Cardinality of automorphisms of $\mathbb C $

Question

I'm studing the proof that $\mathbb C $ has infinite automorphisms involving Zorn's lemma: basically one takes an automorphism $\eta $ of a subfield $F $ of $\mathbb C $, and considers the set $V $ of automorphisms $\phi :E \to E$, with $F\le E $, such that $\phi|_F =\eta$. If one orders this set of automorphisms like this: $\alpha \lt \beta $ iff $\beta |_A = \alpha$, where $\alpha $ is an automorphism of $A\ge F $ and $\beta $ is a automorphism of $B\ge A $. Then since every totally ordered subset of $V $ has a upper bound, there must be a maximal element $\gamma$ in $V $. Clearly this automorphism is defined over all $\mathbb C $ (if not, we could extend it to an automorphism $\gamma_1 \ge \gamma $). However I don't understand how to show that the cardinality of such automorphisms is at least the cardinality of $\mathbb C $: how should I proceed? Thanks

Answer 1

Pick a transcendental number $\alpha\in\mathbb C$. The field $\mathbb Q(\alpha)$ and also the algebraic closure of it $\overline{\mathbb Q(\alpha)}\le\mathbb C$ are countable, so you can pick a new transcendental number $\beta\in\mathbb C$ outside of this algebraic closure, i.e. algebraically independent with $\alpha$ over $\mathbb Q$. Any such choice of $\beta$ gives you a field $\mathbb Q(\alpha, \beta)\le\mathbb C$ and an automorphism $\varphi_{\alpha, \beta}$ which maps $\alpha$ to $\beta$ (and vice versa, by the way). As per your construction, this automorphism then extends to an automorphism of $\mathbb C$ that also maps $\alpha$ to $\beta$.

So, keeping $\alpha$ fixed, and choosing different $\beta$'s, gives you different automorphisms of $\mathbb C$. As said before, $\beta$ can be picked from $\mathbb C\setminus\overline{\mathbb Q(\alpha)}$, which is of the same cardinality as the whole $\mathbb C$ (because the said algebraic closure is countable).