# Fields with trivial automorphism group

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.

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Extension by any single root of a general polynomial has trivial automorphism group. –  Ryan Thorngren Apr 28 '10 at 20:48
@Georges: by "general" I think he just means something like "irreducible and separable, of degree at least 3, and the Galois group of the splitting field is the full symmetric group". So for example over Q this produces a lot of examples. –  Kevin Buzzard Apr 28 '10 at 22:05
I just want to point out that one of the OP's questions remains wide open: can every field be embedded in a field with trivial automorphism group? I don't even myself know the answer for $\mathbb{C}$. –  Pete L. Clark Apr 29 '10 at 6:09
One conceptual difference between $\mathbb{Q}(i)$ and $'\mathbb{Q}( \sqrt{2})$ is that a field in which $-1$ is a square cannot be ordered. Since $'\mathbb{Q}( \sqrt{2})$ has real embeddings it can obviously be ordered. There is a somewhat related invariant called the level of of the field, which if I remember correctly is the least number of summands needed to express -1 as a sum of squares. In the cases above the levels are 1 and $\infty$ respectively. If we had chosen $\sqrt{-2}$ instead the level would have been 2. –  K.J. Moi May 3 '10 at 8:11
For the question Pete points out, see the references in mathoverflow.net/questions/61058 –  dke Apr 8 '11 at 17:24

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

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).

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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.

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Aren't all but two automorphisms of $\mathbb{C}$ hard to see as well by the naked eye? –  Zsbán Ambrus Apr 29 '10 at 8:57
Very much so! And in a Solovay model $L(\mathbb{R})$, there are only two automorphisms of $\mathbb{C}$. But this is a model in which the Axiom of Choice fails. I find it interesting that it is consistent that $ZFC$ that the nontrivial automorphisms of $\prod_{\mathcal{U}} \mathbb{F}_{p}$ are not just hard to see ... they aren't even there! –  Simon Thomas Apr 29 '10 at 15:22

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

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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 $K$.

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Could you sketch a proof that the p-adics have trivial automorphism group? I'm guessing it's the same reason as the reals, but how do you show that any automorphism must "respect" the valuation? –  Vipul Naik Apr 28 '10 at 21:27
One shows, by fair means or foul, that an automorphism of $\mathbb{Q}_p$ must be continuous. For instance it suffices to prove that an automorphism preserves $\mathbb{Z}_p$. –  Robin Chapman Apr 28 '10 at 21:32
The point is that various subsets of Q_p are algebraically defined. For example 1+pZ_p is precisely the elements of Q_p which have n'th roots for all n prime to p (or 1+4Z_p if p=2). Hence 1+p^nZ_p is algebraically-defined, so p^nZ_p is algebraically-defined, so "small" is algebraically-defined, and now you're home. –  Kevin Buzzard Apr 28 '10 at 22:03