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A question inspired by the concept of constructible numbers.

Let $K$ be a field, $\Omega$ some algebraic closure of $K$ and $p$ a prime number. Let $L$ be the set of all elements $\alpha \in \Omega$ for which $\dim_K(K[\alpha])$ is a power of $p$. Is $L$ a subring/subfield of $\Omega$ in general? If not, under what conditions on $p$ or $K$ is it?

The answer is positive for the constructible numbers ($p = 2, K = \mathbb{Q}$).

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  • $\begingroup$ Perhaps the constructible field is the compositum of all fields gotten as towers of quadratic extensions. You’ll have to show that if $K$ and $L$ are both at the top of such towers, so is $KL$. It should be a direct verification. $\endgroup$ – Lubin Feb 22 '17 at 22:32
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$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$$\newcommand{\C}{\mathbb{C}}$$\newcommand{\Q}{\mathbb{Q}}$Let $K = \Q$, $\Omega = \C$, $p = 3$, and $\omega \in \C$ a primitive third root of unity.

Then $\sqrt[3]{2}, \omega \sqrt[3]{2} \in L$, but $\omega = (\sqrt[3]{2})^{-1} \cdot \omega \sqrt[3]{2} \notin L$, as $\Size{\Q[\omega] : \Q} = 2$

So $L$ is not a subfield in this case, nor a subring, as $(\sqrt[3]{2})^{-1} \in \Q[\sqrt[3]{2}]$.

You can do the same, mutatis mutandis, for any odd prime $p$. The reason the argument fails for $p = 2$ is of course that $\Size{\Q[-1] : \Q} = 1 = 2^{0}$.

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  • $\begingroup$ Interesting answer, thanks! Could you give a (sketch of a) general proof in the case $p = 2$? All proofs I have found seem to somehow rely on geometric constructions rather than pure algebra. $\endgroup$ – Bib-lost Feb 21 '17 at 17:37
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    $\begingroup$ @Bib-lost, on re-reading your post, I believe there might be a misconception. The result is not true for $p = 2$ either. Just consider a polynomial $f \in \mathbb{Q}[x]$ of degree $4$, its splitting field $E$, and assume the Galois group is $S_{4}$. Then $E$ is generated by the four roots of $f$, and for each $\alpha$ of them one has $\lvert \mathbb{Q}[\alpha] : \mathbb{Q} \rvert = 4$, but then $E$ contains elements $\beta$ such that $\lvert \mathbb{Q}[\beta] : \mathbb{Q} \rvert = 3$. $\endgroup$ – Andreas Caranti Feb 22 '17 at 12:05
  • $\begingroup$ In that case I need to work on my understanding of the constructible numbers. Thanks for pointing this out. $\endgroup$ – Bib-lost Feb 22 '17 at 15:22

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