I am stuck trying to understand a proof in Asymptotic Differential Algebra and Model Theory of Transseries by L. van den Dries, J. van der Hoeven and M. Aschenbrenner.
The result is the following:
Let $K$ be a field, let $A$ be a local ring of $K$ integrally closed in $K$ with maximal ideal $m_A$. Let $L / K$ be a normal extension, and let $B$ denote the integral closure of $A$ in $L$. If $q,q'$ are maximal ideals of $B$ such that $q \cap A = q' \cap A = m_A$, then there is $\sigma \in Aut(L |K)$ such that $\sigma(q) = q'$.
I understand the proof, except the start, when they write that one can assume $[L : K]$ is finite. Does anyone see how this case yields the general one?
I think we could argue as follows: suppose you prove the claim for some finite normal subextension $\;L'/K\;$ of $\;L/K\;$ . Now, take an element $\;\sigma\in\text{Aut}\,\left(L'/K\right)\;$ fulfilling $\;\sigma q= q'\;$ .
Since $\;L/K\;$ is algebraic (this is a "hidden" assumption, I think, since "normal extension", as far as I know, is only applied to algebraic extensions), we can extend $\;\sigma\;$ to a $\;K\,-$ embedding $\;\Sigma:L\to \overline K\;$ , where the last field is some (the) algebraic closure of $\;K\;$ which of course contains $\;L\;$, too, and since $\;L/K\;$ is normal then in fact this embedding is an automorphism of $\;L/K\;$ , so in fact $\;\Sigma\in\text{Aut}/\left(L/K\right)\;$ , and of course $\;\Sigma\uparrow_{L'}=\sigma\;$, so still $\;\Sigma q=q'\;$.