Automorphism of Complex field

Question

How we can get automorphism $Aut(C/Q)$ which fixes $Q$ ? I know that since transcendence basis for C/Q is infinite from this how can we say order of $Aut (C)$ is infinite because if this is the case then $Aut (R)$ is also infinite

Answer 1

Fix a transcendence basis $\mathscr{B}$ of $\mathbb{C}$ over $\mathbb{Q}$. Then $\mathscr{B}$ is infinite (actually of cardinality $2^{\aleph_0}$).

If $\alpha\colon\mathscr{B}\to\mathscr{B}$ is any bijection, this extends to an automorphism of $\mathbb{C}$ because of the properties of transcendence bases and the fact $\mathbb{C}$ is algebraically closed. If $\hat{\alpha}\colon\mathbb{C}\to\mathbb{C}$ is the endomorphism defined by extending $\alpha$, then $\mathbb{C}$ is algebraic over the image of $\hat{\alpha}$ hence equal to it. Hence $\hat{\alpha}$ is an automorphism.

Any infinite set has infinitely many permutations.

However, this has no consequence on the group of automorphisms of the reals, which is trivial: the only field automorphism of $\mathbb{R}$ is the identity. If you follow the same path as before, you cannot conclude that the image of $\hat{\alpha}$ is $\mathbb{R}$. See Is an automorphism of the field of real numbers the identity map? for more on this topic.