# Geometrically connected affinoid cover of geometrically connected smooth rigid space

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.

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.
• 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 $|P|\le r$. Nov 20, 2023 at 10:18
• 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. Nov 22, 2023 at 14:12