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Let $\overline{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$. Let $\alpha \in \overline{\mathbb{Q}}\setminus \mathbb{Q}$ and let $K \subset \overline{\mathbb{Q}}$ be a maximal extension of $\mathbb{Q}$ in respect to not containing $\alpha$ (so $\alpha \notin K$, but $\alpha$ in every nontrivial extension of $K$). Let $G$ be the Galois group of $\overline{Q}$ over $K$. Show that either $G = \mathbb{Z}/2\mathbb{Z}$ or $G = \mathbb{Z}_p$ ($p$-adic integers) for some prime $p$.

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    – Lord_Farin
    Jan 17, 2015 at 22:24

2 Answers 2

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We first take care of the case $[\overline{\mathbb Q}: K]<\infty$. In this situation, $K$ is a field which is not algebraically closed, but whose algebraic closure is a finite extension. By the Artin-Schreier Theorem, the degree of this finite extension must be $2$, so the desired result follows in this case.

From this point onwards we assume that $\overline{\mathbb Q}$ is an infinite extension of $K$, and the aim will be to prove that the Galois group is some $\mathbb Z_{p}$. If we manage to show that the finite extensions of $K$ in $\overline{\mathbb Q}$ form a tower of cyclic extensions of degrees $p^{n}$, $n\ge 0$ $($one for each $n$$)$ for some prime $p$, then we are done: it will follow from this that the Galois group of $\overline{\mathbb Q}/K$ is the projective limit of the groups $\mathbb Z/p^{n}\mathbb Z$, which is precisely $\mathbb Z_{p}$.

First we prove that $K(\alpha)/K$ is cyclic of degree $p$ for some prime $p$. This extension is Galois: let $\alpha=\alpha_{1},\dots,\alpha_{n}$ be the roots of the minimal polynomial $f$ of $\alpha$ over $K$. Each $K(\alpha_{i})$ contains $K$ strictly, so, by the maximality assumption on $K$, it must contain $K(\alpha)$ as well. Since all $K(\alpha_{i})/K$ have degree $n$, they all coincide, which means that $K(\alpha)$ is the splitting field of $f$, and hence a normal extension of $K$. It is also finite and separable, of course, so the extension is indeed Galois. If its Galois group were not cyclic of prime order, it would have a nontrivial proper subgroup. The fixed field of this subgroup would then be an extension of $K$, lying strictly between $K$ and $K(\alpha)$. Again, this contradicts the maximality assumption on $K$, and finishes the proof of the fact that $K(\alpha)/K$ is cyclic of degree $p$.

Now we show that all finite extensions of $K$ in $\overline{\mathbb Q}$ have order $p^{n}$ for some $n\ge 0$. Assume the contrary. Then we can find a Galois extension $F/K$ whose degree is not a power of $p$. The fixed field of a Sylow $p$-subgroup of $\text{Gal}(F/K)$ has degree greater than $1$ and coprime to $p$ over $K$. Hence it is a field strictly larger than $K$ which cannot contain $\alpha$, which is a contradiction.

The final step of the proof consists in showing that for each $n$ there is precisely one extension of $K$ in $\overline{\mathbb Q}$ of degree $p^{n}$, which is cyclic. By our assumption that $\overline{\mathbb Q}/K$ is infinite, it follows that there are extensions of $K$ of degree $p^{n}$ for any $n$. Choose an extension $F/K$ of degree $p^{n}$, which we may assume Galois $($otherwise replace it with a Galois extension of even larger degree$)$. $G=\text{Gal}(F/K)$ is a group of order $p^{n}$, which by elementary group theory has subgroups of all possible orders $($i.e. all divisors of $p^{n}$$)$. Moreover, a subgroup of $G$ of order $p^{m}$, say, is part of a chain of subgroups of orders $1,p, \dots,p^{m-1},p^{m}, \dots,p^{n}$. The fixed field of a subgroup of $G$ of order $p^{n-1}$ has degree $p$ over $K$. Since there is only one such field, $K(\alpha),\ G$ has only one subgroup $H$ of order $p^{n-1}$. All subgroups of $G$ of order $\le p^{n-2}$ are thus subgroups of $H$ by the preceding discussion, so we can repeat the procedure with $H$ instead of $G$ and conclude that $G$ has precisely one subgroup of order $p^{m}$ for every $m\le n$, which, in turn, implies that $G$ is cyclic. The existence and uniqueness of cyclic extensions of $K$ of degrees $p^{n}$, $n\ge 0$ follow from this.


We remark that our proof immediately generalizes to arbitrary perfect fields $F$ in place of $\mathbb{Q}$; indeed, we used separability at one point. If we took some element $\alpha\in \overline{F}\setminus F^{\text{sep}}$ with $F$ nonperfect, then there was some field $K$ satisfying the hypotheses of the problem such that $F^{\text{sep}}\subseteq K$; therefore $\text{Gal}(\overline{F}/K)\subseteq \text{Gal}(\overline{F}/F^{\text{sep}})=0$. Of course, this gives just one more possibility in the general case.

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One thing I see here is that if $\alpha\in\bar{\mathbb{Q}}\setminus\mathbb{R}$ then $\mathbb{R}\cap\bar{\mathbb{Q}}$ does not contain $\alpha$ and thus $K\supseteq \mathbb{R}\cap\bar{\mathbb{Q}}$. By Artin-Schreier, and by basic Galois theory, this means that $$\mathbf{Gal}(\bar{\mathbb{Q}}/K)\subseteq \mathbf{Gal}(\bar{\mathbb{Q}}/(\mathbb{R}\cap\bar{\mathbb{Q}}))=\mathbb Z/2\mathbb Z,$$ and since $K$ is not algebraically closed the LHS is not trivial and hence this is an equality.

Still not sure yet why the $p$-adic part should hold if $\alpha\in\mathbb R$ (if it does, could be that there are more cases to examine)

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