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I'm working on the following problems from Dummit and Foote Abstract Algebra (Exercise 14.7.13) and I am stuck.

Let $F$ be a subfield of real number $\mathbb{R}$. Let $a\in F$ and $K=F(\sqrt[n]{a})$ where $\sqrt[n]{a}$ denotes a real $n$th root of $a$. Prove that if $L$ is any Galois extension of $F$ contained in $K$ then $[L:F]\leq 2$.

What I have tried so far:

Suppose $[K:F]=m\leq n$ and $[L:F]=d$. Then if we consider the norm, we have $$ N_{K/L}(\sqrt[n]{a})=\prod_{\sigma\in\operatorname{Hom}_F(K,\overline{L})}\sigma(\sqrt[n]{a})= \zeta (\sqrt[n]{a})^{m/d}\in L $$ since $|\operatorname{Hom}_F(K,\overline{L})|=[K:L]$ given that $K=L(\sqrt[n]{a})$ and $\sqrt[n]{a}$ is separable over $L$, and $\zeta$ is some $n$th root of unity since $\sigma$ takes a root of the polynomial $x^n-a$ to another root of this polynomial. If $m=n$, then I will be able to prove the claim by showing the irreducibility of the polynomial $x^d-a$. However, $m=n$ need not be true. So I am stuck.

Any help will be appreciated. Thank you so much!

llf
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Tianyi Wang
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  • This problem seems deceptively difficult. I referred to the following in order to formulate my answer. If you like you can look only at the sources, and give another crack at the problem. Dummit & Foote Exercise 14.7.4: https://math.stackexchange.com/questions/1785837 Keith Conrad's notes on simple radical extensions (Theorem 5.2 in particular): https://kconrad.math.uconn.edu/blurbs/galoistheory/simpleradical.pdf – llf Sep 16 '21 at 04:48

1 Answers1

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Write $\alpha := \sqrt[n]{a} \in \mathbb{R}$, $[K : F] = m \le n$, $[L : F] = d$. The norm $N_{K/L}(\alpha)$ is an element of $L$ and the product of some $[K : L] = m/d$ elements of the form $\zeta \alpha$, where the $\zeta$ are $n$th roots of unity. The set of roots of unity (in the splitting field) is multiplicatively closed, and $L \subseteq \mathbb{R}$. It follows that $N_{K/L}(\alpha) = \pm \alpha^{m/d}$. So $\alpha^{m/d} \in L$.

In fact, $L$ is equal to $F(\alpha^{m/d})$. The set $\{1,\alpha^{m/d}, \alpha^{2m/d}, \dots, \alpha^{(d-1)m/d}\} \subseteq L$ is a subset of a basis for $K/F$, for example $\{1,\alpha,\alpha^2,\dots,\alpha^{m-1}\}$, so it's $F$-linearly independent.

Finally, consider the $F$-minimal polynomial of $\alpha^{m/d}$, call it $f$. Since $\alpha^{m/d}$ generates $L$, $L$ is Galois over $F$ $\iff$ $f$ splits completely in $L$. But note that $f$ is a divisor of $X^d - \alpha^m$ (in $K[X]$), and $X^d - \alpha^m$ only has $\le 2$ real roots. Also, when equality to $2$ holds, the roots are $\pm \alpha^{m/d}$). We conclude that $f$ splits completely in $L$ $\iff$ $\deg f = [L : F] \le 2$.

llf
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