Smooth affinoid rigid-analytic spaces over $\mathbb C_{p}$

Question

Are there any examples of smooth affinoid rigid-analytic spaces $X$ over $\mathbb C_{p}$, which are not the base-change $X=Y_{\mathbb C_{p}}$ of a smooth rigid-analytic variety $Y$ over a finite extension of $\mathbb Q_{p}$?

Answer 1

Surprisingly the answer to this question is no. This is a folkloric result -- not because it's hard (given hard input), but because I don't know a precise place where this is written down. But it follows easily from Elkik's results (I could be overcomplicating things).

Theorem([Elk, Theorem 7 and Remark 2]): Let $\mathrm{Spf}(A)\to \mathrm{Spf}(\mathcal{O}_{\mathbb{C}_p})$ be topologically of finite type, flat, and assume that $\mathrm{Spa}(A[\tfrac{1}{p}])\to \mathrm{Spa}(\mathbb{C}_p)$ is smooth. Then, there exists a finite type $\mathcal{O}_{\mathbb{C}_p}$-algebra $B$ such that $\mathrm{Spec}(B[\tfrac{1}{p}])\to \mathrm{Spec}(\mathbb{C}_p)$ is smooth, and $A$ is isomorphic to $\widehat{B}$ as topological $\mathcal{O}_{\mathbb{C}_p}$-algebras.

This is often referred to as Elkik's algebraization theorem.

Theorem ([Elk, Lemme 6 and Remark 2]): Let $A$ be a topologically of finite type, flat, complete $\mathcal{O}_{\mathbb{C}_p}$-algebra such that $\mathrm{Spa}(A[\tfrac{1}{p}])\to\mathrm{Spa}(\mathbb{C}_p)$ is smooth. Write $A=\mathcal{O}_{\mathbb{C}_p}\langle x_1,\ldots,x_n\rangle/(f_1,\ldots,f_k)$. Then, there exists some $m\geqslant 0$ such that if $(f_1',\ldots,f_k')$ is in $\mathcal{O}_{\mathbb{C}_p}\langle x_1,\ldots,x_n\rangle$, and $$(f_1,\ldots,f_k)=(f_1',\ldots,f_k')\mod p^m,$$ then, $A\cong \mathcal{O}_{\mathbb{C}_p}\langle x_1,\ldots,x_n\rangle/(f_1',\ldots,f_k')$.

I think of this as a sort of multi-dimensional version of Krasner's lemma from classical algebraic number theory.

In any case, these results directly address your question.

Claim: Let $\mathrm{Spa}(R)\to \mathrm{Spa}(\mathbb{C}_p)$ be a smooth morphism. Then, there exists some finite extension $K/\mathbb{Q}_p$ and some smooth morphism $\mathrm{Spa}(R_0)\to\mathrm{Spa}(K)$ such that $\mathrm{Spa}(R_0)\times_{\mathrm{Spa}(K)}\mathrm{Spa}(\mathbb{C}_p)$ is isomorphic to $\mathrm{Spa}(R)$ as a $\mathbb{C}_p$-space.

Proof: Write $R=\mathbb{C}_p\langle x_1,\ldots,x_n\rangle/I$. Consider then $A:= \mathcal{O}_{\mathbb{C}_p}\langle x_1,\ldots,x_n\rangle/I'$ where $I'=I\cap \mathcal{O}_{\mathbb{C}_p}\langle x_1,\ldots,x_n\rangle$. Then, evidently $A$ is a topologically of finite type flat (as it embeds into $R$ and so is $\mathcal{O}_{\mathbb{C}_p}$-torsionfree). So, as $R=A[\tfrac{1}{p}]$, we see from the first theorem of Elkik that there exists a finite type $\mathcal{O}_{\mathbb{C}_p}$-algebra $B$ with smooth generic fiber such that $\widehat{B}=A$. In particular, this means that without loss of generality we may assume that $I'=(f_1,\ldots,f_k)$ where $f_i$ are acutally polynomials over $\mathcal{O}_{\mathbb{C}_p}$. But, let $m$ be as in the second theorem of Elkik. As $\mathcal{O}_{\overline{\mathbb{Q}}_p}$ is dense in $\mathcal{O}_{\mathbb{C}_p}$, we can choose polynomials $f_1',\ldots,f_k'$ with coefficients in $\mathcal{O}_{\overline{\mathbb{Q}}_p}$ such that $f_i'=f_i \mod p^m$ for $i=1,\ldots,k$. In particular, we see that $A\cong \mathcal{O}_{\mathbb{C}_p}\langle x_1,\ldots,x_n\rangle/(f_1',\ldots,f_k')$. But, if $K$ denotes the subfield of $\overline{\mathbb{Q}}_p$ generated by the coefficients of $f_1',\ldots,f_n'$, then it's evident that $A$ is defined over $\mathcal{O}_K$, and thus that $R$ is defined over $K$ as desired. $\blacksquare$

To see why this is somewhat surprising, one can consider the fact that the answer to your question is affirmative if you replace 'affinoid' by 'affine'.

Example: Let $E$ be an elliptic curve over $\mathbb{C}_p$, with $j(E)\notin\overline{\mathbb{Q}}_p$ (e.g., $$E:= V(y^2z-x^3-\frac{36}{j_0-1728}xz^2-\frac{1}{j_0-1728}z^3),$$ with $j_0\notin \overline{\mathbb{Q}}_p$). Let $x$ be any point of $E(\mathbb{C}_p)$. Then, $E-\{x\}$ is a smooth affine $\mathbb{C}_p$-curve which admits no model over a finite extension of $\mathbb{Q}_p$.

Proof: That $E-\{x\}$ is affine is classical (e.g. see this), and it is obviously smooth. Suppose that $X$ is a model of $E-\{x\}$ over a finite extension $K$ of $\mathbb{Q}_p$. Up to enlarging $K$ by a finite extension, we may assume that $X(K)$ is non-empty. Evidently $X$ is smooth, geometrically connected, affine, and its unique smooth compactification $\overline{X}$ is a form of $E$, and so genus $1$. As $\overline{X}(K)\ne \varnothing$ we have that $\overline{X}$ must be an elliptic curve. But, we then see that $j(E)=j(\overline{X})\in K\subseteq \overline{\mathbb{Q}}_p$, which is a contradiction. $\blacksquare$

This just highlights that affines are much more rigid than affinoids.

References:

[Elk] Elkik, R., 1973. Solutions d'équations à coefficients dans un anneau hensélien. In Annales scientifiques de l'École normale supérieure (Vol. 6, No. 4, pp. 553-603).