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

Attempt:

I proved that the norm is in $E$ and if $E\supset\mathbb{Q}(\sqrt[d]{a})$ I could prove that $E=\mathbb{Q}(\sqrt[d]{a})$, but I can't prove that $E\supset\mathbb{Q}(\sqrt[d]{a})$.

user26857
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Gaston Burrull
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1 Answers1

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Let us use multiplicativity of the norm. Write $z=N_{K/E}(\root n\of a)\in E$. Then $$z^n=N_{K/E}((\root n\of a)^n)=N_{K/E}(a)=a^{n/d},$$ because $N_{K/E}(y)=y^{n/d}$ for all $y\in E$. But $E$ is a subfield of the reals, and in the reals the only solutions of $z^n=a^{n/d}$ are $(\pm) \root d\of a,$ where the minus sign is a live possibility only when $n$ is even.

So $z=\pm \root d\of a$. This implies that $\Bbb{Q}(\root d\of a)\subseteq E$. It sounds like you know how to complete the argument.

Jyrki Lahtonen
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    Thanks! I think the rest is: If $[E:Q(\sqrt[d]{a})]>1$ then $[Q(\sqrt[n]{a}):Q(\sqrt[d]{a})]>n/d$ but $\sqrt[n]{a}$ satisfies a $n/d$ grade polynomial over $Q(\sqrt[d]{a})$: $x^{n/d}-\sqrt[d]{a}$ a contradiction, am I right? – Gaston Burrull Nov 24 '13 at 20:34