# 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.

• Finite index subfields of algebraically closed fields are actually quite rare: they never exist in positive characteristic, and in characteristic 0 every finite index subfield of an algebraically closed field has index 2. This is known as the Artin-Schreier theorem (actually, the theorem is a bit stronger and says more about what such subfields look like in the characteristic 0 case). – Eric Wofsey Jan 11 '17 at 3:40

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.)

• Re the last part, we should ask Andreas Blass, which is a user here. :) – Pedro Tamaroff Jan 11 '17 at 3:37
• Why does $k$ have nontrivial automorphisms? How does your argument work when, say, $K = \mathbb C$? – algorithmshark Jan 11 '17 at 3:40
• $k$ has nontrivial automorphisms since it is a nontrivial Galois extension of $k_0(B)$, and so it has $[k:k_0(B)]$ different automorphisms that fix $k_0(B)$. For a "concrete" example with $K=\mathbb{C}$, let $B$ be a transcendence basis for $\mathbb{C}$ over $\mathbb{Q}$, and let $k=\mathbb{Q}(B)(\sqrt{2})$. Then $k$ has an automorphism that fixes $\mathbb{Q}(B)$ and sends $\sqrt{2}$ to $-\sqrt{2}$. – Eric Wofsey Jan 11 '17 at 3:43

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