Is $\text{Gal} (\overline{\Bbb Q}/\Bbb Q)$ countable or uncountable? It seems like it should be countable (because the algebraic closure of $\Bbb Q$ is countable and there are countably many permutations of the irrational algebraic numbers, and a countable union of countable sets is countable. However, I've seen references that seem to imply it is uncountable. What is the answer, and why/how?
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It's uncountable. There are uncountably many permutations of the algebraic numbers. – Qiaochu Yuan Apr 21 '16 at 06:04
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4@QiaochuYuan, there are uncountably many permutations of the complex numbers and yet the Galois group of $\mathbb C$ over $\mathbb R$ is finite! – Mariano Suárez-Álvarez Apr 21 '16 at 06:10
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1@Mariano: yes, of course this isn't an argument, I just wanted to point out that the OP's claim that there are countably many such permutations is false. – Qiaochu Yuan Apr 21 '16 at 06:14
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3There are no countably infinite Galois groups. – Lubin Apr 21 '16 at 19:48
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@Rob: related: http://math.stackexchange.com/questions/260223/aut-mathbbq-overline-mathbbq-is-uncountable?rq=1 – Watson Aug 29 '16 at 16:05
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Let $I\subseteq \Bbb N$ be any subset and let $K_I=\Bbb Q(\{\sqrt{p_i}\}_{i\in I})$ where $p_i$ is the $i^{th}$ prime. Then there are precisely $2^{\Bbb N}$ in fact the Galois group of the compositum of this extension is exactly isomorphic to
$$\prod_{i\in\Bbb N}\Bbb Z/2\Bbb Z$$
and this of course indicates there are uncountably many elements in $\text{Gal}(\overline{\Bbb Q}/\Bbb Q)$
You can also do a simple cardinality argument using inverse limits, but that technology is a bit stronger.
Adam Hughes
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Let me elaborate on Lubin's comment: there are no countably infinite Galois groups. This follows from a general statement about topological spaces: If $X$ is compact, Hausdorff with no isolated points, then $X$ is uncountable. For the proof, see here: https://proofwiki.org/wiki/Compact_Hausdorff_Space_with_no_Isolated_Points_is_Uncountable/Lemma
Kostas Kartas
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In general, if $E/F$ is an Galois extension of fields and $d$ is the dimension of $E$ as an $F$-vector space, then the cardinality of $\mathrm{Gal}(E/F)$ is exactly $2^d$ ($={\aleph_0}^d$) if $d$ is infinite (it is well known that it is $d$ if $d$ is finite). Since $[\overline{\mathbb{Q}}:\mathbb{Q}]=\aleph_0$, the cardinality of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ is exactly $2^{\aleph_0}$, the continuum.
The outline of a proof using transfinite recursion can be found here, and the proof has been checked by a computer using the proof assistant Lean. (The proof concerns $F$-embeddings of $E$ into $\overline{E}$ instead, which are in bijection with elements of $\mathrm{Gal}(E/F)$ when $E/F$ is normal.)
Junyan Xu
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