# Automorphism group of an infinite field.

My primary question is: is there an infinite field $F$ with a finite automorphism group $\text{Aut}(F)$?

So I tried fields with characteristic $2$, say $F_2(\pi)$. But it's still hard to make a positive conclusion since $\pi$'s image is seemingly arbitrary.

So is my conjecture right or not? If it's correct, could you give me any example?

• Sure, $\mathbb Q[\sqrt 2]$ has an automorphism group of size $2$. – Thomas Andrews Dec 23 '15 at 18:06
• Wait, are you talking field automorphisms? How is the automorphism group of the rationals infinite? – Thomas Andrews Dec 23 '15 at 18:06
• @ThomasAndrews Yeah I was talking about field/ring automorphisms. Sorry I seemed to confuse them with group automorphisms when I made that reasoning. I will edit. Thanks for pointing out. – Vim Dec 23 '15 at 18:09
• What field? Be specific. There is only one field automorphism of $\mathbb Q$, so you must be talking about some other sort of automorphism. – Thomas Andrews Dec 23 '15 at 18:10
• Also, it really really doesn't make sense to write $\mathbb F_2(\pi)$. $\pi$ is a real number, and can't be used as a generic indeterminate. As it turns out, $\mathbb Q(\pi)\cong\mathbb Q(x)$, the ring of rational functions if an indeterminate $x$, but that doesn't mean that $F(\pi)$ makes sense for all fields $F$. – Thomas Andrews Dec 23 '15 at 18:15

Aut$(\mathbb{R})$ is trivial. You can prove by induction that any automorphism $\phi$ fixes the natural numbers. Then using the definition of homomorphism you can show that it fixes the integers, then the rational numbers. Also, $\phi$ is order preserving since if $x>y$, then $x-y=r^2$ for some nonzero $r$, so $\phi(x)-\phi(y)=\phi(r^2)$. Finally, if $\phi(x) \neq x$, then wlog $\phi(x)<x$, so choose a rational $q$ with $\phi(x)<q<x$, and then we have $\phi(q)<\phi(x)$, but $\phi(q)=q>\phi(x)$, a contradiction. So $\phi$ is the identity.

Sorry, I should note that the first part of the argument shows that Aut$\mathbb{Q}$ is also trivial.

• The "order-preserving" argument is spectacular. Thanks and plus one! – Vim Dec 23 '15 at 18:22
• That's right, thanks! – Noah Olander Dec 23 '15 at 18:38
• Note that order is crucial - once we pass to $\mathbb{C}$, the automorphism group becomes huge. – Noah Schweber Dec 23 '15 at 18:43
• – lhf Dec 23 '15 at 19:32
• Almost verbatim, if $F$ and $G$ are isomorphic subfields of $R$, where $F$ contains the square roots of all its positive members, and if $\psi:F\to G$ is a field-isomorphism ,then $\psi = id_F$ and $F=G$. – DanielWainfleet Dec 23 '15 at 20:31

The group $\mathrm{Aut}(\mathbb Q)$ is actually trivial and therefore finite! Take an automorphism $$\sigma:\mathbb Q\to\mathbb Q$$

Suppose $\frac ab\in \mathbb Q^\times$ with $a,b\in \mathbb N$. Then \begin{align} \sigma(\frac ab)&=\frac{\sigma(a)}{\sigma(b)}\\&=\frac{\sigma(\overbrace{1+\cdots +1)}^{a\text{ times}}}{\sigma(\underbrace{1+\cdots +1)}_{b\text{ times}}}\\&=\frac{\overbrace{\sigma(1)+\cdots +\sigma(1)}^{a\text{ times}}}{\underbrace{\sigma(1)+\cdots +\sigma(1)}_{b\text{ times}}}\\&=\frac ab&&\text{since }\sigma(1)= 1. \end{align}

• Nice example. Thanks! – Vim Dec 23 '15 at 18:20

The $p$-adic numbers $\Bbb Q_p$ also have trivial automorphism group. The proof starts out in the same way as for $\Bbb R$, and then you have to show $\varphi(z)=z$ for $z\in\Bbb Q$ implies the same for $z\in\Bbb Q_p$. For this, you have to show that any automorphism $\varphi$ of the field is automatically continuous. I suppose there are many methods of doing this, but one way is to look at the set $S$ defined in the following purely algebraic way: an element $s\in\Bbb Q_p$ is in $S$ if and only if for every $m$ prime to $p$, $1+s$ has an $m$-th root in $\Bbb Q_p$.

Notice that such $s$ must necessarily be in $\Bbb Z_p$, the integers of the field. Furthermore, such $s$ can not be a unit of $\Bbb Z_p$, because either $1+s\in p\Bbb Z_p$, and has only finitely many roots in $\Bbb Q_p$, or $1+s$ is also a unit (necessarily then $p>2$), and doesn’t have a $(p-1)^2$-th root in $\Bbb Q_p$, because its class modulo $p$ doesn’t even have such a root.

On the other hand if $s\in p\Bbb Z_p$, then $1+s$ is a principal unit, and has an $m$-th root in $1+p\Bbb Z_p$ for all $m$ prime to $p$. Thus $S=p\Bbb Z_p$, whose powers are a neighborhood base at zero for the $p$-adic topology. These sets are consequently preserved by the automorphism $\varphi$, so $\varphi$ is continuous.