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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.

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This is not correct in general. There are in fact two examples in Bosch's Lectures on Formal and Rigid Geometry p.61-63. Let me sketch the first one.

While it uses rigid-analytic spaces, it can be easily transferred to adic spaces: Weierstraß subdomains can be seen as special cases of rational subdomains (in adic spaces).

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 $R$.

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    $\begingroup$ 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. $\endgroup$ Apr 7, 2021 at 10:31
  • $\begingroup$ 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 $\mathbb Q_p$) $\endgroup$ Apr 7, 2021 at 11:24
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    $\begingroup$ 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. $\endgroup$ Apr 8, 2021 at 7:29
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    $\begingroup$ @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. $\endgroup$ Apr 8, 2021 at 14:40
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    $\begingroup$ @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 $T_2=c$. $\endgroup$ Apr 8, 2021 at 14:46

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