# When is a infinite transcendence-degree rigid fields fixed by a finite extension?

A field is rigid if it has no nontrivial automorphisms. Let $$F$$ be a rigid field which has infinite transcendence degree over $$\mathbb{Q}$$, and let $$E$$ be a finite extension of $$F$$. Then my question is, under what conditions does every automorphism of $$E$$ fix $$F$$?

Are any nontrivial conditions known which guarantee that?

Not much of a criterion, but here is a family of examples: this holds for $$F=\mathbb Q_p$$ and $$E$$ any finite extension. To see that $$F$$ is rigid and any automorphism of $$E$$ fixes it, it is enough to note that any automorphism of $$E$$ is continuous, since $$\mathbb Q$$ will then be pointwise fixed by an automorphism, and its closure is $$\mathbb Q_p$$.
To see any automorphism of $$E$$ is continuous, first we show that its ring of integers $$O_E$$ is first-order definable in $$E$$. Indeed, let $$e$$ be the ramification degree of $$E/F$$. I claim that there exist $$a,b,c\in\mathbb N$$ such that for $$x\in E$$ we have $$x\in O_E$$ iff $$1+p^ax^b=y^c$$ has a solution $$y\in E$$.
First note that if $$x\in O_E$$, the polynomial $$f(T)=T^a-(1+p^bx^c)$$ satisfies $$|f(1)|=|p^bx^c|\leq |p^b|=p^{-b}$$, while $$|f'(1)|=|a|$$ (throughout, the absolute value is the $$p$$-adic one). If $$p^{-b}<|a|^2$$, then from a suitable version of Hensel's lemma we get that $$f$$ has a root.
On the other hand, if $$x\not\in O_E$$, then $$|x|=p^{d/e}$$ for some $$d\geq 1$$. If $$c>be$$, then $$|p^bx^c|=p^{-b}|x|^c\geq p^{-b}p^{c/e}>1$$, which shows $$|1+p^bx^c|=|p^bx^c|=p^{-b+cd/e}=p^{(cd-be)/e}$$. If we arrange that $$cd-be$$ is never divisible by $$a$$, then this will restrict $$|1+p^bx^c|$$ from being an $$a$$-th power. All of these conditions can be fulfilled by taking $$a$$ to be a prime not dividing $$e$$, $$b$$ sufficiently large and not divisible by $$a$$, and $$c$$ a multiple of $$a$$ larger than $$be$$, and this gives us the Diophantine description of $$O_E$$ in $$E$$.
By above, any automorphism of $$E$$ must preserve $$O_E$$, and hence also every set $$p^nO_E,n\in\mathbb N$$. Since the latter form a basis of neighbourhoods of $$0$$ in $$E$$, this gives the desired continuity.
This can be extended to some finite extensions $$F/\mathbb Q_p$$ and extensions $$E/F$$ such that no automorphism of $$E/\mathbb Q_p$$ is identity on $$F$$. Those can be without much difficulty constructed using Galois theory.