# Is the absolute Galois Group of $\Bbb Q$ countable?

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?

• It's uncountable. There are uncountably many permutations of the algebraic numbers. Apr 21 '16 at 6:04
• @QiaochuYuan, there are uncountably many permutations of the complex numbers and yet the Galois group of $\mathbb C$ over $\mathbb R$ is finite! Apr 21 '16 at 6:10
• @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. Apr 21 '16 at 6:14
• There are no countably infinite Galois groups. Apr 21 '16 at 19:48
• @Rob: related: math.stackexchange.com/questions/260223/… Aug 29 '16 at 16:05

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.

• Precisely $2^{\mathbb{N}}$ what, exactly? Apr 21 '16 at 6:13
• @Rob: elements of the Galois group of this subextension. Apr 21 '16 at 6:15
• Excellent answer. Thanks for the easy to understand explanation! Apr 21 '16 at 6:16

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