# Open immersion of affinoid adic spaces

If $$R$$ and $$S$$ are complete Huber rings with $$\varphi: R \to S$$ a continuous map, then is it true in general that if $$\mathrm{Spa}(S, S^\circ) \to \mathrm{Spa}(R, R^\circ)$$ is an open immersion of adic spaces (here $$S^\circ$$ and $$R^\circ$$ are the power-bounded subrings) then $$\mathrm{Spec}(S) \to \mathrm{Spec}(R)$$ is injective?

For example, this is true if $$R$$ and $$S$$ both have the discrete topology, because if $$\frak p$$ and $$\frak q$$ are two prime ideals in $$S$$ which are equal after restricting to $$R$$ then $$(\frak p, |\cdot|_{\rm triv})$$ and $$(\frak q, |\cdot|_{\rm triv})$$ (trivial valuations), which are both points in $$\mathrm{Spa}(S,S)$$, restrict to the trivial valuation on $$R/\varphi^{-1}(\frak p)$$.

But I'm not sure how generally to expect that this is true. If it makes it easier, we can assume that $$R$$ and $$S$$ are Tate, so the adic spaces are analytic.

Take $$(R,R^\circ)=(K,\mathcal{O}_K)$$ a non-archimedean field and let $$D=Spa(R,R^\circ )$$ be the unit disk. Choosing a $$c\in K$$ such that $$0 < |c| < 1$$, we can look at the subspace $$U=\{x\in D \mid |T(x)(T(x)-1)|\leq |c|\}.$$ Then $$U=Spa(S,S^\circ)$$ and $$S\cong K\langle \frac{T(T-1)}{c}\rangle$$, which in turn is isomorphic to $$S_0\times S_1:=K\langle\frac{T}{c}\rangle\times K\langle\frac{T-1}{c}\rangle.$$ One way to see this is that as rigid-analytic spaces we have $$\{x\in D \mid |T(x)(T(x)-1)|\leq |c|\}=\{x\in D \mid |T(x)|\leq |c|\}\coprod \{x\in D \mid |(T(x)-1)|\leq |c|\}.$$ But now $$Spec(S)\cong Spec(S_1)\coprod Spec(S_2)\to Spec(R)$$ is not injective: Both the generic point of $$S_1$$ and $$S_2$$ map to the generic point of $$R$$.
• Thanks Louis! I had a brief look in BGR but forgot to check Bosch's lecture notes. In fact, I'm wondering whether non-connectedness is the essential problem in this example, and non-discreteness is the essential problem (cf. Example 21 in Bosch's notes) in the second example. If your affinoid spaces are connected and you work with rigid spaces over $K= \mathbb{Q}_p$, then I wonder if it's injective. Apr 7, 2021 at 10:31
• In fact for my use case I only care about when things are connected, and things are pseudorigid spaces over $\mathbb Z_p$ (which one can basically think of as rigid spaces over $\mathbb Q_p$) Apr 7, 2021 at 11:24
• Dear Ashwin, having thought about your problem, I suggest the following example (to be given more careful thought though): $R=R^\circ=\mathbb{Z}_p\langle T \rangle$ and $S=\mathbb{Z}_p\langle X,Y \rangle/(XY-p)$. This should be the Laurent domain $\{\frac{1}{p}\leq |x| \leq 1\}$. But then $Spec(S)\to Spec(R)$, say via $T\to X$, is not injective, e.g. $(p,Y-1)$ and $(p,Y+1)$ should both map to $(p,T)\in Spec(R)$. That is, if you go to the special fiber, you are contracting one line. Apr 8, 2021 at 7:29
• @LouisJaburi What you wrote is not a Laurent domain because the common vanishing locus of $X$ and $p$ is non-empty. It does not give rise to an open immersion. Apr 8, 2021 at 14:40
• @LouisJaburi On the other hand, I agree that it should be possible to hide your disconnected example inside a connected one. I guess one could take the domain $\{\lvert a\rvert \le \lvert T_1(T_1-1) \rvert \le \lvert T_2 \rvert\}$ of the two-dimensional closed unit disc (with variables $T_1,T_2$) and then argue in the Zariski-closed subset $T_2=c$. Apr 8, 2021 at 14:46