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Let $F_1$, $F_2$ be subfields of $K$ and $\sigma\colon F_1\to F_2$ be a field isomorphism. Then I would intuitively guess that there should be a nice one-to-one correspondence between the following two sets:

Set 1. The set of all the automorphisms $\phi$ of $K$ such that the following diagram commutes:

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Set 2. The set of all automorphisms $\psi$ of $K$ keeping $F_1$ fixed.

However, I am unsuccessful in my attempts to find any such correspondence (and getting frustrated now!). Is there any?

Atom
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2 Answers2

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If you can find a single such $\phi$, you have your correspondence: Add another column $F_1\hookrightarrow K$ to the left of your diagram, and attach it to your square via the identity map $F_1\to F_1$ and the map $\psi:K\to K$ being an element from your set 2. Observe that the composition of the entire bottom row $\phi\circ\psi:K\to K\to K$ is a corresponding element of set 1. It is not difficult to show that this correspondence is bijective.

As for the existence of $\phi$, well, there lies the problem. Consider, for instance, the case of rational functions over the reals: $$ K=F_1=\Bbb R(x)\\ F_2=\Bbb R(x^2) $$ Here $\sigma$ takes any rational function and simply doubles all the exponents of every $x$ that appears in it. For example, $$f=\frac{x^3-2}{x^4+x^2-x}\implies \sigma(f)=\frac{x^6-2}{x^8+x^4-x^2}$$ There can be no $\phi$ in this scenario, because by commutativity of the diagram and $K=F_1$ we would have $\phi=\sigma$, but this is not surjective. Your set 1 is therefore empty, while your set 2 contains the identity.

Arthur
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you cannot prove it, because it is not true. Take for example the nontrivial automorphism of $\mathbb{Q}(\sqrt{2})$ induced from $\sqrt{2} \mapsto -\sqrt{2}$. This cannot be extended to an automorphism of $\mathbb R$, so your first set is empty. However, the identity is an automorphism of $\mathbb R$ fixing $\mathbb{Q}(\sqrt{2})$, so your second set is nonempty.

Jonas Linssen
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  • Why can't it be extended to an automorphism of $\Bbb R$? I haven't looked too hard at it, but it smells to me like the axiom of choice (or more specifically Zorn's lemma) would let you. – Arthur Apr 25 '23 at 19:04
  • In my interpretation we fixed the embeddings, in both cases the standard embedding. As far as I know $\mathbb R$ has no nontrivial field automorphisms, so we can only choose the identity. But then commutativity forces $\sqrt{2}=-\sqrt{2}$, which is absurd. – Jonas Linssen Apr 25 '23 at 19:17
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    Turns out, this fact I recalled is not as trivial as I thought. See https://math.stackexchange.com/questions/449404/is-an-automorphism-of-the-field-of-real-numbers-the-identity-map – Jonas Linssen Apr 25 '23 at 19:19
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    Yes, of course. I feel like I should've caught that myself. $\sqrt 2$ is a square in $\Bbb R$ but $-\sqrt 2$ is not. So no automorphism can send one to the other. That does make me wonder whether you can make an automorphism of $\Bbb C$ with $\sqrt2\mapsto -\sqrt2$, but that's irrelevant to this post. – Arthur Apr 25 '23 at 19:21