18.727, Topics in Algebraic Geometry (rigid analytic geometry) Kiran S. Kedlaya, fall 2004 Tate algebras (or, Commutative algebra revisited)

We will now talk a bit about Tate algebras, which play a role like that of the polynomial rings over a field in ordinary algebraic geometry.

As usual, K is a complete ultrametric field, which you may assume is discretely valued if you prefer, and k is its residue field. Reminder: I write 𝔬_{K} and 𝔪_{K} for the valuation subring of K and its maximal ideal, rather than the bizarre K^{o} and K^{oo} used in [FvdP].

Convention: when I write ∑_{I} c_{I}x^{I}, the sum will be running over tuples (normally of nonnegative integers) I = (i_{1}, …, i_{n}), and x^{I} = x_{1}^{i_{1}} ⋯ x_{n}^{i_{n}}.

References: The main reference is [FvdP, Section 3.1], but I have lots of issues with the presentation, so I’ve supplemented from [BGR, 5.1]. Also, I’ll cite Lang’s Algebra (third edition; numbering may differ in the current edition) in the exercises as [L].

Tate algebras

The Tate algebra (or standard affinoid algebra)

T_{n} = T_{n,K} = K⟨x_{1}, …, x_{n}⟩

is the subring of the ring of formal power series K⟦x_{1}, …, x_{n}⟧ consisting of sums ∑_{I} c_{I}x^{I} such that |c_{I}| → 0 as I → ∞; what that really means is that for any ϵ > 0, there are only finitely many tuples I such that |c_{I}| > ϵ. Such series are sometimes called strictly convergent power series (as in [BGR]).

Define the Gauss norm on T_{n} by the formula

∥∑_{I} c_{I}x^{I} ∥ = max_{I}{|c_{I}|};

note that the max is really a max and not a supremum, since the |c_{I}| tend to 0. This is in fact a norm, under which T_{n} becomes a Banach algebra over K of countable type (since the monomials x^{I} have dense span); in fact, it’s isomorphic as a Banach space to the space c_{0} of null sequences. (Trivial but handy consequence of this definition: the image of the Gauss norm is the same as the image of K under its norm. This will let us do some “normalization” arguments.)

As usual for normed rings, I write 𝔬_{T_{n}} to mean the subring of T_{n} consisting of elements of norm ≤ 1, and 𝔪_{T_{n}} to mean the ideal of 𝔬_{T_{n}} consisting of elements of norm < 1. Then 𝔬_{T_{n}}/𝔪_{T_{n}} = k[x_{1}, …, x_{n}]; given f ∈ 𝔬_{T_{n}}, I’ll write \overline{f} for its image in k[x_{1}, …, x_{n}] and call it the reduction of f. (That’s why I don’t use the overbar for algebraic closures!)

Here are a few basic facts about Tate algebras.

Lemma 1. (a) A series ∑ c_{I}x^{I} ∈ K⟦x_{1}, …, x_{n}⟧ belongs to T_{n} if and only if the sum ∑ c_{I}α_{1}^{i_{1}} ⋯ α_{n}^{i_{n}} converges in K for any α_{1}, …, α_{n} ∈ 𝔬_{K}.

(b) Suppose that the residue field k = 𝔬_{K}/𝔪_{K} is infinite. Then given a series ∑ c_{I}x^{I} ∈ T_{n}, there exist α_{1}, …, α_{n} ∈ 𝔬_{K} such that

∥∑_{I} c_{I}x^{I} ∥ = |∑_{I} c_{I}α_{1}^{i_{1}} ⋯ α_{n}^{i_{n}} | .

(c) The Gauss norm is fully multiplicative: for all f, g ∈ T_{n}, ∥fg∥ = ∥f∥·∥g∥. (Remember, a Banach algebra is only required to have ∥fg∥ ≤ ∥f∥ · ∥g∥.)

Proof. (a) This is clear; you need only check α_{1} = ⋯ = α_{n} = 1.

(b) We may as well assume ∑_{I} c_{I}x^{I} ≠ 0. Let P (x_{1}, …, x_{n}) be the sum of c_{I}x^{I} over all tuples I for which |c_{I}| is maximal. Then P is a polynomial in x_{1}, …, x_{n}, ∥P ∥ = ∥ ∑_{I} c_{I}x^{I}∥, and

∥P (x_{1}, …, x_{n}) − ∑_{I} c_{I}x^{I}∥ < ∥P ∥.

It thus suffices to prove the claim for P instead of the original series. But P can be written as the product of some c ∈ K with a polynomial P_{0} which has coefficients in 𝔬_{K} but not all in 𝔪_{K}. Since k is infinite, the reduction \overline{P_{0}} does not vanish everywhere. Pick α_{1}, …, α_{n} ∈ 𝔬_{K} so that P_{0}(α_{1}, …, α_{n}) ∉ 𝔪_{K}; then these are a good choice.

(c) Write f = ∑_{I} c_{I}x^{I}, g = ∑_{J} d_{J}x^{J}, and fg = ∑_{H} e_{H}x^{H}. Let I be the tuple maximizing |c_{I}| which is first in lexicographic order. (That is, you compare tuples by first comparing their first components, then their second if the firsts are tied, then their thirds, and so on.) Likewise, let J be the tuple maximizing |d_{J}| which is first in lexicographic order. Then e_{I+J} equals c_{I}d_{J} plus some other terms of the form c_{I′}d_{J′}, where I′ and J′ are two other tuples adding up to I + J. But that means that either I′ must precede I in lexicographic order, in which case |c_{I′}| < |c_{I}| and |d_{J′}| ≤ |d_{J}|, or J′ must precede J in lexicographic order, in which case |c_{I′}| ≤ |c_{I}| and |d_{J′}| ≤ |d_{J}|. In any case, we see that |e_{I+J} − c_{I}d_{J}| < |c_{I}d_{J}|, so |e_{I+J}| = |c_{I}d_{J}. It follows that ∥fg∥ ≥ ∥f∥ · ∥g∥; since we already know the other inequality, we have ∥fg∥ = ∥f∥ · ∥g∥.

It is worth noting what the units are in T_{n}; since we can normalize in T_{n}, we just treat units of norm 1. (This is [BGR, Proposition 5.1.3/1].)

Lemma 2. For f ∈ T_{n} with ∥f∥ = 1, the following are equivalent.

(a) f is a unit in 𝔬_{T_{n}}.

(b) f is a unit in T_{n}.

(c) \overline{f} is constant (i.e., is a unit in k[x_{1}, …, x_{n}]).

(d) |f(0)| = 1 and ∥f − f(0)∥ < 1.

Proof. Note that (a) and (b) are equivalent because ∥ · ∥ is fully multiplicative. Clearly (a) implies (c), and (c) and (d) are equivalent. Finally, given (d), the series f(0) ∑_{i}(1 − f/f_{0})^{i} converges to a reciprocal of f in 𝔬_{T_{n}}, so (a) follows.

Also, note that there is an isomorphism

K⟨x_{1}, …, x_{m}⟩ˆ⊗K⟨y_{1}, …, y_{n}⟩ ≅ K⟨x_{1}, …, x_{m}, y_{1}, …, y_{n}⟩.

The Tate algebra is meant to be the “ring of functions on the closed unit polydisc”. Its properties are a blend between properties of polynomial rings and formal power series rings. This hybrid nature makes it possible to do analytic geometry using Tate algebras using a lot of our intuition from algebraic geometry.

Weierstrass preparation

We say f ∈ 𝔬_{T_{n}} is (normalized) distinguished (in x_{n}) of degree d if

\overline{f} = c_{0} + c_{1}x_{n} + ⋯ + c_{d}x_{n}^{d}

with c_{d} ∈ k^{∗} and c_{i} ∈ 𝔬_{T_{n}} for i = 0, …, d − 1. Note that ∥f∥ = 1, whence the “normalized”; if you allow something of this form times an element of K^{∗}, you get elements which are distinguished in the terminology of [BGR]. However, for this lecture, all my distinguished elements will also be normalized, so I won’t keep saying “normalized”. (Terminology rant:

[FvdP] use “regular” for my “normalized distinguished”, but the word “regular” will come up later with a more useful meaning.)

Then one has the following results; see [FvdP, Theorem 3.1.1] and [BGR, 5.2]. (Caution:

[FvdP] incorrectly applies the moniker “preparation” to (c) instead of (b).)

Theorem 3. (a) (Division algorithm) Suppose f ∈ 𝔬_{T_{n}} is distinguished in x_{n} of degree d. Then any g ∈ T_{n} can be uniquely written as qf + r such that q ∈ T_{n} and r ∈ T_{n−1}[x_{n}], where the degree of r in x_{n} is less than d. Moreover, ∥g∥ = max{∥q∥, ∥r∥}.

(b) (Preparation) If f ∈ 𝔬_{T_{n}} is distinguished in x_{n} of degree d, then there is a unique way to write f = gh with g ∈ 𝔬_{T_{n−1}}[x_{n}] monic of degree d (and hence distinguished) and h ∈ 𝔬_{T_{n}}^{∗}.

(c) (Distinction) If f_{1}, …, f_{m} ∈ 𝔬_{T_{n}} all have norm 1, then there exists an automorphism τ of T_{n} (preserving Gauss norms) such that f_{1}^{τ}, …, f_{m}^{τ} are distinguished in x_{n}.

The last parenthetical is actually moot, as all automorphisms of T_{n} will preserve the Gauss norm, but this will only become obvious a bit later.

Proof. (a) We first verify uniqueness. If qf + r = q′f + r′ are two different decompositions of the same g, then (q − q′)f = r′ − r. By the full multiplicativity of the Gauss norm, this means ∥q − q′∥ = ∥r′ − r∥; pick some c ∈ K with |c| = ∥q − q′∥^{−1}. Then c(q−q′)f = c(r′−r) and so the same is true with bars everywhere; but that contradicts uniqueness in the ordinary division algorithm for polynomials.

We next verify the claim about the norm. If g = qf + r, then on one hand

∥g∥ ≤ max{∥qf∥, ∥r∥} = max{∥q∥, ∥r∥},

and we know we have equality if ∥q∥ ≠ ∥r∥. But if ∥q∥ = ∥r∥ > ∥g∥, we can choose c ∈ K with |c| = ∥q∥^{−1}, and then cg = cqf + cr. But the reduction of cg is zero, so by the uniqueness in the ordinary division algorithm, cq and cr would also have to have zero reductions, contradiction. So even in case ∥q∥ = ∥r∥ we must have ∥g∥ = max{∥q∥, ∥r∥}.

Now for existence. We first check this in case f = f_{0} = c_{0} + c_{1}x_{n} + ⋯ + c_{d}x_{n}^{d} with each c_{i} ∈ 𝔬_{T_{n−1}}; this forces c_{d} ∈ 𝔬_{K}^{∗}. Put g = ∑ d_{I}x^{I}, and apply the ordinary ordinary division algorithm to write x^{I} = q_{I}f +r_{I} with q_{I}, r_{I} ∈ T_{n−1}[z_{n}], with the degree of r_{I} in z_{n} being less than d. By what we already showed, we have max{∥q_{I}∥, ∥r_{I}∥} = ∥x^{I}∥ = 1; thus the series

q = ∑_{I} d_{I}q_{I}, r = ∑_{I} d_{I}r_{I}

converge in T_{n} and T_{n−1}[z_{n}], respectively (the latter because I’ve bounded the degrees, so I really get a polynomial in z_{n} and not a series). By design, g = qf + r.

Now for the general case; write f = f_{0} − D where f_{0} is as in the previous case and ∥D∥ < 1. Given g, we put g_{0} = g; given g_{i}, apply what I just did to write

g_{i} = q_{i}f_{0} + r_{i} = q_{i}f + r_{i} + q_{i}D

and put g_{i+1} = q_{i}D. Then q = ∑_{i} q_{i} and r = ∑ r_{i} converge to limits satisfying g = qf + r, and r is again a polynomial in z_{n} of degree less than d.

(b) We first check existence. Apply division to obtain q′, r′ such that x_{n}^{d} = q′f + r′, and put q = x_{n}^{d} − r′; then q ∈ 𝔬_{T_{n−1}}[x_{n}] is monic of degree d and q′f = q. On reductions, we have \overline{q} = \overline{q′f}, and \overline{q} and \overline{f} are polynomials of the same degree. Hence \overline{q′} is a unit, and so q′ is a unit by Lemma 2. We can thus factor f = gh with g = q and h = (q′)^{−1}.

We next check uniqueness. If f = gh is a presentation of the desired form, we have

x_{n}^{d} = h^{−1}f + (x_{n}^{d} − g),

and this is what you get from an application of (a). Thus specifying f uniquely determines x_{n}^{d} − g, and hence determines g and h.

(c) I’ll let you find the easy proof for k infinite for yourself; instead, I’ll give the slightly more elaborate argument that also works for k finite. Write f_{l} = ∑ c_{I,l}t^{I} for l = 1, …, m, and choose integers e_{1}, …, e_{n−1} ≥ 0 such that whenever I ≠ J are among the finitely many tuples with |c_{I,l}| = |c_{J,l}| = 1 for some l, we have

e_{1}i_{1} + ⋯ + e_{n−1}i_{n−1} + i_{n} ≠ e_{1}i_{j} + ⋯ + e_{n−1}j_{n−1} + j_{n}.

(For instance, you can choose the e_{i} to be successive powers of a bigger than any value of i_{1}, …, i_{n} that shows up in the tuples.) Let τ be the automorphism which substitutes x_{i} + x_{n}^{e_{i}} in place of x_{i} for i = 1, …, n − 1 (and fixes x_{n}). Then

\overline{f_{l}^{τ}} = ∑ \overline{c_{I,l}}(x_{1} + x_{n}^{e_{1}})^{i_{1}} ⋯ (x_{n−1} + x_{n}^{e_{n−1}})^{i_{n−1}}x_{n}^{i_{n}};

if you pick out the unique tuple i_{1}, …, i_{n} maximizing e = e_{1}i_{1}+⋯+e_{n−1}i_{n−1}+i_{n}, then in the reduction, the unique term of highest degree that you see is x_{n}^{e}. That means each f_{l}^{τ} is distinguished.

More properties of Tate algebras

Weierstrass preparation immediately yields the analogue of the Hilbert basis theorem for Tate algebras; in the case of K discretely valued, this is a theorem of Fulton. (Yes, that Fulton! See: A note on weakly complete algebras, Bull. Amer. Math. Soc. 75 (1969), 591–593.)

Proposition 4 (Hilbert basis theorem). The ring T_{n} is noetherian.

Proof. Induction on n. Given a nonzero ideal I of T_{n}, choose f ∈ I nonzero, and apply distinction to find an automorphism τ of T_{n} such that f^{τ} is distinguished in x_{n} of some degree d. Using division, we see that I^{τ} is generated by f^{τ} together with I^{τ} ∩ T_{n−1}[z_{n}]. By the induction hypothesis, T_{n−1} is noetherian, as then is T_{n−1}[z_{n}] by the usual Hilbert basis theorem. Thus I^{τ} is finitely generated, as then is I.

Proposition 5. The ring T_{n} is a unique factorization domain.

Proof. Again, induct on n. Given f ∈ T_{n}, suppose f = g_{1} ⋯ g_{m} = h_{1} ⋯ h_{n} are two factorizations of f into irreducibles; by pushing scalars around, we may reduce to the case where ∥g_{i}∥ = ∥h_{j}∥ = 1 for all i, j. By distinction, there exists an automorphism τ of T_{n} such that f, the g_{i}^{τ}, and the h_{j}^{τ} are all distinguished. By preparation, we can write each g_{i}^{τ} = P_{i}u_{i} with P_{i} ∈ 𝔬_{T_{n−1}}[x_{n}] monic and u_{i} ∈ 𝔬_{T_{n}}^{∗}, and likewise write h_{j}^{τ} = Q_{j}v_{j} with Q_{j} ∈ 𝔬_{T_{n−1}}[x_{n}] monic and v_{j} ∈ 𝔬_{T_{n}}^{∗}. Then P_{1} ⋯ P_{m} equals Q_{1} ⋯ Q_{n} times a unit, but both sides are monic polynomials in x_{n} over 𝔬_{T_{n−1}}, necessarily of the same degree (since that’s true on the reductions). Thanks to the uniqueness of the preparation of P_{1} ⋯ P_{m}, we must in fact have P_{1} ⋯ P_{m} = Q_{1} ⋯ Q_{n}. Moreover, each P_{i} and Q_{j} is irreducible in T_{n}, hence also in T_{n−1}[x_{n}]; thus the factorizations agree up to units, by the unique factorization theorem for polynomials over a UFD (i.e., unique factorization over a field plus “Gauss’s lemma”). That proves that the original factorizations of f^{τ}, and hence of f, agree up to units.

Proposition 6. The Krull dimension of T_{n} is n.

Proof. The sequence of prime ideals

(0) ⊂ (x_{1}) ⊂ (x_{1}, x_{2}) ⊂ ⋯ (x_{1}, x_{2}, ⋯, x_{n})

shows that the Krull dimension is at least n. On the other hand, for any irreducible f ∈ T_{n}, by distinction plus preparation, T_{n}/(f) is finite over T_{n−1}, and so has Krull dimension n − 1. (Remember that making a finite ring extension of a noetherian ring cannot increase its Krull dimension.) Hence T_{n} has Krull dimension at most n, yielding the claim.

Affinoid algebras and Noether normalization

An affinoid algebra is a K-algebra A of the form T_{n}/𝔞 for some ideal 𝔞. By Fulton’s theorem, A is noetherian.

Note that there are a couple of minor discrepancies between this definition and the one in [FvdP, 3.1]. For one, they use the term “Tate algebra” to mean any affinoid algebra; I have seen this elsewhere, but I still think it is nonstandard (e.g., [BGR] does not do this). For another, they define affinoid algebras as integral extensions of Tate algebras. This amounts to using the business end of the Noether normalization theorem (see below), and strikes me as bizarre.

Proposition 7 (Noether normalization). Let 𝔞 be an ideal of T_{n}, and let A = T_{n}/𝔞 be the corresponding affinoid algebra. Then there exists a finite injective map T_{d} → A for some d; moreover, the Krull dimension of A is equal to d.

Proof. We first prove the existence of the map, by induction on n. We may as well assume 𝔞 is a nontrivial ideal; by distinction and preparation, after applying an automorphism of T_{n} we may assume that 𝔞 contains a monic polynomial f ∈ T_{n−1}[x_{n}]. Then T_{n}/(f) is finite over T_{n−1}; if we put 𝔟 = 𝔞 ∩ T_{n−1}, then T_{n}/𝔞 is finite over T_{n−1}/𝔟. By the induction hypothesis, T_{n−1}/𝔟 is finite over some T_{d}, yielding the claim.

For the statement about the Krull dimension, we need only recall that for A → B a finite injective homomorphism of noetherian rings, the rings A and B have the same Krull dimension, and that the Krull dimension of T_{d} is d by Proposition 6.

Warning: unlike in the polynomial situation, an affinoid algebra can have an affinoid subalgebra of greater Krull dimension! See [FvdP, Exercises 3.2.2].

Note that the Nullstellensatz for Tate algebras falls out as a consequence.

Corollary 8. For any maximal ideal 𝔪 of T_{n}, the field T_{n}/𝔪 is finite over K.

Proof. A field has Krull dimension 0, so by Noether normalization, there must exist a finite map K = T_{0} → T_{n}/𝔪.

Affinoid algebras are Banach algebras (and canonically so!)

Recall the following lemma (Lemma 1 from “functional analysis, part 2”).

Lemma 9. Let A be a Banach algebra over K which is noetherian as a ring. Let M be a Banach module over A which is module-finite over A. Then any A-submodule of M is closed.

In particular, since T_{n} is noetherian, any ideal of T_{n} is closed. Hence any affinoid algebra A inherits from a presentation T_{n} → A a quotient norm, under which it becomes a Banach algebra.

It turns out that the topology of an affinoid algebra is uniquely determined by its Kalgebra structure, and all K-algebra homomorphisms of affinoid algebras. To see this, we need to back up and do a little more functional analysis; the following is [BGR, Proposition 3.7.5/2].

Proposition 10. Let B be a noetherian Banach algebra over K, and suppose there exists a collection S of ideals of B such that:

(a) for any 𝔟 ∈ S, dim_{K} B/𝔟 < ∞;

(b) ∩_{𝔟∈S}𝔟 = (0).

Then any K-algebra homomorphism of a noetherian Banach algebra A over K into B is continuous (hence bounded).

Proof. Let f : A → B be a K-algebra homomorphism; we will apply the closed graph theorem to the graph of f. Namely, we need to show that if {x_{n}} is a null sequence in A and f(x_{n}) → y in B, then y = 0.

Pick an ideal 𝔟 ∈ S; note that 𝔟 is closed, by Lemma 9. Put 𝔞 = f^{−1}(𝔟), so that 𝔞 is closed in A, each of A/𝔞 and B/𝔟 inherits a quotient norm, and A/𝔞 → B/𝔟 is injective. Since B/𝔟 is finite dimensional over K, so is A/𝔞, and any linear map between finite dimensional K-vector spaces is continuous. (Remember that there is only one equivalence class of norms on a finite dimensional K-vector space!) Thus we must have y ∈ 𝔟 for each 𝔟; since the 𝔟 have trivial intersection, we have y = 0.

Corollary 11. Any K-algebra homomorphism between affinoid algebras is continuous. In particular, for a fixed affinoid algebra A, the quotient norms induced by different presentations T_{n} → A are all equivalent.

Now I can clarify my remark from earlier about norm-preserving automorphisms of T_{n}.

Corollary 12. Any automorphism of T_{n} preserves the Gauss norm.

Proof. This follows from the previous corollary and the fact that we can recover the Gauss norm from the topology of T_{n}. Namely, an element f ∈ T_{n} satisfies ∥f∥ ≤ 1 if and only if for any null sequence {c_{n}} in K, {c_{n}f} is null in T_{n}.

In other words, the Gauss norm is a “canonical” norm for T_{n}. There is an analogous function on an arbitrary affinoid algebra, but it’s only a seminorm in general; we’ll talk more about it soon.

Exercises

These exercises are from Section 6 of my preprint “Full faithfulness for overconvergent F - isocrystals”, which you can grab from the arXiv if you get stuck.

1. Using distinction, prove that the action of GL_{n}(T_{n}) on n-tuples which generate the unit ideal is transitive. (Hint: compare to a proof of the analogous statement for polynomials; see [L, Theorem XXI.3.4].)

2. Prove that every finitely generated module over T_{n} has a finite free resolution. (Hint: this time, see [L, Theorem XXI.3.6].)

3. Prove the analogue of the Quillen-Suslin theorem for T_{n}: every finitely generated projective module over T_{n} is free. (Hint: use [L, Theorem XXI.2.1] to show that a finite projective is stably free, that is, its direct sum with some free is free. Then compare [L, Theorem XXI.3.6].)