18.727, Topics in Algebraic Geometry (rigid analytic geometry) Kiran S. Kedlaya, fall 2004 Subsets of the projective line, part 1 (of ??)

Before proceeding with more general theory, let’s look at some concrete examples of affinoid spaces. We’ll also look at some spaces that by all rights should be admissible in rigid analytic geometry but which are not affinoid.

Reference: [FvdP, Chapter 2].

Affinoid subsets of ℙ^{1}

As usual, ℙ^{1} will denote the projective line over Spec ℤ, which you may then base-change to any base scheme. Here I’m going to work with a complete ultrametric field K. [FvdP] work with the set of K-valued points ℙ^{1}(K), but this is not really the right way to look at this; I will instead define the set ℙ on which I work to be the set of closed points of the scheme ℙ_{K}^{1}. However, when it’s convenient to do so, I will represent a closed point by a point of ℙ^{1}(K^{alg}) contained in it; I hope this won’t cause too much confusion.

Also, let Γ be the divisible closure in ℝ_{>0} of the group |K^{∗}|; that is, Γ = |(K^{alg})^{∗}|.

A closed disc in ℙ is a subset of one of the forms

{x ∈ K^{alg} : |x − a| ≤ r}

or

{x ∈ K^{alg} : |x − a| ≥ r} ∪ {∞}

for some a ∈ K and r ∈ Γ. An open disc is the same thing but with strict inequalities.

A connected affinoid subset of ℙ is the complement of a nonempty finite union of open discs. An affinoid subset of ℙ is a finite union of connected affinoid subsets.

We want to say that the affinoid subsets are “affinoid spaces” in some natural sense, but we can’t yet say that because we haven’t abstractly defined an affinoid space. The best we can do for now is “naturally” identify them with the maximal spectra of certain affinoid algebras.

Example: the closed disc |x| ≤ 1 is just Max K⟨x⟩. Keep this example firmly in mind as we go along!

0.1 Holomorphic functions

Let F be an affinoid subset of ℙ, and let R(F) ⊂ K(x) be the set of rational functions whose poles do not lie in F. Let ∥ · ∥_{F} denote the supremum norm on R(F):

∥f∥_{F} = sup_{a∈F}{|f(a)|}.

Let A_{F} be the completion of R(F) with respect to ∥ · ∥_{F}.

Theorem 1.

(a) The ring A_{F} is a reduced affinoid algebra.

(b) The natural map F → Max A_{F} is a bijection.

(c) Under the map in (b), the supremum norm corresponds to the spectral norm on A_{F}.

Proof. We can decompose F uniquely as a disjoint union of connected affinoid subsets (see exercises), so it suffices to do the case where F is connected. Also, by performing a fractional linear transformation, I can reduce to the case where F is contained in the closed unit disc (so in particular ∞ ∉ F).

Suppose F is defined by the equations

|x − a_{i}| ≤ r_{i} (i = 1, …, m), |x − b_{j}| ≥ s_{j} (j = 1, …, n).

Choose positive integers e_{1}, …, e_{m}, f_{1}, …, f_{m} and elements ρ_{1}, …, ρ_{m}, σ_{1}, …, σ_{n} ∈ K with |ρ_{i}| = r_{i}^{e_{i}} and |σ_{j}| = s_{j}^{f_{j}}. Let I be the ideal of K⟨x, y_{1}, …, y_{m}, z_{1}, …, z_{n}⟩ generated by

ρ_{i}y_{i} − (x − a_{i})^{e_{i}} (i = 1, …, m), σ_{j} − (x − b_{j})^{f_{i}}z_{j} (j = 1, …, n).

Then one checks (using the hypothesis on the placement of F) that A_{F} is the maximal reduced quotient of

K⟨x, y_{1}, …, y_{m}, z_{1}, …, z_{n}⟩/I

and that the other properties hold. Actually the way to do this is in reverse: first observe (by the Nullstellensatz) that Max A_{F} = F, so the spectral norm coincides with the supremum norm, then use the theorem from last time that A_{F} is complete for the supremum norm.

Exercises

1. (from [FvdP, Exercise 2.1.1]) Prove that every affinoid subset can be written uniquely as a disjoint union of connected affinoid subsets.

2. Suppose that in the proof of Theorem 1, the e_{i} and f_{j} are chosen to be as small as possible. Prove that the quotient

K⟨x, y_{1}, …, y_{m}, z_{1}, …, z_{n}⟩/I

is already reduced, so it is actually equal to A_{F}.

3. Suppose that K is spherically complete. Give an explicit example of an affinoid algebra A which is an integral domain, but whose spectral seminorm is not fully multiplicative. (Hint: consider power series in x and x^{−1} which converge on a suitable annulus; your motivation should functions on a complex annulus which have their suprema on opposite boundary components. We’ll look more at this geometric situation in the next few lectures.)