Non-Archimedian analysished29573/vort/rigid.pdf · 2015. 8. 3. · Non-Archimedian analysis University of Ratisbon Winter term 2014/15 Daniel Heiß: §4: Tatealgebras October 30th,

Non-Archimedian analysis

University of RatisbonWinter term 2014/15

Daniel Heiß:

§4: Tate algebras

October 30th, 2014

Non-Archimedian Analysis §4: Tate algebras

Abstract

In this talk we will further study properties of the Tate algebra Tn like it is Noetherian, factorialand Jacobson. In the second part we will have a closer look at the ideals of Tn seeing that everyideal a ( Tn is complete and strictly closed.Mainly we will follow [Bos08].

Notation. If not otherwise specified K denotes a field that is complete with respect to its non-Archimedian absolute value | • | and K its algebraic closure. As any field extension L/K admitsa unique extension of the absolute value, we will by abuse of notation denote this extension alsoby | • |.Throughout this talk n denotes a natural number. Further ζi denotes an indeterminate for anypositive integer i ∈ N>0 = {1, 2, . . .} and by Tn := K 〈ζ1, . . . , ζn〉 ⊆ K Jζ1, . . . , ζnK we mean theTate algebra of strictly convergent power series, that means the K-algebra of power series thatconverge on Bn(K) :=

{(xi)i ∈ K

n∣∣∣ |xi| ≤ 1 ∀i

}.

On Tn there is the Gauß norm which will be denoted by ‖•‖.Any ring R is assumed to be commutative and unitary. The set of prime ideals of a commutativering R is denoted by Spec(R) while MaxSpec(R) denotes the set of maximal ideals.

I RecollectionThis section is a recollection of facts we have already seen and will need during this talk.

Definition 1.1. A strictly convergent power series g =∑∞i=0 giζ

in ∈ Tn with gi ∈ Tn−1 is called

ζn-distinguished of order s ∈ N iff gs ∈ T ∗n−1, ‖gs‖ = ‖g‖ and ‖gs‖ > ‖gi‖ for all i > s.

Lemma 1.2. Any non-zero element f ∈ Tn can be assumed to be ζn-distinguished of someorder s ∈ N. More precisely: For any finitely many non-zero elements f1, . . . , fr ∈ Tn there is acontinuous automorphism σ : Tn

∼−→ Tn such that σ(fi) are ζn-distinguished and ‖σ(f)‖ = ‖f‖for any f ∈ Tn.

Pr∞f: See [Bos08, Lemma 1.2.7]. 2

Theorem 1.3 (Weierstraß Division). Let g ∈ Tn be ζn-distinguished of order s. Then forany f ∈ Tn there exist unique determined q ∈ Tn, r ∈ Tn−1[ζn] with deg(r) < s and f = qg + r.Furthermore we have ‖f‖ = max{‖q‖ · ‖g‖ , ‖r‖}.

Pr∞f: See [Bos08, Thm 1.2.8]. 2

Definition 1.4. A Weierstraß polynomial in ζn is a monic polynomial ω ∈ Tn−1[ζn] with‖ω‖ = 1.

Remark 1.5. A Weierstraß polynomial ω ∈ Tn−1[ζn] is ζn-distinguished of order deg(ω).

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Lemma 1.6 (Weierstraß Preparation Theorem). Let g ∈ Tn be ζn-distinguished of order s.Then g is associated to a Weierstraß polynomial ω ∈ Tn−1[ζn] with deg(ω) = s.

Pr∞f: See [Bos08, Corollary 1.2.9]. 2

Lemma 1.7. The map Bn(K) −� MaxSpec(Tn), x 7−→ mx :={f ∈ Tn

∣∣ f(x) = 0}is sur-

jective.

Pr∞f: See [Bos08, Corr. 1.2.12]. 2

Lemma 1.8 (Maximum Principle). For any f ∈ Tn we have |f(x)| ≤ ‖f‖ for all x ∈ Bn(K)and further there exists an x ∈ Bn(K) with |f(x)| = ‖f‖.

Pr∞f: See [Bos08, Prop 1.2.5]. 2

Lemma 1.9 (Noether Normalization). Let a ( Tn be an ideal. Then for d := dim(Tn/a)there exists a monomorphism Td ↪−→ Tn such that Td ↪−→ Tn −� Tn/a is an finite monomor-phism.

Pr∞f: See [Bos08, Corr 1.2.10]. 2

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II Some ring-theoretic properties of Tate algebrasIn this section we study the Tate algebra Tn from a ring-theoretic view. We will see things likeit is Noetherian and Jacobson.

Definition 2.1. A ring R is called a Noetherian ring if every ideal a ⊆ R is finitely generated.

Proposition 2.2. The Tate algebra Tn is Noetherian.

Pr∞f: By means of induction we may assume that Tn−1 is Noetherian (note that T0 = K

is Noetherian). Let (0) 6= a ⊆ Tn be some ideal and 0 6= g ∈ a an element of a, which byLemma 1.2 we may consider to be ζn-distinguished. So by Theorem 1.3 the Tn−1-moduleTn/(g) is finitely generated (in fact every x ∈ a is equivalent modulo g to some r ∈ Tn−1[ζn]of degree less then s where s denotes the order of the ζn-distinction of g) and hence (sinceTn−1 is Noetherian by induction hypothesis) Noetherian. As a/(g) ⊆ Tn/(g) is a sub-modulewe have that a/(g) is finitely generated as Tn−1-module and thus a ⊆ Tn is finitely generatedas Tn-module: In fact let {x1, . . . , xr} ⊆ a/(g) be a system of generators of a/(g), then{x1, . . . , xr, g} is a generating system of a. To see this let x ∈ a be arbitrary chosen. Thenx ≡

∑i αixi (mod g) for some αi ∈ Tn−1 and thus x−

∑i αixi = `g for some ` ∈ Tn. 2

Lemma 2.3. Let R be a factorial domain. Then R is normal (that means integrally closed inQuot(R)).

Pr∞f: Let x ∈ Quot(R) be integral over R. Let y, z ∈ R be co-prime with yz = x. By

assumption there are elements r0, . . . , rn−1 such that

(y

z

)n+ rn−1

(y

z

)n−1+ . . .+ r0 = 0 ·zn

=⇒ yn + rn−1zyn−1 + . . .+ r0z

n = 0.

This gives the equation yn = −rn−1zyn−1 − . . . − r0z

n yielding that z divides yn. So anyprime divisor of z is also a prime divisor of y and since y and z are assumed to be co-primez is unit, hence x ∈ R. 2

Proposition 2.4. The Tate algebra Tn is factorial and hence normal.

Pr∞f: Due to Lemma 2.3 we only have to show the factoriality. Again we may proceed byinduction as T0 = K is factorial. So we may assume Tn−1 is factorial and hence by Lemmaof Gauß the polynomial ring Tn−1[ζn] is factorial. Let f ∈ Tn be neither a unit nor zero.By Lemma 1.2 we again my assume f to be ζn-distinguished and by Lemma 1.6 we furtherassume f to be a Weierstraß polynomial. So there is a factorization f = ω1 · · ·ωr with primeelements ωi ∈ Tn−1[ζn].As f is monic, we may assume the same for ωi yielding ‖ωi‖ ≥ 1 for all i. On the other handsince 1 = ‖f‖ = ‖

∏i ωi‖ =

∏i ‖ωi‖ we have ‖ωi‖ = 1 for all i, so all the ωi are Weierstraß

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polynomials.Anyway it remains to verify that the ωi ∈ Tn−1[ζn] ⊆ Tn are prime in Tn.This is true because for any Weierstraß polynomial ω ∈ Tn−1[ζn] with deg(ω) = s the canon-ical morphism Tn−1[ζn]/(ω) −→ Tn/(ω) is an isomorphism.This is true since both – the domain and the co-domain – are Tn−1-modules with basis{ζ0n, . . . , ζ

s−1n } by Euclid’s division and Weierstaß’ division (Theorem 1.3). 2

Definition 2.5. A commutative ring R is called Jacobson iff every prime ideal is an inter-section of maximal ideals.

Example 2.6. (i) Every field (or zero-dimensional ring) is Jacobson as every prime ideal ismaximal.

(ii) Every principal ideal domain R with infinitely many maximal ideals is Jacobson (asdim(R) = 1 the only thing to show is that (0) is an intersection of maximal ideals, but theintersection of all maximal ideals equals (0) as there are infinitely many of them).

(iii) A local domain R with dim(R) > 0 is not Jacobson as the only maximal ideal does notequal (0).

Proposition 2.7. The Tate algebra Tn is Jacobson.

Pr∞f: (Case 1) We claim that Spec(Tn) 3 (0) =⋂

m∈MaxSpec(Tn) m. (Clear for n = 0).For this let f ∈

⋂mm. To any maximal ideal m there corresponds by Lemma 1.7 an x ∈ Bn(K)

such that f ∈ m is equivalent to f(x) = 0. So f ∈⋂

mm is equivalent to f(x) = 0 for allx ∈ Bn(K). But then |f(x)| = 0 for all x ∈ Bn(K) and hence Lemma 1.8 implies that‖f‖ = 0, so f = 0 and we are through.

(Case 2) If p ∈ Spec(Tn) is a maximal ideal, there is nothing to show.(Case 3) Let (0) 6= p ∈ Spec(Tn) be a non-zero prime ideal that is not maximal. The aim

is to show that Jac(Tn/p) :=⋂

MaxSpec(Tn/p) equals the zero ideal. Then by the one-to-one order preserving correspondence between prime ideals of Tn/p and prime ideals of Tncontaining p yields p =

⋂ {m ∈ MaxSpec(Tn)

∣∣ p ⊆ m}and thus the claim.

As p is not maximal Tn/p is an integral domain that is not a field thus d := dim(Tn/p) ≥ 1.By Lemma 1.9 there is a finite morphism ι : Td ↪−→ Tn/p.By the Cohen-Seidenberg theorems and those consequences we know that for every maximalideal m ∈ MaxSpec(Td) there is some maximal idealM∈ MaxSpec(Tn/p) with ι−1(M) = m.So it follows that

Td ∩ Jac(Tn/p) = ι−1(Jac(Tn/p)) =⋂Mι−1(M)

see above⊆

⋂m∈MaxSpec(Td)

m = (0). (∗)

The last equation follows like in the proof of the first case (except d = 0 but then the equationis clear as T0 = K).

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Now choose f ∈ Jac(Tn/p) arbitrarily. As Tn/p is finite over Td there is a polynomialp = Xr + ar−1X

r−1 + . . .+ a0 ∈ Td[X] such that p(f) = 0 and p is of minimal degree amongthe polynomials that vanish for f . So we get:

p(f) = 0 =⇒ a0︸︷︷︸∈Td

= −f r − . . .− a1f︸︷︷︸∈Jac(Tn/p)

=⇒ a0 ∈ Td ∩ Jac(Tn/p) (∗)= (0).

So p = Xr + ar−1Xr−1 + . . . + a1X = X(Xr−1 + ar−1X

r−2 + . . . + a1) =: X · p. Asdeg(p) < deg(p) we have by the minimality of p that p(f) 6= 0.But now in the integral domain Tn/p we have 0 = p(f) = f · p(f) hence f = 0.Since f ∈ Jac(Tn/p) was arbitrary chosen we have Jac(Tn/p) = (0) as desired. 2

Daniel Heiß page 5


III Normed ringsIn this section we will develop some theory concerning rings with a ring norm.

Recollection 3.1. Let K be a field with absolute value | • |. Then – as one easily derivesfrom the properties of absolute values – R :=

{x ∈ K

∣∣ |x| ≤ 1}⊆ K is a subring of K,

the valuation ring. It is a local domain with maximal ideal m :={x ∈ R

∣∣ |x| < 1}: That

m ⊆ R is indeed an ideal follows from the properties of absolute values. That it is maximalfollows from the fact that R \ m = R∗ (in fact: Let x ∈ R \ m, meaning |x| = 1. So observe1 = |1| = |xx−1| = |x| · |x−1|, hence |x−1| = 1 and x−1 ∈ R).To see that it is a domain let x, y ∈ R with xy = 0. Then 0 = |0| = |xy| = |x| · |y| in R, so wemay assume |x| = 0 yielding x = 0.

Definition 3.2. Let R be a ring. A ring norm on R is a map | • | : R −→ R≥0 such that forall a, b ∈ R we have

(i) |a| = 0 ⇐⇒ a = 0,

(ii) |ab| ≤ |a| · |b|,

(iii) |a+ b| ≤ max{|a|, |b|},

(iv) |1| ≤ 1.

The norm is called multiplicative if there holds equality in (ii).

Remark 3.3. If R is non-zero and | • | is a ring norm on R, then we have |1| = 1.

Pr∞f: Let r := |1|. Then r = |1 · 1| ≤ |1| · |1| = r2 and by (iv) we have r ∈ [0, 1]. But ofcourse x2 < x for any x ∈ (0, 1), so we have r ∈ {0, 1} and r = r2, but since 1 6= 0 as R 6= {0}it follows r = 1 as claimed. 2

Remark 3.4. Every non-zero ring R with an multiplicative norm | • | is a domain. This canbe seen as follows: Let x, y ∈ R such that xy = 0. Then 0 = |0| = |xy| = |x| · |y| holds in R, sowe have |x| = 0 or |y| = 0 yielding that either x = 0 or y = 0.

Example 3.5. Let K be a field with absolute value | • |. Then the absolute value induces amultiplicative ring norm on the valuation ring R of (K, | • |).

Definition 3.6. Let R be a ring with multiplicative ring norm | • | such that |x| ≤ 1 for allx ∈ R. Then

• R is called a B-ring, if{x ∈ R

∣∣ |x| = 1}⊆ R∗,

• R is called bald if SR := sup{|x|

∣∣∣ x ∈ R with |x| < 1}< 1.

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Example 3.7. The valuation ring of a valuated field is a B-ring. If it is a discrete absolutevalue, the corresponding discrete valuation ring is a bald B-ring.

Example 3.8. (i) Let p ∈ Z be some prime number and | • | be some absolute value on Fp.Then | • | is trivial, hence (Fp, | • |) is bald.

(ii) Let (R, | • |) be a multiplicatively normed ring with |x| ≤ 1 for all x ∈ R and containing Zas a subring. Then (Z, | • |) is bald.

Pr∞f: (i) It is well-known that the group of units F∗p is cyclic. Let x ∈ F∗p be a primitiveelement. Now 1 = |1| = |xp−1| = |x|p−1.So the real number |x| fulfills |x|p−1 = 1, hence we necessarily have |x| = 1. Now let0 6= y ∈ Fp be an arbitrary element. Then (for some r ∈ N) we have |y| = |xr| = |x|r = 1r = 1.This shows that | • | is trivial. It follows SFp = 0 < 1 and hence the claim.

(ii) Let I :={x ∈ Z

∣∣ |x| < 1}. One immediately verifies that I ⊆ Z is an ideal. Since Z is

a PID, there exists a ∈ Z with I = (a).Now let x ∈ I be some element. Then x = k · a with k ∈ Z, so |x| = |k| · |a|. Now |k| ≤ 1 ask ∈ Z ⊆ R and it follows |x| ≤ 1 · |a|, so we have SZ ≤ |a| < 1 (by definition of a). So Z isbald as claimed. 2

Lemma 3.9. Let R be the valuation ring of (K, | • |) and T ⊆ R be some bald subring. Thenthe set S :=

{x ∈ T

∣∣ |x| = 1}⊆ T is multiplicatively closed and the localization S−1T is a

bald B-ring.

Pr∞f: By Remark 3.3 we have 1 ∈ S and since |xy| = |x| · |y| the first assertion follows.By construction we obtain that S−1T is a B-ring, so the only thing left to show is that thebaldness survives localization.Now let x = y

z ∈ S−1T be some element with |x| < 1. Observe |x| = |yz−1| = |y| · |z−1| = |y|,

but y ∈ T , so we have |x| = |y| ≤ ST < 1 (since T is bald). Since x was arbitrarily chosenwe have the baldness of S−1T .Note that |z−1| = 1 since 1 = |1| = |zz−1| = |z| · |z−1| and |z| = 1 as z ∈ S. 2

Remark. Note that in the situation above T is bald iff S−1T is bald. (analogue proof).

Lemma 3.10. Let (R, | • |) be some bald ring, then the completion R is bald. If further R wasa B-ring then R is also a B-ring.

Pr∞f: (Baldness) Let x ∈ R be some element with |x| < 1. As R ↪−→ R is dense there isa sequence (xi)i∈N of elements in R such that xi

i→∞−→ x. As |x| < 1 the convergence suppliessome i0 ∈ N such that for all i ≥ i0 we have |xi| < 1 and therefore |xi| ≤ SR < 1 (as R isbald). Now the continuity of | • | yields |x| = | limi xi| = limi |xi|︸︷︷︸

≤ε

≤ ε and the claim follows.

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(B-ring) Let x ∈ R be some element with |x| = 1. As R ↪−→ R is dense there is a sequence(xi)i of elements in R that converge to x. Since all elements r ∈ R with |r| < 1 fulfill|r| ≤ SR < 1 this convergence implies that there is some i0 such that |xi| = 1 for all i ≥ i0.But as R is a B-ring we have xi ∈ R∗ for all these xi, yielding x ∈ (R)∗. 2

Lemma 3.11. Let R be an integral domain. Then the smallest subring S ⊆ R equals eitherZ or Z/pZ for some prime number p ∈ Z.

Pr∞f: It is well known that there always exists a ring-homomorphism ϕ : Z −→ R. Its kernelis an ideal in Z and hence generated by some element x ∈ Z.Since the image im(ϕ) ⊆ R is a subring and R is an integral domain, it follows that alsoim(ϕ) ∼= Z/ ker(ϕ) = Z/(x) is an integral domain, hence x ∈ Z is prime or x = 0. Anywaywe obtain an injective homomorphism Z/(x) ↪−→ R and the claim follows. 2

Proposition 3.12. Let K be a field with absolute value | • | and valuation ring R. Let furtherS ⊆ R be some bald subring and a ∈ R with |a| = 1. Then S[a] ⊆ R is a bald subring.

Pr∞f: By localization with respect to{x ∈ S

∣∣ |x| = 1}⊆ S we may assume that S is a

B-ring (cf. Lemma 3.9). Let m ⊆ S denote the maximal ideal of elements with | • | < 1 andT := S/m the residue class field. Further letM ⊆ R be the maximal ideal of the valuationring R and k := R/M be its residue class field.There are two cases: Either the residue class a ∈ k is algebraic over T or transcendental.

(Transcendental case) Let x =∑ri=0 cia

i ∈ S[a] be some element with |x| < 1. Hence in Twe have 0 = x =

∑i ci · ai. But since a is transcendental over T it follows ci = 0 and hence

|ci| < 1 for all i. So we have |x| =∣∣∑

i ciai∣∣ ≤ max

{|ci| |a|i︸︷︷︸

=1

}= max{|ci|} ≤ SS .

Since x ∈ S[a] was arbitrary chosen we have SS[a] ≤ SS < 1, so S[a] is bald.(Algebraic case) Again let x =

∑ri=0 cia

i ∈ S[a] be some element with |x| < 1. Setf :=

∑ri=0 ciX

i ∈ S[X], so we have f(a) = x.Let g ∈ S[X] be some monic polynomial with minimal degree whose reduction is a non-zeropolynomial g ∈ T [X] that annihilates a ∈ k.As 0 = g(a) = g(a) we have |g(a)| < 1. Now define ε := max{SS , |g(a)|} < 1. We claim that|x| ≤ ε and by this we are through.Euclid’s division supplies q, r ∈ S[X] with f = qg + r and m := deg(r) < deg(g). If|(qg)(a)| ≥ |r(a)| we have |f(a)| ≤ |(qg)(a)| = |q(a)|︸︷︷︸

≤1

· |g(a)|︸︷︷︸≤ε

≤ ε and this was the claim (notethat |q(a)| ≤ 1 as |a| = 1 and q ∈ S[X]).If |r(a)| > |(qg)(a)| we have |x| = |f(a)| = |r(a)| so we need to show that |r(a)| ≤ ε.Let r =

∑mi=0 diX

i with di ∈ S. If in the first case we have |di| < 1 for all i then|r(a)| =

∣∣∑i dia

i∣∣ ≤ max{|di|} < 1 and we are through. On the other hand if there are di

with |di| = 1 then in T [X] we have r 6= 0. But this contradicts the construction of g since

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deg(r) < deg(g) and still r(a) = 0 since 0 = f(a) = q(a) · g(a)︸︷︷︸=0

+r(a) = r(a). 2

Proposition 3.13. Let K be a field with a absolute value and valuation ring R. Let further(ai)i∈N be a sequence in R with limn→∞ |an| = 0. Then the smallest subring S ⊆ R containingall the ai is bald.

Pr∞f: By Lemma 3.11 the smallest subring T of the integral domain (Recollection 3.1) R iseither Z or Z/pZ for p ∈ Z prime. In both cases T is bald (cf. Example 3.8).Since limn |an| = 0 there are only finitely many ai with |ai| = 1.Substitute T by T

[ai∣∣∣ |ai| = 1

]. By Proposition 3.12 this is still a bald ring. So we may

assume that there is an ε < 1 such that |ai| ≤ ε for all i ∈ N.Now let x ∈ T [ai

∣∣ i ∈ N], then x =∑ν∈N` cνa

ν11 · · · a

ν`` . Then |x| ≤ max{|cν | · |aν1

1 | · · · |aν`` |} ≤

max{|cν | · ε|ν|}.As |cν | ≤ 1 for any ν ∈ N` (since cν ∈ T ) we have |cν | · ε|ν| ≤ ε|ν| < 1 unless ν = 0 but in thiscase we have |x| = |cν | ≤ ST < 1 since |x| < 1. 2

Remark 3.14. In the situation of Proposition 3.13 we can localize S at{x ∈ S

∣∣ |x| = 1}to

obtain a bald B-ring. If further R is complete, we can pass from S to its completion S toobtain some complete bald B-ring.

Pr∞f: Apply Lemma 3.9 and Lemma 3.10. 2

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IV Normed vector spacesTate algebras over K are both, rings and K-vector spaces. After we have studied normed rings,we proceed with a closer look at normed vector spaces during this section. Both will lead us tosome theory for ideals of the Tate algebra Tn in the next section.

Definition 4.1. Let V be a K-vector space. A norm on V is a map ‖•‖ : V −→ R≥0 suchthat for any x, y ∈ V and c ∈ K

(i) ‖x‖ = 0 ⇐⇒ x = 0,

(ii) ‖x+ y‖ ≤ max{‖x‖ , ‖y‖},

(iii) ‖cx‖ = |c| · ‖x‖.

Example 4.2. If we forget about multiplication in the Tate algebra Tn we obtain a normedK-vector space (with the Gauß norm).

Definition 4.3. Let V be a complete normed K-vector space. A system (xn)n∈N of elementsin V (with N at most countable) is called a topological orthonormal basis of V if thefollowing conditions are satisfied:

(i) ‖xn‖ = 1 for all n ∈ N ,

(ii) For each x ∈ V there are elements (cn)n∈N in K such that x =∑n∈N cnxn,

(iii) For each equation x =∑n∈N cnxn we have ‖x‖ = maxn |cn|.

Remark 4.4. In the Definition 4.3 (iii) the maximum maxn |cn| is well-defined, because as∑n cnxn converges we necessarily have limn |cn| = 0. Furthermore this condition implies

that the coefficients cn are unique: In fact let x =∑n cnxn and x =

∑n dnxn. Then we have

0 = x−x =∑n(cn−dn)xn, so 0 = ‖0‖ = ‖

∑n(cn − dn)xn‖ = maxn |cn−dn| yielding |cn−dn| = 0

for all n ∈ N and hence cn = dn for all n.

Example 4.5. The monomials (ζν11 · · · ζνn

n )ν∈Nn form an orthonormal basis of the K-vectorspace Tn with the Gauß norm. This follows immediately from the definition of the Gauß norm.

Definition 4.6. Let V be a normed K-vector space. Then we set V ◦ :={x ∈ V

∣∣ ‖x‖ ≤ 1}

and V := V ◦/ {

x ∈ V∣∣ ‖x‖ < 1

}.

Theorem 4.7. Let V be a complete normed K-vector space with a given orthonormal basis(xn)n∈N . Let further (R,m) denote the valuation ring of K and k := R/m the residue class field.Now consider a system of elements ym =

∑n∈N cmnxn ∈ V ◦ (with m ∈ M) where the smallest

subring of R containing all the cmn is bald. Assume that the residue classes ym ∈ V form ak-basis of V , then the elements ym form an orthogonal basis of V .

Daniel Heiß page 10


Pr∞f: We prove this statement for the case where R is a discrete valuation ring and cmn ∈ Rfor all m ∈M,n ∈ N . For the more general assertion see [Bos08, Thm 1.3.6].Let m = (π). We want to show that every element v ∈ V can be written uniquely asv =

∑m∈M λmym (note that xn clearly form an orthonormal basis of V , so the cardinality of

N and M coincides yielding that M is at most countable!) with λm ∈ R and limm λm = 0.(Uniqueness) Let

∑m λmym =

∑m µmym, then of course

∑m λmym =

∑m µmym and since

(ym) form an orthonormal basis of V , Remark 4.4 yields λm = µm, but this does not yieldλm = µm, we only obtain λm − µm ∈ m = (π).

(Existence) Let v ∈ V be arbitrary chosen. If v = 0 the assertion is clear, so assume v 6= 0.It suffices to show the assertion for elements v ∈ V ◦ as any element v ∈ V yields an element

1‖v‖v ∈ V

◦.As we assumed v 6= 0 there exists some integer s1 ≥ 0 such that v ∈ πs1V ◦ \ πs1+1V ◦ (notethat clearly V ◦ is an R-module!).Then the residue of π−s1v can be written as a finite sum π−s1v =

∑i λiyi with λi ∈ k.

That means there are elements n1 :=∑i λiyi ∈

⊕nRyn and m1 ∈ πs1+1V ◦ such that

v = πs1n1 +m1.Now either m1 = 0 and we are through as v is then the finite sum v =

∑i π

s1λiyi, orm1 6= 0, but then by the same arguments there exist a finite sum n2 =

∑i λiyi ∈

⊕nRyn

and m2 ∈ πs2+1V ◦ such that m1 = πs2n2 +m2 for some s2 > s1.If this procedure ends at some finite step, then we are through (v is then a finite sum of finitesums ni). If on the other hand this process does not stop, then we obtain v =

∑∞i=1 π

sini =∑i π

s1∑j λijyi =

∑i

(∑j π

siλij)yi where limi

(∑j π

siλij)

= 0 as claimed. 2



V Ideals in the Tate algebraIn this section we can apply this technical Theorem 4.7 to ideals in a Tate algebra.

Corollary 5.1. Let a ⊆ Tn be some ideal. Then there are elements a1, . . . , ar ∈ a such that

(i) ‖ai‖ = 1 for all 1 ≤ i ≤ r,

(ii) for each f ∈ a there are elements f1, . . . , fr ∈ Tn with f =∑i fiai and ‖fi‖ ≤ ‖f‖.

Pr∞f: Let (R,m) be the valuation ring of (K, |•|). Then consider the reduction a of a, mean-ing the image of a ∩ R 〈ζ1, . . . , ζn〉 under the reduction map R 〈ζ1, . . . , ζn〉 −� k[ζ1, . . . , ζn](note that the co-domain is – in fact – a polynomial ring, as all elements of the domainconverge, so the coefficients form a zero sequence, hence the reduction modulo m vanishes atsome point).As k[ζ1, . . . , ζn] is Noetherian there are elements a1, . . . , ar ∈ a with ‖ai‖ = 1 for all i suchthat (ai)1≤i≤r generate the ideal a ⊆ k[ζ1, . . . , ζn]. So the k-vector space a is generated by thesystem of elements (ζν1

1 · · · ζνnn ai)1≤i≤r, ν∈Nn . This supplies a system of elements (ym)m∈M ′ of

the form ζν11 · · · ζνn

n ai for ν ∈ Nn and 1 ≤ i ≤ r (note that M ′ is countable!) such that theirreduction ym form a k-basis of a. Now enlarge the system (ym)m∈M ′ by some elements ofthe form ζν1

1 · · · ζνnn for some ν ∈ Nn to a system (ym)m∈M such that their reduction form

a k-basis of k[ζ1, . . . , ζn]. As (by Example 4.5) the system (ζν11 · · · ζνn

n )ν∈Nn is an orthogonalbasis of Tn we deduce that M is still countable.To write the ym as a linear combination of the orthogonal basis (ζν1

1 · · · ζνnn )ν∈Nn we only need

the coefficients a1, . . . , ar and those clearly form a zero sequence, so the smallest subring of Rcontaining the ai is bald by Proposition 3.13. Therefore Theorem 4.7 supplies that (ym)m∈Mis an orthogonal basis of Tn.

We now show that the elements a1, . . . , ar are as claimed: Let f ∈ a be arbitrary cho-sen. Then – as (ym)m∈M is an orthogonal basis of Tn – we have f =

∑m∈M cmym with

cm ∈ K and of course |cm| ≤ ‖f‖ (by definition of the Gauss norm). Now observe the ele-ment f ′ :=

∑m∈M ′ cmym. By construction of the elements ym for m ∈M ′ we can also write

f ′ =∑ri=1 fiai for some fi ∈ Tn with ‖fi‖ ≤ ‖f‖. Especially we see f ′ ∈ a and hence we are

through if we can prove f = f ′. To do this observe f − f ′ =∑m∈M\M ′ cmym ∈ a.

If there is some µ ∈ M \M ′ such that cµ 6= 0, then c−1µ (f − f ′) ∈ a (as a ⊆ Tn is an ideal)

and the reduction of that element is non-zero contradicting the construction of (ym)m∈M ′ ask-basis of a. 2

Remark 5.2. The proof of Corollary 5.1 shows more precisely:

(i) Every collection of elements a1, . . . , ar with ‖ai‖ ≤ 1 satisfy the assertion of Corollary 5.1as soon as their reduction ai generate the ideal a ⊆ k[ζ1, . . . , ζn].

(ii) The sub-system of elements (ym)m∈M ′ of the orthonormal basis (ym)m∈M is an orthonormalbasis of a ⊆ Tn.



(iii) Any convergent series∑m∈M ′ cmym with cm ∈ K give rise to an element of a.

Lemma 5.3. Let Y be some complete space with non-archimedian absolute value | • | andX ⊆ Y a subspace. Then X is complete if and only if every convergent series

∑i∈N xi (in Y )

with xi ∈ X converges in X.

Pr∞f: (⇐) This is clear as the sequence of partial sums forms a Cauchy sequence.(⇒) Let (xi)i ∈ X be a Cauchy sequence. Let y0 := x0 and yi := xi − xi−1 for all i ≥ 1.

Then we clearly have xn =∑ni=0 yi

n→∞−→ x for some x ∈ X. Note that the series converges inY as (yi)i form a zero sequence and | • | is non-archimedian. 2

Corollary 5.4. Let a ⊆ Tn be some ideal. Then a is complete and thus a ⊆ Tn is closed.

Pr∞f: Choose some generators a1, . . . , ar ∈ a like in the assertion of Corollary 5.1. Now letf =

∑n∈N fn be some series that converges in Tn (note that Tn is complete) with elements

fn ∈ a. Now by Corollary 5.1 we have fn =∑ri=1 fn,iai with fn,i ∈ Tn with ‖fn,i‖ ≤ ‖fn‖.

But now we have f =∑n∈N fn =

∑ri=1 (

∑n∈N fn,i) ai ∈ a as claimed (note that fn is a

zero sequence since∑n fn converges, so as ‖fn,i‖ ≤ ‖fn‖ also fn,i is a zero sequence and

everything is fine). 2

Definition 5.5. Let R be a normed ring and a ⊆ R be some ideal. Then a ⊆ R is strictlyclosed if for all f ∈ R there is some a0 ∈ a such that ‖f − a0‖ = infa∈a ‖f − a‖.

Corollary 5.6. Let a ⊆ Tn be some ideal. Then a ⊆ Tn is strictly closed.

Pr∞f: Let f ∈ Tn be some arbitrary chosen element. Now in the situation of the proof ofCorollary 5.1 we use the orthonormal basis (ym)m∈M to write f =

∑m∈M cmym for some

cm ∈ K. Now consider a0 :=∑m∈M ′ cmym and we claim that this a0 is as claimed. First we

have a0 ∈ a (cf. Remark 5.2 (iii)).Now let a ∈ a be arbitrary chosen. Then – as (ym)m∈M ′ ⊆ a is an orthonormal basis (cf.Remark 5.2) – we have a =

∑m∈M ′ dmym for some dm ∈ K. Observe

‖f − a‖ = max{

maxm∈M ′

{|cm − dm|}, maxm∈M\M ′

{|cm|}}≥ max

m∈M\M ′{|cm|} = ‖f − a0‖ .

So the assertion follows as a ∈ a was arbitrarily chosen. 2

Lemma 5.7. Let (Xi,dXi) (for 1 ≤ i ≤ n) be some complete metric spaces. Then X :=∏ni=1Xi is complete with respect to the maximum metric dX .

Pr∞f: Let (xk)k∈N ⊆ X be some Cauchy sequence. Write xk = (x1k, . . . , x

nk) (with xik ∈ Xi).

Then for any 1 ≤ i ≤ n the sequence (xik)k∈N ⊆ Xi is a Cauchy sequence. In fact: Letε > 0, then there is some N ∈ N such that for all k, ` ≥ N we have ε > dX(xk, x`) =



max1≤i≤n dXi(xik, xi`) ≥ dXi(xik, xi`).Now Xi is complete by assumption, so there is some yi ∈ Xi such that xik

k→∞−→ yi. Settingy := (y1, . . . , yn) we claim limk→∞ xk = y. To see this, let ε > 0. Then for all 1 ≤ i ≤ n thereexists Ni ∈ N such that dXi(xik, yi) ≤ ε for all k ≥ Ni. Now set N := maxiNi and observethat for all k ≥ N we have dX(xk, y) = maxi dXi(xik, yi)︸︷︷︸

≤ε

≤ ε yielding the claim. 2

Eventually we generalize Corollary 5.1 for modules instead of ideals:

Corollary 5.8. Let N ⊆ T sn be a Tn-sub-module of the direct sum of finitely many copies ofTn and let ‖•‖ denote the maximum norm on T sn induced by the Gauss norm on Tn. Then thereare elements x1, . . . , xn ∈ N that generate N as Tn-module and satisfy

(i) ‖xi‖ = 1 for all i,

(ii) For each x ∈ N there are elements f1, . . . , fr ∈ Tn such that x =∑ri=1 fixi with ‖fi‖ ≤ ‖x‖.

Pr∞f: All in all this proof resembles those of Corollary 5.1. Let R denote the valuation ring ofK and consider the reduction map (R 〈ζ1, . . . , ζn〉)s −� (k[ζ1, . . . , ζn])s. Let further N denotethe image of N ∩ (R 〈ζ1, . . . , ζn〉)s under this map. Then we see that N is a k[ζ1, . . . , ζn]-sub-module of (k[ζ1, . . . , ζn])s and hence – since the latter is Noetherian – it is finitely generated.So there are elements x1, . . . , xr ∈ N with ‖xi‖ = 1 such that their reduction xi generateN as k[ζ1, . . . , ζn]-module. Therefore we can choose a system of elements (ym)m∈M ′ of theform ζν1

1 · · · ζνnn xi for some ν ∈ Nn and i ∈ {1, . . . , r} such that the reduction (ym) of these

elements form a k-basis of the k-vector space N .Now let ei denote the unit vectors in T sn. Then we can enlarge the system (ym)m∈M ′ to asystem (ym)m∈M by adding some elements of the form ζν1

1 · · · ζνnn ei for ν ∈ Nn and 1 ≤ i ≤ s

so that the reduction of this system forms a k-basis of (k[ζ1, . . . , ζn])s.On the other hand we have an orthonormal basis, namely Z := (ζν1

1 · · · ζνnn ei)1≤i≤s, ν∈Nn and

this basis induces a k-basis of (k[ζ1, . . . , ζn])s.To write the x1, . . . , xr in terms of Z using converging linear combinations with coefficientsin K we need finitely many zero sequences in K. They can be fused into a single zerosequence and hence the smallest subring R′ ⊆ R of R containing these elements is bald byProposition 3.13.Now we obtain the elements ym (withm ∈M ′) by multiplication of the xi with some elementsζν1

1 · · · ζνnn (with ν ∈ Nn), so we have ym ∈

∑z∈ZR

′z for all m ∈ M , hence by Theorem 4.7the (ym)m∈M form an orthonormal basis and the rest of the assertion follows as in the proofof Corollary 5.1. 2



References[Bos08] Siegfried Bosch. Lectures on Formal and Rigid Geometry. Lecture notes, University of

Münster, 2008.

Daniel Heiß page ii

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Non-Archimedian analysished29573/vort/rigid.pdf · 2015. 8. 3. · Non-Archimedian analysis University of Ratisbon Winter term 2014/15 Daniel Heiß: §4: Tatealgebras October 30th,