Fields with trivial automorphism group

Question

Is there a nice characterization of fields whose automorphism group is trivial? Here are the facts I know.

1. Every prime field has trivial automorphism group.
2. Suppose L is a separable finite extension of a field K such that K has trivial automorphism group. Then, if E is a finite Galois extension of K containing L, the subgroup $Gal(E/L)$ in $Gal(E/K)$ is self-normalizing if and only if L has trivial automorphism group. (As pointed out in the comments, a field extension obtained by adjoining one root of a generic polynomial whose Galois group is the full symmetric group satisfies this property).
3. The field of real numbers has trivial automorphism group, because squares go to squares and hence positivity is preserved, and we can then use the fact that rationals are fixed. Similarly, the field of algebraic real numbers has trivial automorphism group, and any subfield of the reals that is closed under taking squareroots of positive numbers has trivial automorphism group.

My questions:

1. Are there other families of examples of fields that have trivial automorphism group? For instance, are there families involving the p-adics? [EDIT: One of the answers below indicates that the p-adics also have trivial automorphism group.]
2. For what fields is it true that the field cannot be embedded inside any field with trivial automorphism group? I think that any automorphism of an algebraically closed field can be extended to any field containing it, though I don't have a proof) [EDIT: One of the answers below disproves the parenthetical claim, though it doesn't construct a field containing an algebraically closed field with trivial automorphism group]. I suspect that $\mathbb{Q}(i)$ cannot be embedded inside any field with trivial automorphism group, but I am not able to come up with a proof for this either. [EDIT: Again, I'm disproved in one of the answers below]. I'm not even able to come up with a conceptual reason why $\mathbb{Q}(i)$ differs from $\mathbb{Q}(\sqrt{2})$, which can be embedded in the real numbers.

ADDED SEP 26: All the questions above have been answered, but the one question that remains is: can every field be embedded in a field with trivial automorphism group? Answering the question in general is equivalent to answering it for algebraically closed fields.

Answer 1

As Robin as pointed out, for all primes $p$, $\mathbb{Q}_p$ is rigid, i.e., has no nontrivial automorphisms. It is sort of a coincidence that you ask, since I spent much of the last $12$ hours writing up some material on multiply complete fields which has applications here:

Theorem (Schmidt): Let $K$ be a field which is complete with respect to two inequivalent nontrivial norms (i.e., the two norms induce distinct nondiscrete topologies). Then $K$ is algebraically closed.

Corollary: Let $K$ be a field which is complete with respect to a nontrivial norm and not algebraically closed. Then every automorphism of $K$ is continuous with respect to the norm topology. (Proof: To say that $\sigma$ is a discontinuous automorphism is to say that the pulled back norm $\sigma^*|| \ ||: x \mapsto ||\sigma(x)||$ is inequivalent to $|| \ ||$. Thus Schmidt's theorem applies.

In particular this applies to show that $\mathbb{Q}_p$ and $\mathbb{R}$ are rigid, since every continuous automorphism is determined by its values on the dense subspace $\mathbb{Q}$, hence the identity is the only possibility. (It is possible to give a much more elementary proof of these facts, e.g. using the Ostrowski classification of absolute values on $\mathbb{Q}$.)

At the other extreme, each algebraically closed field $K$ has the largest conceivable automorphism group: $\# \operatorname{Aut}(K) = 2^{\# K}$: e.g. Theorem 80 of

http://alpha.math.uga.edu/~pete/FieldTheory.pdf.

There is a very nice theorem of Bjorn Poonen which is reminiscent, though does not directly answer, your other question. For any field $K$ whatsoever, and any $g \geq 3$, there exists a genus $g$ function field $K(C)$ over $K$ such that $\operatorname{Aut}(K(C)/K)$ is trivial. However there may be other automorphisms which do not fix $K$ pointwise.

There is also a sense in which for each $d \geq 3$, if you pick a degree $d$ polynomial $P$ with $\mathbb{Q}$-coefficients at random, then with probability $1$ it is irreducible and $\mathbb{Q}[t]/(P)$ is rigid. By Galois theory this happens whenever $P$ is irreducible with Galois group $S_d$, and by Hilbert Irreducibility the complement of this set is small: e.g. it is "thin" in the sense of Serre.

Addendum: Recall also Cassels' embedding theorem (J.W.S. Cassels, An embedding theorem for fields, Bull. Austral. Math. Soc. 14 (1976), 193-198): every finitely generated field of characteristic $0$ can be embedded in $\mathbb{Q}_p$ for infinitely many primes $p$. It would be nice to know some positive characteristic analogue that would allow us to deduce that a finitely generated field of positive characteristic can be embedded in a rigid field (so far as I know it is conceivable that every finitely generated field of positive characteristic can be embedded in some Laurent series field $\mathbb{F}_q((t))$, but even if this is true it does not have the same consequence, since Laurent series fields certainly have nontrivial automorphisms).

Answer 2

This is not a family but instead an interesting example ... Let $\mathcal{U}$ be a nonprincipal ultrafilter over the set $\mathbb{P}$ of primes and consider the corresponding ultraproduct $K = \prod_{\mathcal{U}}\mathbb{F}_{p}$

of the fields of prime order $p$. If $CH$ holds, then $K$ always has $2^{2^{\aleph_{0}}}$ automorphisms ... but none of the nontrivial ones is easily seen by the naked eye. There is a good reason for this. Shelah has recently shown that it is consistent that there exists a nonprincipal ultrafilter $\mathcal{U}$ such that $K = \prod_{\mathcal{U}}\mathbb{F}_{p}$ has no nontrivial automorphisms.

Answer 3

There are examples involving the $p$-adics: for instance $\mathbb{Q}_p$ itself has trivial automorphism group. Indeed as $\mathbb{Q}(i)$ embeds in $\mathbb{Q}_p$ when $p\equiv1$ (mod 4) then $\mathbb{Q}(i)$ does embed in a field with trivial automorphism group. Indeed this is the case for all number fields (finite extensions of $\mathbb{Q}$).

Now for an example of an algebraically closed field, $K$, an extension $L$ of $K$ and an automorphism $\tau$ of $K$ not extending to one of $L$. Let $K$ be the algebraic closure of $\mathbb{Q}$, considered as a subfield of $\mathbb{C}$ and let $\tau$ be complex conjugation. Let $L=K(x,\sqrt{x^3+ax+b})$ be the function field of an elliptic curve $E$ over $K$. Each automorphism of $L$ takes $K$ to itself. Suppose the $j$-invariant of $E$ is $i$ (considered as an element of $K$). Then any automorphism of $L$ taking $K$ to itself must fix $i$, and so cannot restrict to $\tau$ on $K$.

Answer 4

In the paper below Shelah, among many other things, gives constructions of real closed fields with no nontrivial automorphisms that are not subfields of the reals.

S. Shelah, Models with second order properties. IV. A general method and eliminating diamonds -- Annals Pure and Applied Logic 25 (1983) 183-212