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!

  • $\begingroup$ 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. $\endgroup$ Jul 7, 2023 at 10:15
  • $\begingroup$ 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! $\endgroup$
    – Tom Adams
    Jul 7, 2023 at 14:17

1 Answer 1


If your space has several connected components, then a rational domain may well isolate one of them and you could get a non-injective map.

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.


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