Suppose we consider a rigid extension field $F$, i.e., $\text{Aut}(F) = 1$ over the complex numbers $\mathbb C$. What is the minimal cardinality of $F$? In particular it should hold that in this case $|F| > |\mathbb C|$.


Moreover, if we replace $\mathbb C$ with any other algebraically closed field, what can one say in this case? 


Any comment, reference, or pointer is highly appreciated.


All the best,
Sebastian



What is the size of the smallest rigid extension field of the complex numbers?
Firstly, it may be worth linking to [this related question](https://mathoverflow.net/questions/22897/) on MO. 


Pröhle proved that all fields of characteristic 0 can be embedded in a rigid field - see ["Does a given subfield of characteristic zero imply any restriction to the endomorphism monoids of fields?"](http://www.ams.org/mathscinet-getitem?mr=862178) for his particular construction.


Later, Dugas and Göbel showed in 
["All infinite groups are Galois groups over any field"](http://www.ams.org/mathscinet-getitem?mr=0906820) that for ${\mathbb C}$ it can be done in a field of cardinality the successor cardinal of $2^{\aleph\_0}$ which is as good as you could expect.


In follow-up papers 
["Automorphism groups of fields I"](http://www.ams.org/mathscinet-getitem?mr=1305739) and ["Automorphism groups of fields II"](http://www.ams.org/mathscinet-getitem?mr=1481565), they show that for any group $G$ and any field $K$, there exists an extension with automorphism group isomorphic to $G$ and cardinality $\aleph\_0|K||G|$.


