If k is a nondiscrete topological field, we can define an analytic space over k just like complex analytic spaces over $\mathbb{C}$. If you replace "complex analytic space" and "complex algebraic variety" with "analytic space over $k$" and "algebraic variety over $k$", respectively, under what conditions on $k$ does GAGA, or weaker similar results, hold? Presumably $k$ must be algebraically closed, but I'm wondering whether this is enough, or whether more conditions must be added, or whether this really only works for $k=\mathbb{C}$.

2$\begingroup$ See mathoverflow.net/questions/121187/…. $\endgroup$– Jérôme PoineauOct 12, 2016 at 14:08
1 Answer
If $k$ is a field that is complete with respect to some ultrametric valuation, then there is the "GAGR" (i.e. géométrie algébrique et géométrie rigide) theorem. A succinct explanation (in French, without proof) is:
Jarraud, Pierre. À propos de G.A.G.R.. Groupe de travail d'analyse ultramétrique 11 (19831984): 14.
The original source for the result, listed as $[4]$ in Jarraud's note, seems to be:
Köpf, Ursula. Über eigentliche Familien algebraischer Varietäten über affinoïden Räumen, Schriftenreihe Univ. Münster, 2. Serie, Heft 7 (1974).

1$\begingroup$ All this time I thought it was "est" instead of "et". $\endgroup$– arsmathOct 12, 2016 at 10:25