Can it happen that a proper subfield of \(\mathbb{C}\) is isomorphic to \(\mathbb{C}\)? How?
Solution
Yes.
Choose a transcendence basis \(B\) of \(\mathbb C/\mathbb Q\). Then \[ |B|=\operatorname{trdeg}_{\mathbb Q}\mathbb C=2^{\aleph_0}. \] Pick \(t\in B\) and put \[ B_0=B\setminus\{t\}. \] Since \(B\) is infinite, \[ |B_0|=|B|. \] Let \(K\) be the algebraic closure of \(\mathbb Q(B_0)\) inside \(\mathbb C\), i.e. the set of elements of \(\mathbb C\) algebraic over \(\mathbb Q(B_0)\).
The field \(K\) is proper because \(t\) is transcendental over \(\mathbb Q(B_0)\), so \(t\notin K\). On the other hand, \(K\) is algebraically closed of characteristic \(0\) and \[ \operatorname{trdeg}_{\mathbb Q}K=|B_0|=|B|=\operatorname{trdeg}_{\mathbb Q}\mathbb C. \] By the classification of algebraically closed fields by characteristic and transcendence degree, \[ K\cong\mathbb C. \] Thus \(\mathbb C\) has a proper subfield isomorphic to itself.
Equivalently, a bijection \(B\to B_0\) extends to an isomorphism \[ \mathbb C\xrightarrow{\sim}K\subsetneq\mathbb C, \] giving an injective but non-surjective field endomorphism of \(\mathbb C\).