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I'm looking for extensions of the quadratic closure of $\mathbb{Q}$ of

degree $3$,
degree $5$.

Furthermore,

Does there exist an extension of degree $4$?

What I know:
I know that the quadratic closure of $\mathbb{Q}$ in $\overline{\mathbb{Q}}$ is $$Q^q:=\bigcup_{i=0}^\infty \mathbb{Q}_i,$$ where $\mathbb{Q}_0=\mathbb{Q}$, and $\mathbb{Q}_i=\mathbb{Q}_{i-1}(\sqrt{\mathbb{Q}_{i-1}})$.
However I wouldn't know how to find extensions hereof, especially when they're of a certain degree.
For the second question, I would say, intuitively, that the answer is no. But again I'm not sure why.

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  • $\begingroup$ $\mathbb Q^q(2^{\frac 13}),\mathbb Q^q(2^{\frac 15} )$ $\endgroup$ Commented Apr 22, 2016 at 17:16
  • $\begingroup$ @GeorgesElencwajg Can you maybe show me how I can prove that those are extensions of degree $3$,$5$, respectively? $\endgroup$ Commented Apr 22, 2016 at 18:06
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    $\begingroup$ Show that $T^3-2$ has no root in $\mathbb Q^q$, since every element of $\mathbb Q^q$ has degree over $\mathbb Q$ a power of $2$. $\endgroup$ Commented Apr 22, 2016 at 18:20
  • $\begingroup$ @GeorgesElencwajg Would you mind making an answer out of that? I don't really see it $\endgroup$ Commented Apr 22, 2016 at 20:44
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    $\begingroup$ Do you see that every element of $\Bbb Q^q$ is of degree a power of two over $\Bbb Q$? $\endgroup$ Commented Apr 24, 2016 at 15:27

1 Answer 1

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The question of whether $\Bbb Q^q$ has any extensions of degree $4$ is very interesting. It does, for:

Let $\rho$ be quartic over $\Bbb Q$ with its splitting field $K$ having $S_4$ as Galois group (over $\Bbb Q$). So $[K\colon \Bbb Q]=24$. This $K$ is the normal closure of $\Bbb Q(\rho)$, that is, it’s the intersection of all normal extensions of $\Bbb Q$ that contain $\rho$. In other words, every normal extension $\Omega$ of $\Bbb Q$ that contains $\rho$ must satisfy $K\subset\Omega$. Since $K\not\subset\Bbb Q^q$ (again, by degree considerations), we see that $\rho\notin\Bbb Q^q$.

Now I claim that $\rho$ is still quartic over $\Bbb Q^q$. Let $f$ be the $\Bbb Q$-minimal polynomial for $\rho$. Over $\Bbb Q^q$, $f$ will have no roots, thus will either split into a product of two quadratics, or remain irreducible. If product of two quadratics, the roots of one factor of $f$ will be quadratic over $\Bbb Q^q$, so in that field. Only remains that $f$ is irreducible over $\Bbb Q^q$, in other words $[\Bbb Q^q(\rho):\Bbb Q^q]=4$.

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