There are examples of non-isomorphic algebraic varieties defined over complex numbers whose associated complex-analytic spaces are isomorphic. Are there such examples for varieties defined over $p$-adic fields, if we consider rigid analytification? In the linked answer, the word "Stein" is used so I am not completely sure if those examples work verbatim.
1 Answer
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It seems possible to adapt Kim's example to the p-adic setting, although there is a significant discrepancy in how Picard groups of affine curves behave. We want to construct a non-trivial line bundle $L$ on an affine algebraic curve $X$ such that $L^{an}$ is trivial on $X^{an}$ so that the total space $Tot(L)$ and $X\times\mathbb{A}^1$ is a pair of varieties with the desired property.
Let $K$ be the completion of an algebraic closure of $\mathbb{Q}_p$ and, as usual, denote by $\mathcal{O}_K$ and $k$ the ring of integers and the residue field.
Proposition. If $x,y\in C(\overline{\mathbb{Q}}_p)$ are points on a smooth projective curve $C$ defined over an algebraic extension of $\mathbb{Q}_p$ such that the class $[x]-[y]$ in $Pic^0(C)$ is not torsion, then for $X=C\setminus \{x,y\}$ the canonical map of Picard groups $Pic(X)\to Pic(X^{an})$ is not injective.
Choose local coordinates $t_x,t_y$ on the curve $C$ at the points $x, y$ so that there are affinoid neighborhoods $x\in U_x, y\in U_y$ on which $f_x$ and $f_y$ define isomorphisms with closed disks. For $r\in p^{\mathbb{Q}}$ let $X_r$ be the affinoid subspace of $C^{an}$ defined by $|t_x|\geq r, |t_y|\geq r$. The analytification $X^{an}$ is the direct limit of $X_r$ over $r$ tending to zero.
Van der Put has shown(Prop 3.1 in http://www.numdam.org/item/AIF_1980__30_4_155_0/) that $Pic(X_r)$ is the quotient of $Pic^0(C)$ by the subgroup generated by the classes of degree $0$ divisors supported on $C(K)\setminus X_r(K)$. The same is true in the complex analytic geometry and implies there that $Pic(X_r)$ is trivial because the image of any analytic ball in $C$ generates the group of points of the Jacobian of $C$. But this is not the case in the $p$-adic setting as there are many open subgroups of the rigid-analytic group $Pic^0(C)$! For instance, if $C$ has good reduction over $\mathcal{O}_K\subset K$ then the preimage of the identity in $Pic^0(C_k)$ under the specialization map is such subgroup (it is isomorphic to an open polydisc as a variety).
Anyway, the analytic Picard group is different from the algebraic one. Indeed, the kernel of the map $Pic^0(C)\to Pic(X_r)$ contains the subgroup generated by $[q]-[p]$ hence it contains the closure of this subgroup(because $Pic(X_r)$ is the quotient by an open subgroup which is automatically closed). Using that $Pic(X)=\lim\limits_{\leftarrow}Pic(X_r)$, we get that the kernel of $Pic^0(C)\to Pic(X^{an})$ also contains this closure. If $[q]-[p]$ is not a torsion point, then the closure of the subgroup generated by it is strictly larger than the subgroup itself. Indeed, $[q]-[p]$ lies in the compact group $(Pic^0C)(L)$ for some finite extension $L$ of $\mathbb{Q}_p$ and $(Pic^0 C)(L)$ so if the subgroup $\langle [q]-[p]\rangle$ is closed, it is compact but a (separated) topological group isomorphic to $\mathbb{Z}$ cannot be compact. $\square$
Summarizing, the composition $Pic(C)\to Pic(X)\to Pic(X^{an})$ has larger kernel than the map $Pic(C)\to Pic(X^{an})$ so we can find a line bundle which will drive Kim's construction.
It is probably true that for any number of points $p_1,\dots, p_n$ the analytic Picard group of their complement is the quotient of $Pic(C)$ by the closure of the subgroup generated by $[p_1],\dots ,[p_n]$(the above argument only shows that it is the quotient by the subgroup which contains this closure). Note also that, as opposed to the complex analytic situation, the complement of one point has non-trivial Picard group.
