3 added 203 characters in body; deleted 1 characters in body

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? Do algebraically closed fields have the property? (I think so, because 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.

2 added 410 characters in body; added 105 characters in body; deleted 1 characters in body

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? Do algebraically closed fields have the property? (I think so, because 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)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.
1

# 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.
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?
2. For what fields is it true that the field cannot be embedded inside any field with trivial automorphism group? Do algebraically closed fields have the property? (I think so, because 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). 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. 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.