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I'm interested in the following question: suppose that $F_1,F_2$ are two fields lying in some common field, write $F_1F_2$ as their compositum, that is, field of all elements of the form $$\left\{\dfrac{\Sigma_i a_ib_i}{\Sigma_j a'_jb'_j}: a_i,a'_j\in F_1,b_i,b'_j\in F_2,\Sigma_j a'_jb'_j\neq 0\right\}.$$ Suppose that $\sigma_1\in\mathrm{Aut}(F_1)$, $\sigma_2\in\mathrm{Aut}(F_2)$ have $\sigma_1|_{F_1\cap F_2} = \sigma_2|_{F_1\cap F_2}$, must there exists $\sigma\in \mathrm{Aut}(F_1F_2)$ such that $\sigma|_{F_1} = \sigma_1$, $\sigma|_{F_2} = \sigma_2$? Note that in our assumption $F_1$ and $F_2$ can be rather arbitrary, so I don't think Galois theory has a role here; but I'm not asking to find a relation between $\mathrm{Aut}(F_1F_2)$ and $\mathrm{Aut}(F_1)$ and $\mathrm{Aut}(F_2)$, I'm just interested in the possibility of extension.

Of course, if such $\sigma$ exists, it has a unique way to be defined. The problem only comes when one wants to show the well-definedness and injectivity, and the problem becomes: suppose that $F_1,F_2$ are two fields lying in some common field, $F_1F_2$ their compositum. Suppose that $\sigma_1\in\mathrm{Aut}(F_1)$, $\sigma_2\in\mathrm{Aut}(F_2)$ have $\sigma_1|_{F_1\cap F_2} = \sigma_2|_{F_1\cap F_2}$. If $\Sigma_i a_ib_i = 0$ for $a_i\in F_1, b_i\in F_2$, must we have $\Sigma_i \sigma_1(a_i)\sigma_2(b_i) = 0$?

Edit: well I have to admit that Galois theory has a role here. Specifically, let $F\subset F_1\cap F_2$ be a field fixed by $\sigma_1$ and $\sigma_2$. If one of $F_1$ or $F_2$ is Galois over $F$, then such $\sigma$ exists. Actually, we can always define $F\subset F_1\cap F_2$ be the field fixed by $\sigma_1$ and $\sigma_2$: any field $F_0\subset F_1\cap F_2$ fixed by $\sigma_1$ and $\sigma_2$ will therefore be contained in this $F$, and $F_i/F_0$ being Galois implies $F_i/F$ being Galois because $F$ is an intermediate field.

We need the following result (FT, Proposition 7.15):

Let $E$ and $L$ be field extensions of $F$ contained in some common field. If $E/F$ is Galois, then $EL/L$ and $E/(E\cap L)$ are Galois, and the map $$\sigma\mapsto \sigma|_E: \mathrm{Gal}(EL/L)\to \mathrm{Gal}(E/(E\cap L))$$ is an isomorphism of topological groups.

Suppose that in our setting $F_1$ is Galois over $F$. Using Zorn's lemma we can extend $\sigma_2\in \mathrm{Aut}(F_2/F)$ to $s\in \mathrm{Aut}(\overline{F_1F_2}/F)$, where $\overline{F_1F_2}$ is the algebraic closure of $F_1F_2$ (or any algebraically closed field containing $F_1F_2$). Since $F_1$ is Galois over $F$, it is standard that $s|_{F_1}\in \mathrm{Gal}(F_1/F)$. By $s|_{F_1\cap F_2} = \sigma_2|_{F_1\cap F_2} = \sigma_1|_{F_1\cap F_2}$ we have $\sigma^{-1}_1\circ s|_{F_1}\in \mathrm{Gal}(F_1/(F_1\cap F_2))$. By the lemma above, there exists $e\in \mathrm{Gal}(F_1F_2/F_2)$ such that $\sigma^{-1}_1\circ s|_{F_1} = e|_{F_1}$. Define $\sigma = s|_{F_1F_2}\circ e^{-1}:F_1F_2\to \overline{F_1F_2}$ being an injective homomorphism, it's easy to see $\sigma|_{F_1} = \sigma_1$, $\sigma|_{F_2} = \sigma_2$.

It remains to show that $\sigma\in\mathrm{Aut}(F_1F_2)$. Since $\sigma(F_1) = \sigma_1(F_1) = F_1$, $\sigma(F_2) = \sigma_2(F_2) = F_2$ we have $\sigma(F_1F_2)\subset F_1F_2$. Choose $\sigma':F_1F_2\to \overline{F_1F_2}$ such that $\sigma|_{F_1} = \sigma^{-1}_1$, $\sigma|_{F_2} = \sigma^{-1}_2$, then $\sigma\circ\sigma'$ and $\sigma'\circ\sigma$ fix $F_1$ and $F_2$, so they are identities over $F_1F_2$. This shows that $\sigma$ and $\sigma'$ are mutual inverses.

(This is essentially the proof of Corollary 7.16 below Proposition 7.15 in the link.)

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  • $\begingroup$ Did you play around with any examples? $\endgroup$ Commented Mar 1, 2023 at 1:07
  • $\begingroup$ @CharlesHudgins Sorry I currently cannot think of any examples. Are there some fancy examples for field automorphisms? $\endgroup$ Commented Mar 1, 2023 at 1:17
  • $\begingroup$ I can't either. You'll want a field extension which contains (or is) a compositum of two subfields which have nontrivial intersection. $\endgroup$ Commented Mar 1, 2023 at 9:20

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No, try with $$F_1=\Bbb{R},F_2=\Bbb{Q}((-2)^{1/4}), F_1F_2=\Bbb{C},F_1\cap F_2=\Bbb{Q}$$ $$\sigma_1=Id,\sigma_2((-2)^{1/4})=-(-2)^{1/4}$$

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  • $\begingroup$ Genius! Write $u = 2^{1/4}, v = (-2)^{1/4}$, then $2u - u^2v + v^3 = 0$, but $\sigma_1(2u) - \sigma_1(u^2)\sigma_2(v) + \sigma_2(v^3) = 2u + u^2v - v^3\neq 0$. $\endgroup$ Commented Mar 1, 2023 at 11:13

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