1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.