L/E)$. For each $\sigma$, $\sigma(a^{1/n}) = \xi_\sigma$, where $\xi_\sigma^n = 1$. Thus

$$ \beta = (\Pi_\sigma \xi_\sigma) a^{e/n} $$ Since $\beta \in E \subset \mathbb{R}$ and $a^{1/n} \in \mathbb{R}$, $\Pi_\sigma \xi_\sigma = \pm 1$. Thus $a^{e/n} \in E$. Now, by the Fundamental Theorem of Galois Theory, $e = [\text{Gal}(/E) : \text{Gal}(/K)] = [K : E] = \frac{n}{d}$. Note that $a^{1/n}$ is a root of $x^e = \beta$. This indicates that $[K : \mathbb{Q}(\beta)]=[\mathbb{Q}(\beta, a^{1/n}): \mathbb{Q}(\beta)] \leq e = \frac{n}{d}$. On the other hand, $\mathbb{Q}(\beta) \subset E$. Thus we have $\mathbb{Q}(\beta) = E$.