# $\Sigma_i a_ib_i = 0$ implies $\Sigma_i \sigma_1(a_i)\sigma_2(b_i) = 0$ for $\sigma_i\in\mathrm{Aut}(F_i),i=1,2$ that agree in $F_1\cap F_2$?

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.)

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

No, try with $$F_1=\Bbb{R},F_2=\Bbb{Q}((-2)^{1/4}), F_1F_2=\Bbb{C},F_1\cap F_2=\Bbb{Q}$$ $$\sigma_1=Id,\sigma_2((-2)^{1/4})=-(-2)^{1/4}$$
• Genius! Write $u = 2^{1/4}, v = (-2)^{1/4}$, then $2u - u^2v + v^3 = 0$, but $\sigma_1(2u) - \sigma_1(u^2)\sigma_2(v) + \sigma_2(v^3) = 2u + u^2v - v^3\neq 0$. Commented Mar 1, 2023 at 11:13