18.727, Topics in Algebraic Geometry (rigid analytic geometry) Kiran S. Kedlaya, fall 2004 More on affinoid algebras (amended 4 Oct 04)

Addenda on the spectral seminorm

A norm on a Banach algebra A is power-multiplicative if ∥f^{n}∥ = ∥f∥^{n} for any f ∈ A and any positive integer n. Our proof that the Gauss norm on T_{n} has a topological characterization adapts to show that for any affinoid algebra A, there is at most one power-multiplicative Banach norm on A. We now know that if a power-multiplicative norm exists, it must be the spectral seminorm; hence such a norm exists if and only if A is reduced.

In fact, the spectral seminorm is “minimal” in the following sense [BGR, Corollary 3.8.2/2].

Proposition 1. Let A be an affinoid algebra equipped with a Banach norm ∥ · ∥. Then for all f ∈ A, ∥f∥_{spec} ≤ ∥f∥. In particular, |f(x)| ≤ ∥f∥ for any x ∈ Max A (we already showed this if ∥ · ∥ is a quotient norm from a Tate algebra).

Proof. Apply the formula

∥f∥_{spec} = lim_{n→∞} ∥f^{n}∥^{1/n}

and note that ∥f^{n}∥ ≤ ∥f∥^{n} because ∥ · ∥ is a Banach algebra norm.

This yields the following characterization of nilpotent elements, in the vein of our characterization of power-bounded elements [BGR, Proposition 6.2.3/2].

Proposition 2. For A an affinoid algebra and f ∈ A, the following statements are equivalent:

(a) f is topologically nilpotent (i.e., {f^{n}} is a null sequence in A);

(b) |f(x)| < 1 for all x ∈ Max A;

(c) ∥f∥_{spec} < 1.

Proof. The equivalence of (b) and (c) follows from the maximum modulus principle, and (a) implies (c) by the previous proposition. Given (c), choose c ∈ K and m ∈ ℕ such that |c| > 1 but ∥cf^{m}∥_{spec} ≤ 1. Then cf^{m} is power-bounded (from last time), so c^{−1}(cf^{m}) = f^{m} is topologically nilpotent, as then is f. Thus (c) implies (a), and we are done.

Spectral norms are Banach norms

We now know that the spectral seminorm on a reduced affinoid algebra is a norm. However, more than that is true: it is a Banach norm. (This proof is from [FvdP, Theorem 3.4.9]; the proof in [BGR, Theorem 6.2.4/1] is a bit more intricate.)

We say a bounded map f : A → B of Banach spaces is strict if the bijection between A/ ker(f) with its quotient topology and im(f) with its subspace topology is an isomorphism of normed spaces. Since ker(f) is closed, by the open mapping theorem this is equivalent to saying that im(f) is closed in B.

Note that if A and B are affinoid algebras and f is finite, it is automatically strict, because f(A) is an A-submodule of B, and any submodule of a finitely generated module over an affinoid algebra is closed (see lemma from “p-adic functional analysis 2”). In fact, if A is an affinoid algebra and M_{1}, M_{2} are finitely generated modules over A, then every A-algebra homomorphism M_{1} → M_{2} is strict.

On the other hand, here’s an example of a bounded non-strict map: the homomorphism T_{1} → T_{1} sending x_{1} to cx_{1}, where c ∈ K satisfies |c| < 1.

Lemma 3. Let A ↪ B be a strict inclusion of affinoid algebras. Then the spectral seminorm on B restricts to the spectral seminorm on A.

Proof. Choose a Banach norm ∥ · ∥ on B; it then restricts to a Banach norm on A by strictness (and the fact that any two Banach norms on an affinoid algebra are equivalent), and applying the formula ∥f∥_{spec} = lim_{n→∞} ∥f^{n}∥^{1/n} gives us the claim.

Theorem 4. Let A be a reduced affinoid algebra. Then A is complete under ∥ · ∥_{spec}. In particular, every Banach algebra norm on A is equivalent to the spectral norm.

Proof. The last sentence will follow from what we showed earlier: any two Banach algebra norms on an affinoid algebra are equivalent. So we focus on showing that A is complete.

We first reduce to the case where A is an integral domain. Let 𝔭_{1}, …, 𝔭_{m} be the minimal primes of A. Choose a Banach norm ∥·∥_{A} on A, put A_{i} = A/𝔭_{i} (which is an integral domain), and equip each A_{i} with the quotient norm induced by ∥ · ∥_{A}. Then A_{1} ⊕ ⋯ ⊕ A_{m} becomes a finitely generated Banach module over A under the max norm

(a_{1}, …, a_{m}) = max_{i}{∥a_{i}∥}.

Let ι : A → A_{1} ⊕ ⋯ ⊕ A_{m} be the canonical injection; then ι(A) is an A-submodule of A_{1} ⊕ ⋯ ⊕ A_{m}. By the lemma from “p-adic functional analysis 2”, ι(A) is closed in A_{1} ⊕ ⋯ ⊕ A_{m}, and ι is an isomorphism onto its image by the open mapping theorem. The map ι is isometric for the spectral norms (because Max A = ∪_{i} Max A_{i}), so proving that the spectral norm on each A_{i} is equivalent to the quotient norm proves that the spectral norm on A is equivalent to ∥ · ∥_{A}.

From now on, we assume A is an integral domain. If B is a reduced affinoid algebra containing A and A ↪ B is strict (e.g., if A ↪ B is finite), showing that B is complete under its spectral seminorm implies that A is complete under its spectral seminorm, thanks to the previous lemma. In particular, we can write A as a finite integral extension of T_{d} for some d ≥ 0, and take B to be the integral closure of T_{d} in the normal closure of Frac A over Frac T_{d}. This lets us break the problem into two steps.

(a) Show that if A is finite over T_{d} and Frac A is purely inseparable over Frac T_{d}, then the spectral norm on A is complete.

(b) Show that if B is finite over A, Frac B is Galois over Frac A, and the spectral norm on A is fully multiplicative and complete, then the spectral norm on B is complete.

So we work on these two steps separately.

(a) There is nothing to show unless the characteristic of K is p > 0. Note that for any affinoid algebra A over K and any field extension K′ containing K and complete for a norm extending that on K, the inclusion of A into the completed tensor product Aˆ⊗_{K}K′ is strict. In particular, we may reduce to the case where K is algebraically closed. In that case,

A ⊆ K⟨x_{1}^{1/p^{m}}, …, x_{d}^{1/p^{m}}⟩

for some m, so it’s enough to check the completeness of K⟨x_{1}^{1/p^{m}}, …, x_{d}^{1/p^{m}}⟩ under its spectral norm. But again, this is true because that spectral norm is the Gauss norm.

(b) Put G = Gal(Frac B/ Frac A), and put

Trace(f) = ∑_{g∈G} f^{g};

this gives a map from Frac B to Frac A such that ∥ Trace(f)∥_{A,spec} ≤ ∥f∥_{B,spec}. (This trace is the same as the trace of multiplication by f as a Frac A-linear transformation on Frac B.) From basic algebra, we know that the Frac A-linear pairing

(x, y) ↦ Trace(xy)

on Frac B is nondegenerate.

Choose e_{1}, …, e_{n} ∈ B which form a basis for Frac B over Frac A, and let e_{1}^{∗}, …, e_{n}^{∗} be the dual basis for the trace pairing. We show that Ae_{1} + ⋯ + Ae_{n} is a Banach module under the spectral norm, by showing that the spectral norm is equivalent to the maximum norm

∥f_{1}e_{1} + ⋯ + f_{1}e_{n}∥_{B,spec} = max_{i}{∥f_{i}∥_{A,spec}}.

Choose a_{0} ∈ A such that a_{0}e_{j}^{∗} ∈ 𝔬_{B}^{spec} for j = 1, …, n; now given f_{1}e_{1} + ⋯ + f_{n}e_{n} ∈ Ae_{1} + ⋯ + Ae_{n}, we have

a_{0}f_{j} = Trace(a_{0}e_{j}^{∗} ∑_{i} f_{i}e_{i})
∥ Trace(a_{0}e_{j}^{∗} ∑_{i} f_{i}e_{i})∥_{A,spec} ≤ ∥a_{0}e_{j}^{∗} ∑_{i} f_{i}e_{i}∥_{B,spec}

≤ ∥ ∑_{i} f_{i}e_{i}∥_{B,spec}.

Therefore

∥ ∑_{i} f_{i}e_{i}∥_{B,spec} ≥ ∥a_{0}∥_{A,spec} max_{i}{∥f_{i}∥_{A,spec}}.

Since we also have

∥ ∑_{i} f_{i}e_{i}∥_{B,spec} ≤ max_{i}{∥f_{i}∥_{A,spec}} max_{i}{∥e_{i}∥_{B,spec}},

the spectral norm restricted to Ae_{1} + ⋯ + Ae_{n} is equivalent to the maximum norm, and so is a Banach norm.

For some a ∈ A, aB ⊆ Ae_{1} + ⋯ + Ae_{n}; since the spectral norm on A is multiplicative, the spectral norm on B is thus complete.

Note that this theorem can also be interpreted as follows: a sequence of elements of A converges to zero under some (any) Banach algebra norm if and only if it converges uniformly to zero on Max A.

The reduction of an affinoid algebra

In case you are wondering when the spectral seminorm is not just a norm but is actually fully multiplicative (like the Gauss norm), here is your answer. Recall that for A an affinoid algebra, we defined

𝔬_{A}^{spec} = {f ∈ A : ∥f∥_{spec} ≤ 1}.

Now define

𝔪_{A}^{spec} = {f ∈ A : ∥f∥_{spec} < 1}

and \overline{A}^{spec} = 𝔬_{A}^{spec}/𝔪_{A}^{spec}; we call the latter the reduction of A. Then we have the following [BGR, Proposition 6.2.3/5].

Proposition 5. The spectral seminorm is a fully multiplicative norm if and only if A is reduced and \overline{A} is an integral domain.

Note that A being an integral domain is not enough; see exercises.

Proof. We already know that the spectral seminorm is a norm if and only if A is reduced; also, if A is reduced and the spectral norm is fully multiplicative, then the product of elements of spectral norm 1 again has spectral norm 1, so \overline{A} is an integral domain. Conversely, suppose A is reduced and \overline{A} is an integral domain. Given f, g ∈ A nonzero, there exists an integer n such that ∥f∥_{spec}^{n} and ∥g∥_{spec}^{n} belong to |K^{∗}|, by the maximum modulus principle. (Namely, the spectral seminorm is always the norm of the evaluation of f at some point whose residue field is finite over K.) Choose c, d ∈ K^{∗} with c∥f∥_{spec}^{n} = d∥g∥_{spec}^{n} = 1. Then the product of the images of cf^{n} and dg^{n} in \overline{A} must be nonzero because \overline{A} is an integral domain; that is, ∥cf^{n}dg^{n}∥_{spec} = 1. Hence

1 = ∥cf^{n}dg^{n}∥_{spec} = c∥f∥_{spec}^{n} · d∥g∥_{spec}^{n}

and (by power-multiplicativity of the spectral seminorm) it follows that ∥fg∥_{spec} = ∥f∥_{spec} · ∥g∥_{spec}.

Exercises

1. 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.)

2. Let A be an affinoid algebra, and suppose f ∈ A satisfies ∥f∥_{A,spec} = 1. Prove that the natural map from A to B = A⟨x⟩/(fx − 1) is strict. (Such a B is called a strict localization of A.)