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I'm not requiring this extension to be Galois, that's why I wrote $\text{Aut}$ instead of $\text{Gal}$. I'm not very familiar with infinite extensions nor with profinite groups. I don't know if my question is part of of the inverse Galois problem, since this problem is generally related to finite groups.

To start with the "basic" infinite extension $\Bbb Q(X)$, I know that $\text{Aut}_{\Bbb Q}(\Bbb Q(X)) \cong \text{PGL}_2(\Bbb Q)$, which is far from being $\Bbb Z$. Notice that this question is not really related to mine.

Thanks for your help!

Community
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Watson
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    what about $\Bbb Q(X,\sqrt[3] X, \sqrt[3] {X+1}, \sqrt[3] {X-1}, \sqrt[3] {X+2},\ldots)$ ? – mercio Jul 26 '16 at 10:51
  • @mercio: Might $X\mapsto -X$ extend to a $\Bbb{Q}$-automorphism of that field as well? – Jyrki Lahtonen Jul 26 '16 at 12:48
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    True... :/ :/ .... $\Bbb Q(X,{\sqrt[3]{P(X+k)}, k\in \Bbb Z})$ where $P(X)$ is ... uh... something complicated ? like $P(x) = x^3+2$ ? – mercio Jul 26 '16 at 13:17
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    This is nice and tricksy question. The automorphism $X\mapsto X+1$ extends to no finite extension of $\Bbb Q(X)$, so you’d have to find some big extension that it did extend to, and hope that no other automorphisms got created. You’re certainly right to recognize that your $K$ can’t be a number field. – Lubin Jul 26 '16 at 19:56
  • @Lubin: is it right to say that such an extension $K / \Bbb Q$ can't be Galois, because $\Bbb Z$ is not a profinite group (I'm not sure about this fact)? – Watson Aug 20 '16 at 18:28
  • Maybe we should try an algebraic extension of $\Bbb Q$… – Watson Aug 22 '16 at 12:56
  • @JyrkiLahtonen: do we have at least an example of extension $K/\Bbb Q$ such that $\text{Aut}(K)$ is abelian? Apparently, abelian profinite groups are in duality with locally finite discrete abelian groups, using Pontryagin duality. – Watson Aug 25 '16 at 08:45
  • Answering my question for rings instead of fields of characteristic $0$ is much easier, for instance the ring $\Bbb F_2[X_1, ..., X_n]$ has automorphism group $\Bbb Z^n$ (using the fact that $\Bbb F_2^{\times} = {1}$). – Watson May 12 '18 at 17:59
  • This paper cites the following result : if $M$ is any monoid, then there is a field $F \supset \Bbb Q$ such that $End(F) = M$. – Watson May 13 '18 at 14:53
  • @Watson: It is not true that $\forall M\exists F(\textrm{End}(F)=M)$. The endomorphism monoid of a field consists of injective functions, hence (if endomorphisms act on the left) the endomorphism monoid of any field must satisfy the left cancellation law. In particular, the monoid $\langle {0,1}; \ast\rangle$ satisfying $1\ast 1 = 1$, $0\ast 1 = 1\ast 0 = 0\ast 0=0$ is not isomorphic to the endomorphism monoid of any field. – Keith Kearnes Oct 06 '22 at 07:10
  • @KeithKearnes, yes sorry, the monoid needs to have a "1" and right cancellation. The original paper is https://www.ams.org/journals/tran/1987-304-01/S0002-9947-1987-0906820-7/ – Watson Oct 06 '22 at 11:10

1 Answers1

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This question is answered affirmatively in the paper

Kuyk, Willem The construction of fields with infinite cyclic automorphism group. Canad. J. Math. 17 1965 665-668.

Armand Brumer wrote this in his MathSciNet review of the paper:

``J. de Groot [Math. Ann. 138 (1959), 80-102, 101; MR0119193] has asked whether every group $G$ is the full automorphism group of some field. The author answers this question affirmatively for the infinite cyclic group. In fact, let $\mathbb K$ be an algebraic closure of a purely transcendental extension $\mathbb Q(t_0)$ of the rationals. Choose inductively elements $t_i$ in $\mathbb K$ satisfying $t_i^2=t_{i−1}+1$ for all integers $i$, and set $\Omega = \bigcup_{i\in \mathbb Z} \mathbb Q(t_i)$. Then the automorphism group of $\Omega$ is the infinite cyclic group generated by $t_i\mapsto t_{i+1} (i\in\mathbb Z)$.''

Later Kuyk's result was extended in:

Fried, E.; Kollár, J. Automorphism groups of fields. Universal algebra (Esztergom, 1977), pp. 293-303, Colloq. Math. Soc. János Bolyai, 29, North-Holland, Amsterdam-New York, 1982.

These authors show that for any group $G$ there exists a field $\mathbb F$ such that $\textrm{Aut}(\mathbb F)\cong G$. Moreover, $\mathbb F$ may be required to have any given characteristic different from $2$.

Keith Kearnes
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