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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). 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 $L=K(\xi_n=e^{2\pi i/n})$, splitting field of $x^n-a$. $L$ over $\mathbb{Q}$ has Galois group $<\sigma,\tau>$ 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 fixed field of $<\tau>$ therefore subfield of $K$ must be fixed by a group containing $<\tau>$, and it's not hard to check $<\tau,\sigma^{d}>$ fixes $K=\mathbb{Q}(\sqrt[d]{a})$.

However what I don't understand is the hint given...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?

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    $\begingroup$ At page 582, in exercise 17, you can find the origin of that "mysterious" notation. $\endgroup$ – user26857 May 15 '16 at 8:15
  • $\begingroup$ Thanks...that's indeed not easy to find... $\endgroup$ – user175968 May 15 '16 at 8:22
  • $\begingroup$ For what it's worth, let me mention that I found this exercise very difficult. $\endgroup$ – Georges Elencwajg May 16 '16 at 13:17
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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.

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  • $\begingroup$ How did you go from $a^e$ to $\sqrt[d]{a}$? $\endgroup$ – TheLearningLearner Apr 19 '18 at 3:44
  • $\begingroup$ @Georges Elencwajg , could you explain why e=[K:E]=n/d? K is not galois, how can we justify this statement? $\endgroup$ – Cloud JR K Apr 27 '20 at 16:02

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