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

Thanks in advance

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3 Answers 3

up vote 10 down vote accepted

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 $A$.

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I see, thanks ! –  Zorba le Grec Dec 16 '12 at 20:30

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.

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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 $G_{\mathbf Q}$).

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