Can a nontrivial $\phi \in Gal(\bar K /\bar k )$ preserve all algebraically closed subfields?

Question

The answer is "yes" if $\bar K = \overline{k(x)}$ (in fact, every $\phi \in Gal(\overline{k(x)}/\bar k)$ preserves all algebraically closed subfields -- there's only two of them, so it's not that hard!), but I'm not sure about higher transcendence degree. I suspect that for large enough $\bar K$ the answer must be "no", probably already at $\bar K = \overline{k(x,y)}$.

To clarify, as LSpice points out below, there are plenty of automorphisms $\phi \in Gal(\overline{k(x,y)}/\bar k)$ which do not preserve all algebraically closed subfields -- e.g. automorphisms swapping $x$ and $y$. So it's certainly not the case that all $\phi \in Gal(\bar K / \bar k)$ preserve all algebraically closed subfields. The question is for what $\bar K, \bar k$ there exists some such $\phi$ (other than the identity).

For context, we might perturb the question. There are no nontrivial automorphisms of $k(x,y)/k$ preserving all subfields. Any automorphisms of a vector space preserving all subspaces, is multiplication by a scalar.

I'd really like to ask this question for an arbitrary first-order theory, but perhaps I will save that for a future post.

Answer 1

Let me rephrase your question using a little bit of model theory (actually combinatorial geometry). Since the theory of algebraically closed fields is strongly minimal, the algebraic closure induces a pregeometry. More precisely given $\bar k \subseteq \bar K$ there is a pregeometry $G(\bar K/\bar k) = (\bar K, cl_{\bar k})$ associated to it where $cl_{\bar k}(A)$ is the algebraic closure of the field generated by $\bar k$ and $A$. Now your question becomes if there is a nontrivial automorphism of $\bar K/\bar k$ that fixes setwise all closed sets in $G(\bar K/\bar k)$.

Since we are interested only in the closed sets the right object to look at is the geometry $\underline G$ associated to the pregeometry $G = (A, cl)$. To obtain this we define an equivalence relation on $A$ as $a \sim b$ if $cl(a) = cl(b)$ and look at $(A \setminus cl(\emptyset)) / \sim$. The closure is defined as $cl(B/\sim) = \{a/\sim : a \in cl(B)\}$. Now an automorphism of $G$ induces an automorphism of $\underline G$ and it preserves closed sets of $G$ if and only if the induced automorphism on $\underline G$ is the identity.

To sum up we associate a geometry $\underline G(\bar K/\bar k)$ to the pair $\bar k \subseteq \bar K$. There is a map $\sigma : Aut(\bar K/\bar k) \to Aut(\underline G(\bar K/\bar k))$ and your questions is whether it is injective. This map actually factors through $Aut(\bar K)_{\{\bar k\}}$ - the set of automorphisms of $\bar K$ that fix $\bar k$ setwise. So let $\chi : Aut(\bar K)_{\{\bar k\}} \to Aut(\underline G(\bar K/\bar k))$ be the natural map, (i.e. $\chi(g)(cl_{\bar k}(x)) = cl_{\bar k}(gx)$) and $i : Aut(\bar K/\bar k) \to Aut(\bar K)_{\bar k}$ be the inclusion map. Then $\sigma = \chi \circ i$.

The map $\chi$ is studied by Evans and Hrushovski in The Automorphism Group of the Combinatorial Geometry of an Algebraically Closed Field (Journal of LMS, Volume 52, 1995) under the assumption that the transcendence degree of $\bar K$ over $\bar k$ is at least $5$. One of their main results is that it is surjective. On page 214 they also remark that if the characteristic of $\bar K$ is nonzero, then the kernel of $\chi$ consists of powers of Frobenius and otherwise it is trivial. This then implies that the kernel of $\sigma$ is trivial. Therefore the answer to your question is negative if the transcendence degree of $\bar K$ over $\bar k$ is at least $5$.

Answer 2

Here's an argument that only trivial such automorphisms exist already for $\overline{k(x,y)}/k$ (note that in this post I don't require the automorphism to fix $\bar k$). It may be similar to the argument in the Evans-Hrushovski paper linked to by Levon Haykazyan; I haven't looked closely.

Let $\bar K / k$ have transcendence degree $\geq 2$. Say that $\phi \in Gal(\bar K / k)$ is infinitesimal if it preserves all algebraically-closed subfields. Let $G \subset Gal(\bar K / k)$ be the group of infinitesimal automorphisms. Note that $G$ is normal in $Gal(\bar K / k)$.

Lemma: If $\phi \in G$, then $\phi$ has no fixed points outside of $\bar k$.

Proof. Assume there is $\phi \in G$ with fixed points outside of $\bar k$. Then we can find $s$ a fixed point and $t$ a non-fixed point with $s,t$ algebraically independent. Let $u = \phi(t) - t$. Then

• Because $\phi$ preserves $\overline{k(t)}$, we have $\overline{k(u)} \subseteq \overline{k(t)}$, since $u = \phi(t) - t$;

• Because $\phi$ preserves $\overline{k(s+t)}$ we have $\overline{k(u)} \subseteq \overline{k(s+t)}$, since $u = \phi(t) - t = \phi(s+t) - (s+t)$.

Both these inclusions cannot be equalities, because $t$ and $s+t$ are algebraically indepdent and consqeuently $\overline{k(s+t)} \neq \overline{k(t)}$. But from either inclusion being proper, we conclude that $u \in \bar k$. So $\phi(t) = t + u$ with $u \in \bar k$. Let $1 < d \in \mathbb{N}$ be coprime to the characteristic of $k$. If $t^d$ is not fixed by $\phi$, then the same arguments as before show that $\phi(t^d) = t^d + u'$ with $u' \in \bar k$; if $t^d$ is fixed by $\phi$ then we still have such an equation with $u' = 0$. But we also have $\phi(t^d) = (t+u)^d$, so that $(x+u)^d - x^d - u' = 0$ is a polynomial over $\bar k$ satisfied by $t$. Because $d$ is not 1 or divisible by the characteristic of $k$ and $u \neq 0$, this is not the zero polynomial. So $t \in \bar k$, a contradiction. So there are no fixed points outside of $\bar k$.

Corollary: If $\phi \in G$ and $s \in \bar K \setminus \bar k$, then $\phi$ is uniquely determined by the value of $\phi(s)$.

Proof. If there is $\psi \in G$ with $\psi(s) = \phi(s)$, then $\psi^{-1}\phi$ fixes $s$ and so must be the identity by the lemma.

Corollary: $G$ is the group of Frobenius automorphisms, and in particular the part fixing $\bar k$ is trivial.

Proof. It suffices to show this for $\bar K = \overline{k(s,t)}$. For any automorphism $\alpha \in Gal(\overline{k(s)}/k)$, choose a lift $\beta \in Gal(\overline{k(s,t)}/k)$ which fixes $t$. If $\phi \in G$, then $\beta \phi \beta^{-1} \in G$. We have $\beta \phi \beta^{-1}(t) = \phi(t)$; by the corollary, then, $\beta \phi \beta^{-1} = \phi$. So $\phi(s) = \beta \phi \beta^{-1}(s) = \alpha \phi|_{\overline{k(s)}}\alpha^{-1}(s)$. That is, $\phi|_{\overline{k(s)}}$ lies in the center of $Gal(\overline{k(s)}/k)$, which is known to consist only of the Frobenius automorphisms. Since $\phi \in G$ is determined by $\phi|_{\overline{k(s)}}$, it follows that that $\phi$ is also a Frobenius automorphism.

Actually, I'm not sure I can prove that the center of $Gal(\overline{k(s)}/k)$ is just the Frobenius maps. So here's another argument. When we allow $s,t$ to vary, $Gal(\overline{k(s,t)}/k)$ is generated by elements like $\alpha$ above. So we see that $\phi$ in fact commutes with all of $Gal(\overline{k(s,t)}/k)$. This allows us to extend $\phi$ to an automorphism of $\bar K / k$ for any transcendence degree -- we can define it locally, and we get a homomorphism because we only have to check addition and multiplication, which are binary operations, and knowing how $\phi$ works on degree-2 extensions lets us handle this. Then we can pass to a very large $\bar K$ and conclude that because $\phi$ commutes with all automorphisms, it must be definable, i.e. definable by a polynomial. Since $\phi$ is invertible, the polynomial can only have one root, so it must be a Frobenius map.