I am interested in the following question:

Given a geometrically connected smooth rigid analytic space $X$ over a non-archimedean field $k$, is it always possible to find an affinoid open covering, where each element of the covering is geometrically connected?

Or perhaps more generally: is it true that any connected affinoid open subset of $X$ is geometrically connected? I don't imagine that this is true, but I can't think of a counterexample.

For example, if $X$ were a geometrically connected smooth scheme over a field, then because $X$ is normal, $X$ is actually geometrically irreducible. Therefore any connected open subset is automatically geometrically irreducible, so geometrically connected. In particular, the question would be true in this case.

Any ideas, counterexamples, or insight would be very much appreciated.


1 Answer 1


About your second question (connected affinoid subsets being geometrically connected), you will run into trouble already in the case of the affine line, as soon as $k$ is not separably closed. Indeed, pick a point $x$ whose residue field is a finite extension of $k$, not purely inseparable. Then there exists a finite extension $k'$ of $k$ such that $x$ has several preimages over $k'$. If you take a small enough neighborhood of $x$, it will have several connected components, at least one around each preimage.

Your first question is not very clear to me. Do you want to require some conditions on your affinoids? Do you want to refine a given covering or may the affinoid domains all be very big?

  • $\begingroup$ Thanks for your answer. I'm just trying to understand your example: if X is the affine line over k, and x and k' are as you describe, why must it be true that any small neighbourhood of x has multiple connected components after it is base changed to k'? By `small neighbourhood', I presume you mean a rigid analytic disk D(x,r) with centre x of radius r > 0. But such a disk is geometrically connected (the base change of a tate algebra in one variable over k, from k to k', is the tate algebra over k')? $\endgroup$
    – James
    Nov 19, 2023 at 19:40
  • $\begingroup$ And for the first question: I would just like to find <some> admissible open covering of X which satisfies (1) all elements of the covering are affinoid and (2) all elements of the covering are geometrically connected. $\endgroup$
    – James
    Nov 19, 2023 at 19:43
  • $\begingroup$ A point $x$ of the sort I describe above is not a $k$-rational point, so the neighborhood is not really a disk. For instance, if $x$ is associated to a maximal ideal generated by a polynomial $P$, a typical neighborhood will be given by $|P|\le r$. $\endgroup$ Nov 20, 2023 at 10:18
  • $\begingroup$ OK great, I understand, thanks. I can see why for small r this won't be geometrically connected - is it clear that such a neighbourhood is always connected? $\endgroup$
    – James
    Nov 20, 2023 at 20:54
  • $\begingroup$ If you base change to an algebraically closed field, the preimage of $\{|P|\le r\}$ will be a union of disks centered at the roots of $P$ and exchanged by Galois action. Since disks are connected, you are done. $\endgroup$ Nov 22, 2023 at 14:12

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