2
$\begingroup$

Convention: By APF extension, I mean APF extension of $\mathbb{Q_p}$.
For $\mathbb{Q_p} \subseteq L_1 \subseteq L_2$ where $L_2/L_1$ is finite, we know that $L_1/\mathbb{Q_p}$ is APF iff $L_2/\mathbb{Q_p}$ is APF. With this in mind, I have two questions:
1-Is the composition of a finitely many infinite APF extensions again APF?
2-What about infinite compositions of infinite APF extensions?

$\endgroup$
1
  • 2
    $\begingroup$ APF = arithmetically profinite $\endgroup$
    – Niels
    Commented Dec 21, 2021 at 7:58

0

You must log in to answer this question.

Browse other questions tagged .