1. Ultrametric fields

Transcription

1 1. Ultrametric fields This first chapter is aimed at recalling basic definitions and properties on ultrametric fields. We will examine ultrametric absolute values, valuation rings and residue fields. Particularly, we will define holes of a subset and infraconnected sets that are essential with regards to the behaviour of analytic functions (certain authors improperly call such sets connected sets which makes no sens in topology since there are no connected sets except singletons in an ultrametric field). A major interest of the class of infraconnected sets (among others) is that this is the biggest class of sets of an ultrametric complete algebraically closed field where the famous Krasner Mittag-Leffler theorem. Definitions and notations: Throughout the book, we denote by N the set of integers 0, by Z the ring of relative integers, by Q the field of rational numbers, by R the field of real numbers and by C the field of complex numbers. Given a topological space T and a subset S of T,wedenotebyS its closure (also called adherence) and by S its interior (also called opening). Let E be a field provided with an absolute value. and let log be a real logarithm function of basis θ>1. We call valuation associated to that absolute value the mapping v from E to R defined as v(x) = log x and here, we set Ψ(x) = log x and E = { x x E}. If a set F contains the zero of a ring, we denote by F the set F \{0}. An absolute value is said to be trivial if x =1 x E \{0}. Throughout the book, we will denote by L a field complete with respect to a non-trivial ultrametric absolute value and by K an algebraically closed field complete with respect to a non-trivial ultrametric absolute value. We will denote by. the archimedean absolute value defined on R. Lemma 1.1 is immediate: Lemma 1.1: LetEbeafieldprovidedwithanultrametricabsolutevalue.. The completion of E with respect to that absolute value is provided with an ultrametric absolute value which continues that of E. The set { x x E } is a subgroup of the multiplicative group R +. Definition: Given a field E provided with an ultrametric absolute value., the multiplicative group { x x E } is called the value group of E and the additive group {v(x) x E} is called valuation group of E. Similarly, the set {Ψ(x) x E } is a subgroup of R called valuation group of E. The field E is said to have discrete valuation or to have discrete absolute value if its valuation group is a discrete subgroup of R and hence is isomorphic to Z. Else, the valuation group is dense in R and E is said to have dense valuation or to have a dense absolute value.

2 2 Ultrametric fields Lemma 1.2 is classical and proven in the same way no matter what the absolute value of E. Lemma 1.2: Let E be a field provided with two absolute values whose associated valuations are v and w, respectively. They are equivalent if and only if there exists r>0 such that w(x) =rv(x) whenever x E. Proof. If such a r exists, the two absolute values are seen to be equivalent. Reciprocally, we assume them to be equivalent and take a E such that v(a) 0. It is seen that w(a) 0. On the other hand, for all x E and for all m, n N, we have v( xm v(x) ) > 0 if and only if w(xm ) > 0. Therefore, we see that an an v(a) > n m is equivalent to w(x) w(a) > n. Then, since Q is dense in R, wehavev(x) m v(a) = w(x) w(a) whenever x E and therefore w(x) v(x) = w(a) v(a). Notations: The set of the x E such that x 1 will be denoted by U E and the set of the x E such that x < 1 will be denoted by M E. Then Lemma 1.3 is immediate: Lemma 1.3: U E is a local subring of E whose maximal ideal is M E. Definitions and notations: Henceforth U E is called the valuation ring of E. The maximal ideal M E of U E is called the valuation ideal and the field L = U E M E is called the residue class field of E. For any a E, the residue class of a will be denoted by a. The characteristic of E is named the residue characteristic of E and will be denoted by p. Lemma 1.4: Let F be a subfield of E, lete (resp. F) be the residue class field of E (resp. of F ). Then F is a subfield of E. IfE is algebraically closed and if its valuation is not trivial, it is dense. Proof. The first statement is immediate. Next, given α E such that 0 < α < 1andβ E such that β q = α s we have v(β) = q s v(α) whenever s N and q Z. Lemma 1.5: Let V be a L-vector space of finite dimension provided with two norms. Then the two norms are equivalent. Proof. Let. and. be the two norms on V. We proceed by induction on the dimension of V and assume the equivalence true for subspaces of dimension n<q.letv have dimension q. Lete 1,...e q be a base of V. Let us suppose that the two norms are not equivalent on V. Then there exists a sequence (u n ) n N

3 Ultrametric fields 3 q of the form u n = a j,n e j, with u n 1, such that lim u n =0. Let n j=1 S be the subspace of V generated by {e 1,..., e q 1 }. For every n N, we put q 1 v n = a j,n e j. j=1 First, we suppose that (1) lim a q,n =0. n Since lim u n = 0, we have lim v n = 0. By hypothesis, the restrictions n n of the two norms to S are equivalent, hence we have lim v n = 0. But since n u n 1 for all n N, this contradicts (1). Now, since (1) is not true, there exists a subsequence of the sequence ( a q,n ) n N that admits a strictly positive lower bound and therefore, without loss of generality, we can clearly assume that there exists r>0 such that a q,n r for all n N. Let(x n ) n N be the sequence defined as x n = u n. It is seen that a q,n (2) lim x n =0. n The two norms. and. are equivalent on S and they both are equivalent q 1 to the product norm. defined as b j e j = max b j. Since E is 1 j q 1 j=1 complete, S is complete with respect to., hence S is closed in V with respect to the two norms. and.. Hence by (2) e q belongs to S, which is absurd and finishes the proof. Theorem 1.6: Let F be an algebraic extension of L, provided with two absolute values extending the one L. These absolute values are equal. Proof. Let v, w be the valuations associated to these absolute values. Let a F. By Lemma 1.5, the two absolute values are equivalent on L[a]. Hence by lemma 1.2, there exists r>0 such that w(x) =rv(x) whenever x L[a]. But since v(x) =w(x) whenever x L and since there exists u L such that v(u) 0, we have r =1. Lemma 1.7: Let A be a L-algebra. Let φ be a semi-norm of L-algebra satisfying φ(x n )=(φ(x)) n x A. Thenφ is ultrametric. Proof. Let a, b A satisfy φ(a) ϕ(b). We just have to show that φ(a + b) φ(a). Obviously we have φ((a + b) n )=φ ( n Cn k ak b n k). For each k =0,...n we have φ(c k n ak b n k )= C k n φ(ak b n k ) φ(a) k φ(b) n k φ(a) n hence φ((a + b) n ) (n+1)φ(a) n and therefore φ(a+b) n n +1φ(a) for all n N. Finally we obtain φ(a + b) φ(a). k=0

4 4 Ultrametric fields The most classical example of an ultrametric complete algebraically closed field is the field C p that will be described later. Notations: Consider the field E provided with an ultrametric absolute value. Let a E and let r R +.Wedenotebyd(a, r) the disk {x E x a r}, by d(a, r ) the disk {x E x a <r} and we call circle of center a, of radius r the set C(a, r) =d(a, r) \ d(a, r ). Given r 1 and r 2 such that 0 <r 1 <r 2 we denote by Γ(a, r 1,r 2 ) the annulus {x E r 1 < x a <r 2 } and by (a, r 1,r 2 ) the annulus {x E r 1 x a r 2 }. We know that if b d(a, r) thend(b, r) = d(a, r). Inthesamewayif b d(a, r )thend(b, r )=d(a, r ). Moreover given two disks T and T such that T T then either T T or T T. We denote by δ the distance defined on E by δ(a, b) = a b. Givena E and a subset D of E, wesetδ(a, D) = inf{ x a x D} and given two subsets D, F of E, wesetδ(d, F) = inf{ x y x D, y F }. We set diam(d) =sup{ x y x D, y D} and diam(d) isnamedthe diameter of D. Similarly, we set codiam(d) = sup{ x y x D, y / D} and codiam(d) is named the codiameter. Of course the following three statements are seen to be equivalent : i) d(a, r) =d(a, r ) ii) C(a, r) = iii) r/ E Further, the disks d(b, r ) included in C(a, r) (resp.ind(a, r)) are the disks d(b, r ) such that b C(a, r) (resp. ind(a, r)). They are called the classes of C(a, r) (resp.ofd(a, r)). Henceforth D will denote a subset of the field L. The closure of D (also called adherence of D) is denoted by D and the interior of D (also called opening of D) is denoted by D. Given a point a L we put δ(a, D) = inf{ x a x D}. Then δ(a, D) is named the distance of a to D. Given two subsets D, E of L we put δ(d, E) = inf{ x y x D, u E}. Then δ(d, E) is called the distance between D and E. We will denote by L an extension of L provided with an absolute value that extends that of L. Givena L, r > 0, d(a, r) (resp. d(a, r )) will denote the disk {x L x a r} (resp. {x L x a <r} ). Let D be a subset of L, of diameter R R (resp. + ), whose holes form a family ( d(a i,r i )). Let a D. We will denote by D the set d(a, R) \ ( ( i N d(a i,r i )) ) (D \ D) (resp. L ( \ d(a i,r i )). ) i I i I

5 Ultrametric fields 5 Lemma 1.8: Let d(a, r), d(b, s) be disks such that d(a, r) d(b, s) with r s. Thend(a, r) d(b, s). Let us also notice the following basic lemma: Lemma 1.9: Suppose that the residue class field E of E is finite, of cardinal q. Then for every disk d(a, r) with a E and r E, admits only q classes. Lemma 1.10: D \ D admits a unique partition of the form (Ti ) i I, whereas each T i is a disk of the form d(a i,r i ) with r i = δ(a i,d). Proof. For every a D\D, letr(a) =δ(a, D). Let α and β be two points in D\D such that β α <r(α). It is easily seen that for every x D, wehave x β = x α, and then the family of the disks T (α) =d(α, r(α) )(α D \ D) makes a partition of D \ D because given α and β D \ D, either α β < r(α) and then T (α) =T (β), or α β r(α) andthen α β r(β) hence T (α) T (β) =. Definition and notation: Such disks d(a i,r i ) are called the holes of D. If D is bounded of diameter R we denote by D the disk d(a, R) for any a D. If D is not bounded we put D = L. Example 1: The holes of a disk d(a, r ), with r L, are the classes of C(a, r). Example 2: The only one hole of L \ d(0, 1 )isd(0, 1 ). Example 3: The holes of L \ d(0, 1) are the disks d(a, 1 ) with a d(0, 1). Definitions: A set D is said to be infraconnected [51], [63], if for every a D, the mapping I a from D to R + defined by I a (x) = x a has an image whose closure in R + is an interval. In other words, a set D is not infraconnected if and only if there exist a and b D and an annulus Γ(a, r 1,r 2 )with0<r 1 <r 2 < a b such that Γ(a, r 1,r 2 ) D =. Lemma 1.11 is obvious: Lemma 1.11: If D is infraconnected of diameter R R (resp. + ) then I a (D) =[0,R] (resp. I a (D) =[0, + [ ). The following Lemma 1.12 gives a point of view from a hole of D. Lemma 1.12: Let D be an infraconnected set and let α belong to a hole T of diameter ρ. Theclosureoftheset{ x α x D} is an interval whose lower bound is ρ. Proof. We just have to show that for every r and r such that ρ<r<r < diam(d), there exists β D such that r< β α <r. By definition of the holes there exists b D such that α b <rand then, since D is infraconnected, there exists β D such that r< b β <r. But it is seen that β α = b β.

6 6 Ultrametric fields Given two infraconnected sets A and B we may prove A B to be infraconnected in the following two hypothesis (Th and 1.15). Theorem 1.13: Let A and B be two infraconnected sets such that A B. Then A B is infraconnected. Proof. If A and B are not bounded, the statement is obvious because for every a A, I a (A) = R + and for every a B we have I a (B) = R +. Now we may assume A to be bounded, of diameter R, while B has diameter R R (resp. is not bounded). Then A B has diameter R (resp. is not bounded). Let c A B, leta A B and let us show that I a (A B) =[0,R ](resp. [0, + [). For conveniencewe first assume B to be bounded. Since c A B we see that x a max( x c, c a ) R whenever x A B hence I a (A B) [0,R ]. Hence we just have to show that I a (A B) [0,R ]. Obviously I a (A B) = I a (A) I a (B) =[0,R] I a (B). Hence we have to show that I a (B) [R, R ]. But when x B with x a >R,weseethat x a = x c (because c a R) hence I a (B) ]R, R ]=I c (B) ]R, R ] and finally I a (B) [R, R ] because I c (B) [R, R ]. When B is not bounded, in the same way it is seen that I a (A B) =[0, + [. This finishes showing that A B is infraconnected. Corollary 1.14: LetDbeasubsetofL.TherelationR defined by xry if there exists an infraconnected subset of D that contains x and y, is an equivalence relation. Proof. R is obviously reflexive and symmetric It is transitive by Theorem Definition: The equivalence classes with respect to this relation are called the infraconnected componants. Examples : 1) d(0, 1 ) d(1, 1 ) is infraconnected. Its holes are the disks d(α, 1 ) with α = α 1 =1. 2) Let r ]0, 1[ and let D = d(0, 1 ) d(1,r). ThenD is not infraconnected, its infraconnected components are d(0, 1 )andd(1,r). The holes of D are the disks d(α, 1 ) with α = α 1 = 1 and the disks d(α, α 1 ) with r< α 1 < 1. Theorem 1.15: Let A and B be infraconnected sets such that à = B. Then A B is infraconnected. Proof. Obviously à B = Ã. IfA is bounded let à = d(α, R) and otherwise let à = L. First let us assume A to be bounded. For a A, theset{ x a x A} is dense in [0,R] hence so is the set { x a x A B}. InthesamewayB

7 Ultrametric fields 7 plays the same role hence this still holds for a B. Finally if A is not bounded we just replace [0,R]by[0, + [. That finishes proving Theorem Definition: An infraconnected subset D of L is said to be affinoid if it is of q the form d(a, R) \ d(b k,r k ) with R and r k L k. A subset D of L is said k=1 to be affinoid if it is a finite union of infraconnected affinoid subsets. Proposition 1.16: Let D 1, D 2 be two infraconnected affinoid subsets of L such that D 1 D 2 and set D = D 1 D 2 and E = D 1 D 2. Then both D and E are infraconnected affinoid. Moreover, D is either D1 or D 2 and each hole of D is either a hole of D 1 or a hole of D 2. Proof. By Theorem 1.13 D is infraconnected. Consider now D 1,oftheform d(a, r) \ ( m i=1 d(a i,r i )) and D 2 is of the form d(b, s) \ ( n i=1 d(b i,s i )). Suppose for instance r s and let c D 1 D 2. Then we can check that ( ( E = d(c, r) \ d i=1 (a i,r i )) ( n i=1d(b i,s i ))) which is an infraconnected affinoid again. Since D 1 D 2, wehave D 1 D 2, hence D is either D 1 or D 2, hence diam(d) L. Next, since the holes of both sets are in finite number, each hole of D is either a hole of D 1 or a hole of D 2, so each hole of D has a diameter in L and of course they are in finite number. Definition: We will call empty annulus of D an annulus Γ(a, r 1,r 2 ) such that i) r 1 = sup{ x a x D, x a r 2 } ii) r 2 = inf{ x a x D, x a r 1 } The set d(a, r 1 ) D will be denoted by I D (Γ(a, r 1,r 2 )) while the set (L \ d(a, r2 )) D will be denoted by E D(Γ(a, r 1,r 2 )). When there is no risk of confusion about the set D we will just write I(Γ(a, r 1,r 2 )), (resp. E(Γ(a, r 1,r 2 ))), instead of I D (Γ(a, r 1,r 2 )), (resp. E D (Γ(a, r 1,r 2 )) ). Remark 1: By definition, D is not infraconnected if and only if it admits an empty annulus. Remark 2: By definition {I(Γ(a, r 1,r 2 )), E(Γ(a, r 1,r 2 ))} is a partition of D. Examples: Let r ]0, 1[, let D = d(0,r) d(1, 1 ) and let D = d(0,r ) d(1,r). Then Γ(0,r,1) is an empty annulus of D and also of D. In the same way Γ(1,r,1) is also an empty annulus of D. Notations: Let X (D) be the set of the empty annuli of D. Given Λ 1 and Λ 2 X(D), it is easily seen that I(Λ 1 ) I(Λ 2 )isequivalenttoe(λ 1 ) E(Λ 2 ). We will denote by therelationdefinedonx (D) byλ 1 Λ 2 if I(Λ 1 ) I(Λ 2 ) and we set Λ 1 < Λ 2 if Λ 1 Λ 2 and Λ 1 Λ 2.

8 8 Ultrametric fields The following Lemmas 1.17 and 1.18 are easily seen. Lemma 1.17: The relation is a relation of order on X (D). LetΛ 1 and Λ 2 be two empty annuli of D. The following assertions are equivalent : i) Λ 1 and Λ 2 are not comparable with respect to the order ii) I(Λ 1 ) E(Λ 2 ) iii) I(Λ 2 ) E(Λ 1 ) iv) I(Λ 1 ) I(Λ 2 )=. Lemma 1.18: Let Λ X(D) and let x I(Λ) (resp. x E(Λ)). The infraconnected component of x is included in I(Λ) (resp. in E(Λ)). If Λ X(D) is such that Λ < Λ then I(Λ ) E(Λ). The following Lemma 1.19 is a direct consequence of Lemmas 1.17 and Lemma 1.19: Let Θ be an empty annulus of D. The family of the empty annuli Λ S is totally ordered. Proof. Let Λ 1 and Λ 2 X(D) satisfy Λ 1 Θ, Λ 2 Θ. Then I(Λ 1 ) I(Λ 2 ) I(Θ) hence I(Λ 1 ) is not included in E(Λ 2 ) hence Λ 1 and Λ 2 are comparable. Lemma 1.20: Let Θ be a minimal element of X (D) for the order. Then I(Θ) is an infraconnected component of D. Proof. Suppose that I(Θ) is not infraconnected. By definition I(Θ) is of the form d(a, R) D hence there exists an empty annulus Λ = Γ(α, r 1,r 2 )ofi(θ) with α d(a, R),r 1 <r 2 R and some β I(Θ) such that r 2 α β R. Since Λ d(a, R) weseethatλ D = hence Λ is an empty annulus of D and therefore Λ < Θ. This ends the proof of Lemma Theorem 1.21: D has finitely many infraconnected components if and only if it has finitely many empty annuli. Moreover if so does D then one of the infraconnected components is A 0 = E(Θ) while the others are of the form Θ X (D) A i = I(Λ i ) ( ) E(Θ), with Λ i X(D). Θ<Λ i Proof. We will first assume X (D) to be finite and we will prove that the infraconnected components are in the form A i, above, so that there will be finitely many ones. Let Λ 1,..., Λ n be these empty annuli of D and for every i =0,..., n,let A i be the subsets of D defined from Λ 1,..., Λ n as above. For every x D, for every i =1,..., n, either x I(Λ i )orx E(Λ i ) hence it is easily seen that x n belongs to one of the A i, hence D = A i. We check that A i A j = whenever i=0

9 Ultrametric fields 9 i j. First we assume i =0,j > 0. Hence A 0 E(Λ j ) while A j I(Λ j ) hence A 0 A j =. Now we suppose i>0,j > 0. If Λ i < Λ j then a j E(Λ i ) while A i I(Λ i )andthena i A j =. Hence we may assume that Λ 1 and Λ 2 are not comparable and then by Lemma 1.17 we have I(Λ i ) I(Λ j )= hence A i A j =. Consequently, the family (A i ) 0 i n makes a partition of D. Now we will show that each A i is infraconnected. Suppose that a certain A h is not infraconnected for some h>0(resp. h = 0). Then it admits an empty annulus Λ = Γ(a, r 1,r 2 ). Firstwenoticethatifh =0thenΛ h > Λ because both a, b are centers of Λ h. Now, if h =0(resp. h>0), let Θ X(D) (resp. let Θ X(D) be such that Θ < Λ h ). Since both a, b belong to E(Θ), it is seen that all Λ is included in E(Θ) and therefore is included in A h. This contradicts the hypothesis and finishes proving that A h is infraconnected. Next we check that each A j is maximal in the set of the infraconnected subsets of D. Indeed let B be a subset of D that strictly contains a certain A h and let a B \ A h. If h = 0, there exists Θ X(D) such that a I(Θ), but A h E(Θ) and therefore Θ is included in an empty annulus of B. If h>0, either a belongs to E(Λ h )whereasa h I(Λ h )andthenλ h is included in an empty annulus of B, or there exists Θ X(D) satisfying Θ < Λ h and a I(Θ), but then A h I(Θ) and therefore Θ is included in an empty annulus of B. Thus in each case B is not infraconnected and this finishes showing that each A i is maximal in the set of the infraconnected subsets of D. As a consequence, the infraconnected components of D are the A i. Now conversely, we assume D to have infinitely many empty annuli. First let us suppose that D has a sequence of empty annuli (Λ n ) n N such that Λ n < Λ n+1 (resp. Λ n > Λ n+1 ) for all n N. By Lemma 1.15, for every n N there exists x n E(Λ n ) I(Λ n+1 )(resp. x n I(Λ n ) E(Λ n+1 )) and then the infraconnected component X n of x n satisfies X n E(Λ n ) I(Λ n+1 )(resp. X n I(Λ n ) E(Λ n+1 )) hence X n X m = for all n m, hence D has infinitely many infraconnected components. Finally we may assume that every totally ordered set of empty annuli is finite. Hence there exists a sequence of empty annuli Λ n that are minimal elements for the order on X (D) and then I(Λ n ) I(Λ m ) = whenever n m. By Lemma 1.19, I(Λ n ) is an infraconnected component D n of D such that D n D m = whenever n m. This finishes proving that D has infinitely many infraconnected components and this ends the proof.