12
$\begingroup$

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.

$\endgroup$

1 Answer 1

6
$\begingroup$

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 $R$.

$\endgroup$
6
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.