# Galois group, algebraic closure over maximal extension

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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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.
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)