Fields with trivial automorphism group have been addressed in a question on Mathoverflow (see this). However, I didn't find sufficient information of number fields with such properties.

Q. Let $K$ be a number field whose only field automorphism is trivial one. What properties $K$ must satisfy? For example, can it contain some roots of unity (other than $1,-1$)? Can their degree over $\mathbb{Q}$ be prime power? any composite number?

  • 3
    $\begingroup$ If $K/\Bbb Q$ is a number field there is a Galois closure $L/\Bbb Q$ with Galois group $G$ and unique subgroup $H$ for which $K=L^H$ (the elements in $L$ fixed by all elements in $H$). For $g$ to stabilize $K$, it must preserve the property of being fixed by $h\in H$, so $hgk=gk$ for all $h\in H,k\in K$, or equivalently $(g^{-1}hg)k=k$ which means $g^{-1}hg\in H$ so $g$ normalizes $H$. Note every automorphism of $K$ extends to one of $L$, so this means the automorphism group of $K$ is $N_G(H)/H$. To be trivial, $H$ must be its own normalizer. $\endgroup$ – runway44 Jul 1 at 4:40
  • 1
    $\begingroup$ Under the Galois correspondence between subfields of $L$ and subgroups of $G$, Galois extensions correspond to normal subgroups and Galois closures correspond to normal cores. Thus, for the Galois closure of $K$ to be $L$, the normal core of $H$ must be trivial. In other words, the Galois group $G$ cannot contain any nontrivial normal subgroup itself contained within $H$. $\endgroup$ – runway44 Jul 1 at 4:40

You have a wide range of examples. Obviously, extensiosn of degree $2$ have a non trivial automorphisms, so assume $n\geq 3$.

Assume that $K/\mathbb{Q}$ is an extension of degree $n$ which has no non-trivial subextensions and which is not cyclic of prime degree (this always exists, see below) . Then any automorphism of $K$ is trivial.

Indeed,assume that $\sigma$ is an automorphism.

By Artin's lemma, $K/K^{\langle \sigma\rangle}$ has degree $o(\sigma)$.Now $K^{\langle \sigma\rangle}$ is a subextension. By assumption,$K^{\langle \sigma\rangle}=K$, and in this casee $o(\sigma)=1$ and $\sigma=Id$, or $K^{\langle \sigma\rangle}=\mathbb{Q}$. In this case, it means that it is cyclic, and since there is no nontrivial extensions, then $n$ must be prime.

To construct such $K$, here is how to proceed: take the generic extension $\mathbb{Q}(t_1,\ldots,t_n)/\mathbb{Q}(t_1,\ldots,t_n)^{S_n}$. Since $\mathbb{Q}$ is Hilbertian, you can always specialize to a Galois extension $L/\mathbb{Q}$ of group $S_n$. Now set $K=L^{S_{n-1}}$. Since there is no subgroups of $S_n$ containing $S_n$, we are done ($K$ cannot be Galois because $S_{n-1}$ is not normal in $S_n$.)

It proves that any degree $n\geq 3$ is possible.

Notice that you can replace $\mathbb{Q}$ by any number field $F$ to get an extension of degree $n$ without intermediate subfields.

Using this, I think you can construct new families of examples like this:

Take $F/\mathbb{Q}$ and $K/F$ two extensions without intermediate subfields, and which are not cyclic of prime order. Then I suspect that $K/\mathbb{Q}$ has trivial automorphism group, but I have not checked it works.

  • $\begingroup$ Yes, i forgot the cas of cyclic extensiosn of prime order. I modify my answer right away ! $\endgroup$ – GreginGre Jul 6 at 7:04

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.