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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$]

Here is a solution in MSE.

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

I rewrite the solution in that answer here.

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.

Question: In this line, $$\beta=\Pi _\sigma \sigma (\alpha)=(\Pi _\sigma w_\sigma) \cdot\alpha^e\quad (\operatorname { where } e=[K:E]=\frac nd)$$ $K$ is not necessarily galois extension.$|Aut(K/E)|$ need not be equal to $[K:E]=\frac nd$.

It's there in all the solution I came across. I wonder why it has to be true. Please explain why $e$ in above equation has to be $[K:E]$

user26857
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Cloud JR K
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2 Answers2

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It doesn't need to be a Galois extension. Let $L$ denote the splitting field for $x^n - a$ over $\mathbb{Q}$.

$\sigma$ is running through the set of isomorphisms of $K$ into a fixed algebraic closure of $E$. This is in one to one correspondence with the coset representatives for $\mathrm{Aut}(L/K)$ in $\mathrm{Aut}(L/E)$. Even if $\mathrm{Aut}(L/K)$ is not a normal subgroup inside $\mathrm{Aut}(L/E)$, from the one to one correspondence, $$e = \frac{|\mathrm{Aut}(L/E)|}{|\mathrm{Aut}(L/K)|} = [K:E] = \frac{n}{d}.$$

user26857
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  • Is this a consequence of isomorphism extension theorem? – Cloud JR K Apr 28 '20 at 07:09
  • e is actually Aut(K/E)... – Cloud JR K Apr 28 '20 at 07:10
  • No, e is not a group. Furthermore, $\sigma$ need not come from $Aut(K/E)$. They will come from $Aut(K/E)$ if $Aut(L/K)$ is normal in $Aut(L/E)$. Look at page 582 in Dummit and Foote as well as the Fundamental Theorem of Galois Theory. –  Apr 28 '20 at 13:55
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A setting where a similar problem and answer occur with a more general base field is in my answer at Radical extension. The positivity of $a$ is replaced by the assumption that $n$th roots of unity in the top field are actually in the bottom field. That holds in the setting of your question because your top field has an embedding in the real numbers and therefore can't contain any roots of unity besides $\pm 1$, which are in $\mathbf Q$.

KCd
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