# Injectivity of sheaf restriction maps for wide open neighbourhoods of rational subdomains

Let $$X=\text{Sp}(A)$$ be an affinoid $$K$$ space, where $$K$$ is a $$p$$-adic field. If $$f_0, f_1,..., f_s \in A$$ generate the unit ideal then we can define the rational subdomain $$U= X(f_0, f_1..., f_s) = \{ x \in X: \vert f_i(x) \vert \leq \vert f_0(x) \vert \text{ for } i=1...s \}$$ of $$X$$ with coordinate ring $$\mathcal{O}(U)= A \langle x_1,...,x_s \rangle/(f_1-f_0x_1, ..., f_s-f_0x_s)$$.

If $$\rho \in \sqrt{\vert K^\times \vert}$$ and $$\rho >1$$ then, by definition, there exists a natural number $$n$$ such that $$\rho^n= \vert t \vert$$ for some $$t \in K^\times$$. We can define another rational subdomain $$U(\rho)= X(\rho f_0, f_1..., f_s) = \{ x \in X: \vert f_i(x) \vert \leq \rho \vert f_0(x) \vert \text{ for } i=1...s \}$$ of $$X$$. I think its coordinate ring is $$\mathcal{O}(U(\rho))= A \langle t^{-1}x_1,..., t^{-1}x_s \rangle/(f^n_1-f^n_0 x_1, ..., f^n_s-f^n_0 x_s)$$.

As $$U \subset \subset_{X} U(\rho)$$ we have, in particular, a sheaf restriction map $$r_{U(\rho)U}: \mathcal{O}(U(\rho)) \rightarrow \mathcal{O}(U)$$. I think this is just given by $$t^{-1} x_i \mapsto x_i$$ for all $$i=1...s$$. I have been trying to work out whether this restriction map is necessarily injective. So far, my attempts have been direct, i.e. trying to prove that the kernel is zero, but I haven't got anywhere. Irritatingly, I also cannot find any counterexample to the statement I am trying to prove (I might be being stupid).

Does anybody know whether what I am trying to prove is true? If so, could you please point me towards a proof? I would also be very interested to know whether the more general statement $$U \subset \subset_{X} V$$ (wide open neighbourhood) with $$U$$ an affinoid subdomains of $$X$$ implies that the structure sheaf restriction map $$\mathcal{O}(V) \rightarrow \mathcal{O}(U)$$ is injective. My original question is a special case of this one. I had a go at the special case first because things seemed more explicit!

Thank you so much in advance for any help!

• This is not going to work. If you take an affinoid space with 2 connected components (even 2 points if you want), you can find a rational domain that isolates one but that meets the two when you enlarge the radius. I can write down an example later if it is helpful. Jul 7, 2023 at 10:15
• Thank you. From what you have written, I think I understand why what I am attempting to prove cannot be true. If you have time to write down the example, that would be really helpful. Thanks again! Jul 7, 2023 at 14:17

For an explicit example, you can choose $$A = \mathbb{Q}_p \langle pT \rangle/(T(pT-1))$$. Its spectrum has exactly two points: $$0$$ and $$1/p$$. To find a example of a non-injective map, you can consider the Weierstrass domains $$\{|T| \le 1\}$$ (which contains only $$0$$) and $$\{|T| \le 1/p\}$$ (which is the whole space).
Also, the map you write is wrong. When $$n=1$$, you want to send $$x_i$$ to $$x_i$$ in order to preserve the relations in the quotient.