# Number fields with only trivial field automorphism

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?

• 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. – runway44 Jul 1 at 4:40
• 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 $H$. – runway44 Jul 1 at 4:40

## 1 Answer

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.

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