$\mathrm{Aut}_\mathbb{Q}(\overline{\mathbb{Q}})$ is uncountable

Question

How do you show that $\mathrm{Aut}_\mathbb{Q}(\overline{\mathbb{Q}})$ is uncountable ?

Thanks in advance

Answer 1

Hint: Show there is an infinite set of mutually independent roots of order $2$ all of which are complex, then for every subset $A$ of this set there is a permutation which conjugates all the roots of those in $A$.

Answer 2

Consider any chain of unequal subfields $\mathbb{Q}\subset K_1\subset K_2\subset\ldots$ that are all normal over $\mathbb{Q}$ and whose union is $\overline{\mathbb{Q}}$. Then we can extend any automorphism of $K_2/K_1$ in finitely many (but more than one) ways, and that will extend to any of the (finitely many but more than one) automorphisms of $K_3/K_2$, etc. So we can build automorphisms of the algebraic closure inductively, each one of which is an infinite sequence of finitely many choices, and there are uncountably many such sequences.

Answer 3

In general, a profinite space is either finite or uncountable. (A profinite space is a compact, totally disconnected, Hausdorff space. The proof that such a space is either finite or uncoutable is completely topological; see if you can come up with it!)

But $G_{\mathbf Q}$, which is profinite, cannot be finite, since there are finite Galois extensions of $\mathbf Q$ of arbitrary high degree (and their Galois groups are quotient of $G_{\mathbf Q}$).

Answer 4

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 $\mathrm{Aut}_F(E)$ 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 $\mathrm{Aut}_{\mathbb{Q}}(\overline{\mathbb{Q}})$ 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 $\mathrm{Aut}_F(E)$ when $E/F$ is normal.)