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Let $L$ be a field, and $K$ and $M$ subfields of $L$ which are isomorphic as abstract fields.

I am looking for interesting examples for $L, K$ and $M$ such that $\mathrm{Gal}(L/K)$ and $\mathrm{Gal}(L/M)$ are not isomorphic.

In particular (this might be a stupid question): how about the case $L = \mathbb{C}$, and $K$ and $M$ isomorphic copies of $\mathbb{R}$ ?

My last question makes no sense if all isomorphic copies of the reals are contained in one and the same $\mathrm{Aut}(\mathbb{C})$-orbit, but I am not aware of this possible fact.

And even if the answer on the latter question is that all such Galois groups are isomorphic, is this still true if we do not accept the Axiom of Choice ?

Boccherini
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  • If $K\subset \Bbb{C}$ is isomorphic to $\Bbb{R}$ and $\Bbb{C}/K$ is algebraic then $K(i)$ is algebraically closed, so $K(i)=\Bbb{C}$ and $Gal(\Bbb{C/R})\cong Gal(\Bbb{C}/K)$ and $K$ is the image of $\Bbb{R}$ under some $\sigma\in Aut(\Bbb{C})$. – reuns Jan 23 '23 at 19:45

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The notation $\text{Gal}(L/K)$ should be avoided unless $L/K$ is a Galois extension; if you just want to talk about the automorphism group then you can use $\text{Aut}(L/K)$.

As a simple example you can take $L = K = \mathbb{Q}(x), M = \mathbb{Q}(x^2)$. Here we really do have Galois extensions, and $\text{Gal}(L/K)$ is trivial but $\text{Gal}(L/M) \cong C_2$ generated by $x \mapsto -x$.

$\mathbb{C}$ does in fact admit more than one copy of $\mathbb{R}$ even up to automorphisms of $\mathbb{C}$, because (assuming choice) there are copies of $\mathbb{R}$ of infinite index. For example (again, assuming choice) $\mathbb{C}$ is abstractly isomorphic to the algebraic closure of $\mathbb{R}(x)$, which contains such a copy of $\mathbb{R}$.

In the absence of choice it's consistent with ZF that

  • $\text{Aut}(\mathbb{C})$ consists only of the identity and complex conjugation, and
  • there is exactly one copy of $\mathbb{R}$ in $\mathbb{C}$, namely the usual one.

This is because there are models of ZF in which every homomorphism between Polish groups is automatically continuous (see, for example, Theorem 2.2 here). The infinite index embeddings $\mathbb{R} \to \mathbb{C}$ constructed using choice are not even measurable, and neither are the "wild" automorphisms of $\mathbb{C}$ constructed using choice.

Qiaochu Yuan
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  • If $\widetilde{\mathbb{R}}$ is such a copy of infinite index, do you have any idea what $\mathrm{Aut}(\mathbb{C}/\widetilde{\mathbb{R}})$ is, or $\vert \mathrm{Aut}(\mathbb{C}/\widetilde{\mathbb{R}}) \vert$? – Boccherini Jan 24 '23 at 14:50
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    @Boccherini: with the above example it's the absolute Galois group of $\mathbb{R}(x)$. This is apparently close to a free profinite group on uncountably many generators: https://math.stackexchange.com/questions/318690/absolute-galois-group-of-mathbbrt – Qiaochu Yuan Jan 24 '23 at 21:30