How many automorphisms does the abstract field \(\CC\) have? How can an automorphism of a subfield, for example \(\sqrt2\mapsto-\sqrt2\) on \(\QQ(\sqrt2)\), be extended to an automorphism of \(\CC\)?
Solution
As an abstract field, \[ |\Aut(\CC)|=2^{2^{\aleph_0}}. \] These automorphisms are generally wildly discontinuous; the only continuous field automorphisms are the identity and complex conjugation.
1. Any automorphism of a subfield \(F\subseteq\CC\) extends to an automorphism of \(\CC\).
Let \(\sigma:F\to F\) be a field automorphism. Choose a transcendence basis \(B\) of \(\CC/F\). Extend \(\sigma\) to an automorphism of the purely transcendental field \(F(B)\) by fixing every element of \(B\).
By definition of transcendence basis, \(\CC\) is algebraic over \(F(B)\); since \(\CC\) is algebraically closed, the standard extension theorem for embeddings into algebraically closed fields extends this automorphism of \(F(B)\) to an automorphism of \(\CC\).
In particular, the nontrivial automorphism of \(\QQ(\sqrt2)\) given by \(\sqrt2\mapsto-\sqrt2\) extends to an automorphism of \(\CC\).
2. There are at least \(2^{2^{\aleph_0}}\) automorphisms.
Let \(B\) be a transcendence basis of \(\CC/\QQ\). Since \(|\CC|=2^{\aleph_0}\) and \(\overline\QQ\) is countable, one has \[ |B|=2^{\aleph_0}. \] Every permutation of \(B\) extends to an automorphism of \(\QQ(B)\) and then, by 1, to an automorphism of \(\CC\). The number of permutations of a set of cardinality \(\mathfrak c=2^{\aleph_0}\) is \[ 2^{\mathfrak c}=2^{2^{\aleph_0}}. \]
3. There are at most \(2^{2^{\aleph_0}}\) automorphisms.
Every automorphism is a function \(\CC\to\CC\). Hence \[ |\Aut(\CC)|\le |\CC|^{|\CC|} =\mathfrak c^{\mathfrak c} =2^{\mathfrak c}. \] Combining with 2 gives equality.