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.



Open immersion of affinoid adic spaces
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 $R$.


