Are there any rigid algebraically closed fields? A field $k$ is rigid if its group of automorphisms $\mathrm{Aut}(k)$ is trivial. For example, the reals $\mathbb R$ and the prime fields $\mathbb Q$ and $\mathbb F_p$.
A field $k$ is algebraically closed if every nonconstant polynomial $p(x) \in k[x]$ has a root in $k$. For instance, the algebraic closures of the above: $\bar{\mathbb R} = \mathbb C$, $\bar{\mathbb Q}$, etc.

Is any field both rigid and algebraically closed?

I am inclined to say no, mostly because I couldn't think of any, but also because it feels like in an algebraically closed field, there should always be finite index subfields which can be fiddled around with. Regardless, I have no idea how to prove or disprove this.
 A: Any algebraically closed field has tons of automorphisms.  The key fact is that if $K$ and $L$ are algebraically closed and algebraic over subfields $k\subseteq K$ and $\ell\subseteq L$, then any isomorphism $f:k\to \ell$ extends to an isomorphism $\bar{f}:K\to L$.  The proof is to just extend $f$ to one element of $K$ at a time: given $\alpha\in K\setminus k$, there exists $\beta\in L$ whose minimal polynomial over $\ell$ is the polynomial obtained by applying $f$ to the coefficients of minimal polynomial of $\alpha$.  You can then extend $f$ to an isomorphism $k(\alpha)\to\ell(\beta)$ which sends $\alpha$ to $\beta$.  Repeating this by transfinite induction, you get an extension of $f$ to an isomorphism $\bar{f}$ defined on all of $K$.  The image of $\bar{f}$ must be all of $L$, since it is an algebraically closed field containing $\ell$ and $L$ is algebraic over $\ell$.
So given an algebraically closed field $K$, let $k_0$ be the prime subfield and let $B$ be a transcendence basis for $K$ over $k_0(B)$, so $K$ is algebraic over $k_0(B)$.  Now let $k\subset K$ be some nontrivial finite Galois extension of $k_0(B)$ and let $f:k\to k$ be a nontrivial automorphism.  By the key fact above (with $\ell=k$ and $L=K$), $f$ extends to a nontrivial automorphism of $K$.
(Note that this argument uses the axiom of choice, both in the proof of the key fact (to choose $\alpha$ and $\beta$ at every step of the induction) and in getting a transcendence basis for $K$ over $k_0$.  I don't know whether it is possible for a rigid algebraically closed field to exist if you don't assume the axiom of choice.)
A: This question deals with a similar topic.  Specifically, one answer cites a paper that constructs (at least one) real closed field with no non-trivial automorphisms.  Unfortunately, often fields of this sort are very difficult to describe, and I'm unable to find a .pdf online.
I encourage you to check out the question, but the paper (which I was able to find at sci hub, but won't link) is:

S. Shelah, Models with second order properties. IV. A general method and eliminating diamonds -- Annals Pure and Applied Logic 25 (1983) 183-212

DOI is http://dx.doi.org/10.1016/0168-0072(83)90013-1
