Khóa luận tốt nghiệp Toán tin: Ultrametric spaces and P−Adic numbers fields = Không gian siêu metric và trường số P-ADIC

Applying the obtained results to study field with non-archimedean valuation, particularly in the context of p—adic number fields.. The extension from @ to € is: First, we extend the fiel

MINISTRY OF EDUCATION AND TRAINING

HO CHI MINH CITY UNIVERSITY OF EDUCATION

GRADUATION THESIS

Ho Chi Minh City, May 06th 2024

MINISTRY OF EDUCATION AND TRAINING

HO CHI MINH CITY UNIVERSITY OF EDUCATION

GRADUATION THESIS

Field of science discipline Mathematics

Student: Le Van Manh; Male

Ethnicity: Kinh

Department: Department of Mathematics and Infomatics; Year: 4; Total years of training: 4

Field of study: Pedagogy of Mathematics

Supervisor: Assoc Prof Dr My Vinh Quang

Ho Chi Minh City, May 06th 2024

Ho Chi Minh City, May 6th 2024

The confirmation of the supervisor

Assoc Prof Dr My Vinh Quang

Ho Chi Minh City, May 6th 2024

The confirmation of the chairman of the board

Assoc Prof Dr Tran Tuan Nam

I am also immensely grateful to my algebra teacher from the Department of

Math-ematics - Informatics, whose expertise was instrumental in refining my approach to thesubject matter, Your detailed and constructive comments have helped me to sharpen mythinking and brought my work to a higher level

I must express my very profound gratitude to my family and friends for providing me

with unfailing support and continuous encouragement throughout my years of study and

through the process of researching and writing this thesis This accomplishment wouldnot have been possible without them Thank you

Table of contents

INTRODUCTIO! ii

TT b beet e eee, 5 NNNNNaaHạqạia 3

Part I

INTRODUCTION

1 The reason for choosing the topic

Ultrametric space is a metric space that has many special, extraordinary properties Forexample, in this space, every triangle is equilateral, every sphere has infinitely many

center, The p—adic number fields themselves are also ultrametric spaces with the

p-adic valuation as the metric Therefore, researching and exploring ultrametric spaces

and applying them to the study of the p—adic number fields is an interesting and practicalendeavor That’s why I choose this topic for my graduation thesis

2 Research objective

Describing and researching the properties of ultrametric spaces, especially, focusing onthe properties like compactness, separation and completeness

Applying the obtained results to study field with non-archimedean valuation, particularly

in the context of p—adic number fields

3 Research scope

Ultrametric spaces, fields with non-archimedean valuations, »—adic numbers fields

4 Research method

Refer to specialized book, journal, combining with existing knowledge and skill,

ap-ply the results, tools of calculus and non-archimedean analysis to present and solve the

problem

5 The structure of graduation thesis

This graduation thesis consists of an introduction, two chapters, a conclusion and a

bib-liography

Part II

RESEARCH RESULT

Chapter 1

Fields with non-archimedean valuations

1 Valuations and non-archimedean valuations

1.1 Valuations and valued fields

Definition 1.1.1 Let K is a field A valuation on K is a map ||: IK — IK such that for

all x,y € K, we have:

i) |x| > Ox € KK, lz| = 0 ifand only if = 0

ii) |zw| = |x|\y).

lit) |z + y| < |z| + ly! (the triangle inequality)

The pair (K, ||}, also written as K is called a valued field.

Example 1.1.1 The fields Q, B.C along with the absolute value are all the valued fields

Example 1.1.2 An arbitrary field K always has a valuation on it, that is the trivial

0 ife=0,

|x| = :

1 otherwise

Let (IK, ||) be a valued field, then the map (x,y) => Íz — y| is a metric on K and it

valuation defined by:

yields a topology on K as an usual way

Proposition 1.1.2 Let (K,||} be a valued field We have |\\ = 1, | = z| = ||, |e? | = a|~! and |x — y| > |\2| — |yl| for all x,y € K.

Proof Firstly, we have |1| = |12| = '1/? and |1| 4 0, so that 1| = 1.

Next, for arbitrary z, € K, we have:

|x|* = |z?| = |(—z)?| = | — zl? and they are both not less than 0, therefore |z| = | — z}.

By triangle inequality, we have |y) < jy — «| + 'z| and |x| < [a — | + |yÍ, we may obtain

|z = 0| = llz| = l»|l "

Example 1.1.3 There is just one valuation on a finite field and that is the trivial

valua-tions Indeed, for any finite valued field (F ' |) and for x € F*, we have \x\\""| = |zf”| =

1] = 1, then |x| = 1

Proposition 1.1.3 Let (IK, l) be a valued field, then all the maps (x,y) => x+y, (x,y) =>

ay, # => =# (+, € K) and the map x #~Ì (x € K*) is continuous.

Proof Take xa, yo is arbitrary elements in K and z > 0.

First, there exist 0 < 6 < ¢/2 such that for every pair of elements x,y € K which satisfy

max (|x —zo|,ly— val) < ä, we have |(z + ) — (zo + yo}| < |# — zo| + |ụ — vol < 26 < z.

So (x,y) 6 # + y Is continuous

,

Second, there exist 0 < đ < and § < 1 such that for alla.y 6 K

2max (|20| + 1, |yo| + 1)

which satisfy max {|x — zol lu — 0 ) < ở, we have |x| < |zo| + 1 So that ay — zpyo! =

ry — #0 + 2m — #o@w4| <S |ellw — vol + |walle — zo] < 26 max (\zo] + 1, |wọ| +1) < «.

Because of that, (z /} + 2y is continuous

Third, there exist 0 < § = z such that for all « € K which satisfies |+z — xo! < 6, we have

|(—+) — (—#o)| = |#aạ — 2] < 6 = e The map + ++ —z 1s continuous.

Finally, take xp € K* and e > 0 There exist < ở < min (Izoi/2, e|xq|?/2) such that

for all z € K satisfying |x — zp| < 5, we have |z| > |za| — Í# — xq) > lzol/2 So that

1

|x |= Ly = |x = ra|lzza|”Ì < 26|zạ|~? < e So that z > z~! is continuous "

Take a look at the very first example, note that @' < IR < © The extension from

@ to € is: First, we extend the field of rational numbers @ to the field of real numbers

IK as a completion of Q with respect to the absolute value as a valuation Then, the

field of complex numbers C is the algebraic closure of JR and interestingly, the field C is

complete with respect to the absolute value Hence, we may consider an arbitrary valued

field K and ask whether it has a completion Now, take a look of what actually is a/the

completion of a valued field

Definition 1.1.4 Let (IK.||) be a valued field A completion of IK is a complete field

(L, ||}, together with a dense isometrical field embedding j : K > L That is, a map

3: —+ IL whose range is dense in IL and such that for all +, e KK, we have

J1(+ + y) = 7#) + 7(w).

dav) = 7(z)?0):

|j(#)| = la|l.

With the following result, we may call the completion of a valued field instead of a

completion of a valued field

Lemma 1.1.5 Let (IL,|{} and (L', | I’) be completions of a valued field (IK, |) with

em-beddings j :K — Land j': K - L', respectively Then (L l) and (, | ‘) are

isomor-phic in the following sense There exist a bijective map œ : L —> LJ such that oj = j' and

such that for all x,y © Lowe have a(x + y) = ofa) + aly), d(x) = œ(+)z(w} |m(#)[f = |x]

Proof Consider the rule z : L — LL’ determined by: for any z € IL, there exists a

se-quence (z„) C K such that lim j(za) = 2, we have lƒ(#„) — j’(am)l’ = |za — #mÌ =3(đn} = jÍzm)|, so that 7'(z„} is convergent in L’ Then ø(+} = lim 7'(za)

First, we prove that ø is amap For two sequences (z„} and (2!,) of K such that lim 7(#„} =

lim j(x),) = +, we have |j’(xn)—j'(2,)!' = [gu =#g| = Lien) —J(e%,)| => O when n + +00.

So that lim ’(2,) = lim j'{2!,} and ø is a map.

For x,y in L, there exists sequences (+) („) in K such that J(+„) —> +, J/(wa) —> y By

that we have z{z + y) = lim 7(z„ + „} = lim 7(z;) + lim j/{/„} = a(x) + oly); olay) =

lim 7'(z„w„) = lim 7'(za) lim 7'(w„} = 2(z)ø(w): |ø(œ)|[' = | lim 7{za)|Jf = lim |j"{2_)\' =

lim |z„| = lim |J(„)| = 'z' So that ¢ is a homomorphism

For every x € KK, the sequence (7(z))¿ is convergent to 7(z), so that ¢()(2)) = lim J'(z)

#{+) Therefore a e j = j’

Let y be an arbitrary element of L’, there exists (z„) C K such that j’(a,,) > y It is clear

that 7(#¿) is also convergent to some point + € L and that ø(+) = y It is clear that if the

sequence (z„} € L’ satisfies that //(z„) => 0, then (z„) and j{2,,) are also convergent to

0 and « (0) = 0 Therefore ø is isomorphism "

With this result, from now on, we shall use the term the completion of a valued field

But, does it exist? The following theorem says yes

Theorem 1.1.6 Each valued field has a completion

Proof Let (IK,||) be a valued field Let C be the set of all Cauchy sequences in K, or

equivalently, C = {(z„) C K | lim |z„ = z„| = 0} We defined two operations on Œ as

follows.

For (2) (in) € C, we define

(2p) + (ta) = (#n + ta) and (#a)(Ua) = (®nta).

(i) We show that these two operations are well-defined.

First, we have |(#a + jn) = (#m + Yn) < |ÍŒa = #m)| + [ựa — tới | => 0 when nym => 00 So

that (z„}-+-(y„) € C Second, we have |sn%, ma |[ < |{en—2m)|lYn!+l2m! le —Ym| 3 0when n,m — % as (%,,) and (y,,) are in C so that they are bounded Therefore (+„)(„) €

C.

Set le = (1),0c = (0) (sequence consists of only the clement | and 0, respectively).Take (2a) (un) (2n) € C, we have

((an)} + (Yn)) + (en) = (#a + Yn + Zn) = (tn) + (Qin) + (end):

(n) + (Un) = (#a + ta) = (Ma + Tn) = Cn) + (tn),

(#n)((Wn) + (2n)) = (EuWn + #n#a) = (Tn) Yn) + (En)(za).

((mn) + (Yn) )(2n) = {InZn + Unzn) = {2n)(2n) + (un

)(Zn)-There for C is a ring and its unit is le

(ii) Let N be the set of all null sequences of C, i.e

So N is an ideal of C Morcover, take (r„) € C \ N There exists © > 0 such that

for all K € N, there exists n > 4 such that z„| > ¢ Because (r„} € C, there exists

M € § such that |x, — z„„Í < ¢/2 Take np > Af such that |2,,| > ý We have lzạ| >

l#nal — [en — #uạÌ > £/2 for all + > nạ Set u, defined by 1 ifn < nọ and ~ ifn > ng We

have lưa — Um| = |1/an — 1/2ml = |Ea — #mÌ|#a#m|T! < |#a — #m|4/£? for all n,m > nạ.

Therefore («,,) € C Beside that, u,x, = 1 forall x > nọ, so that 1 — (up ){a,} € N So

1 € (N,z„} (minimal ideal of C containing N and z„) or (N,a,) = C We obtain that

N is a maximal ideal of C and C/N isa field.

(iii) The map j : K - C/N defined by

j(x) = (t)n + N

is obviously an injection of K into C/N Moreover, for all x,y € K, we have j(2 + y) =

(iv) We define a valuation || |) on C/N as follows:

Il(n) +N] = lim |za| ((£„) €

C)-We show what this definition is meaningful For (z„) € C, we have ||z„| — |x|) =

Ey — Xm 1, SỐ that (|+„ |) 1s a Cauchy sequence in & and it is convergent Now, let („) € C

such that (s,,)4+-N = (za}-:V We have (rp —yn) € N and that (|z„| = ưa | < len =a| => 0

when ø > co Therefore lim |x,,| = lim |

With arbitrary elements (z„} + +V, (yn) + N € C/N, it is clear that

|(z„) +N = lim jay) > 0, J|(œ„) + N|Í = 0 if and only if lim za| = 0 or (z;} + N = 0c +N

Il ((£ø) + X) ((w„) + N) || = lim lzaa| = lim lz› | lim |uạ| = Íl(ea) + N||||(u) + X|I.

I ((n) + N) 1 ((ya) + N) lÌ = lim |z„ + ạ| < lim |za| + lim |yn| = ||(z„) + VI] + |Í(wa) + NI.

We obtained that (C/N, | ||) is a valued field.

(v) We shall prove that (Œ/ N, ||||) is complete Take a), a2, a3 is a Cauchy sequence

in C/N, we may write a; as (2¿j); + NV For every & € N*, because (r¿;) € C, there exists

X(k) € Ñ such that |z;; — yj] < 1/k for all ÿ 7ˆ > N(k) Set a = (#ax(„)) + N We

prove that a, - ain C/N

For k € §*, because (a„) is Cauchy, there exists K€ N such that |\a, — „|| < 1/3k

for all n,m > K Take n,m > max (3k, K), there exist p > max (V(r), N({m)) such that

#ụp — #„p| < 1/2m Therefore

#nXN(n) — TnN{m}) < TnNín) — Tnp| + |#np ~ #mp| + |#mp ~ TmN(m)Ì s l/k,

OF |f„V(W) = #m@m)Ì Š 1ý for all n,m > max (3k, ZÝ) and that a € C/N

Moreover, we also have

ltnj ~ #;wrnl = |#n; = TnaN(nl + lan} ~ #/w@øl = 1/3k + 1/k = 4/35,

for all ÿ > N{n) That means

la, — al| < 4/34 for all ø > max (3š K)

We obtain that a, => ain C/N.

(vi) Now, we have to show that 7{IK) is dense in C/V For an arbitrary element (z„) + Ý

in C/N For every ¢ > 0, there exist ng such that |x, — x,,| < ¢ for all n > nọ Then

|I7{zn„) — (an) + N)|| = lim |xnạ — zal < £ "

With this result, we acknowledge that the completion of a valued field depends on

valuation The completions of a field may be different with respect to different

valua-tions But with two valuations which induce the same topology on a valued field, the

complctions shall be the same And that comes to the next part of this subsection

Definition 1.1.7 Two valuations on a field KK are equivalent if they induce the sametopology on K

This definition maybe a little tricky, with the following result, the term equivalent

valu-ations shall be clearer

Theorem 1.1.8 Ler | ¡ and |\2 be equivalent valuations on a field % Then there is a

positive real number ¢ such that ||; = | l5

Proof Suppose that | |; is trivial but | (2 is not, then there exist « € IK* such that |z|z < 1,

then z” is convergent with respect to | |2 but not convergent with respect to |ị This

contradiction shows that | |; is trivial if and only if | |2 is trivial In this case ||; = | Id

Assume that) |¡ is non-trivial Note that for a non-trivial valuation | |, we have «”

conver-gent to 0 if and only if |x| < 1 There exist z € IK* such that |x|; > 1 +” is converconver-gent

to 0 with respect to | |¡, and | |), | (2 are equivalent, so that z—” is also convergent to 0 with

respect to | |» Therefore |x|) > 1 and there exists ¢ > 0 such that |x); = [a{5.

For an arbitrary element € KK", we may assume that |y|; > 1, there exists a > 0 such

that |z|) = |z|ƒ We shall prove that z|¿ = + $ For every rational number mi/n > a, we

have |y" |, = |2|{* < (z1, so that |y"z~”*|¡ < 1 Therefore |y"2~""|2 < 1 or |yl2 < [xg

mn

Similarly we prove |y|2 > |z|¿ “” for all rational number m/n < a.

Then z|š = Ízi5° = |z|† = |u|¡ and we obtain that ||; = | |§ s

Furthermore, there are more equivalent definition of equivalent valuations on a field

showed as follows

Theorem 1.1.9, For two valuations ||; and |\> on a field \K The following conditions

are equivalent

i) |x|) < 1L ifand only if |x|2 < 1 for every + € K

ii) |x|) < 1 # and only if \x\2 < 1 for every x € K.

iti) ||; and | \y are equivalent

iv) ||¡ és positive power of | Ía

Proof i) = ii) Assume that we have i} Let z € IK such that |2|) < 1 If ||) > 1 Then

jx~'|> < 1, so that |z~l(; < 1 Then |x|; > 1 This contradiction proves ii}.

it) > #0 If ||, is trivial, then K = {2 e K | [x]; < 1} = {x e K| |z|z < 1}, so that

| |2 1s trivial and these two valuations are the same Assume that these valuations is

non-trivial Take an arbitrary open set V in (K, | |¡), leta € V, then there exists 1 > 5 > O such

that {a EK|le-al; < 5} C V We can find an element b € IK” such that |b]; < 4, then

{x €K| l(z — a}/b|, < 1} C V By ), we obtain that {x € K | |(x — a)/bl2 < 1} C V,

so that {z € IK | |x — al, < |bl2} C V and V is also open in (IK, | |2).

itt) = tv) 1s proved by the theorem above

iv} = 2) Assume that | |) = | |š fore > 0 We have {2 € K | |z|z < 1} = {z e K| |xlg <1} =

{z e KỊ lzl¡ <1} "

Note 1.1.1 From the theorem above, consider the case the two valuations are both

non-trivial Suppose that x € IK, |x|, < 1 implies + ¿ < 1 From this, it follows already that

||, and | |z are equivalent

Indeed, let 2 € K* such that {2 < 1 I/z|ị > 1, then |x~!|, < 1, it follows that|+—l{ạ < 1

and |z/) > 1 This contradiction shows that |x|; < 1 We obtained that z › < 1 yields

Take an arbitrary open set V in (K | li) let a € V, then there exists 1 > J > 0 such that

{x ¢K| |x — al) < đ} C V We can find an element 6 € K* such that ||; < 3 And we

obtain that

{r< K | l> — ala < Ib|z} Cc {xe K | |(a — a}/bl2 < 1} Cc {re K | |(+ — a} fb], < 1} eV.

and \’ is also open in (K, | |2).

Conversely, take an arbitrary open set U in (IK, | |2), leta € U, then there exists 1 > 5 > 0 such that {+ € K | {x — aj; < 5} CU We can find an element b € IX" such that |b|; < ó.

And we obtain that

{re K | |a — aÌị < lol } = {r<K| |Ír — a} fol, < 1} Cc {reK| l(a — a}/bla < 1} cu,

and UU is also open in (KK | |¡).

We proved that ||; and ||» are equivalent

As promised, we now show that the completions of a field with respect to two

equiv-alent valuations are the same

Theorem 1.1.10 Let | |; and ||» be two equivalent valuations of a field K Let (Ly | Ít)

and (1, || \|2) be the completions of (K, ||:) and (IK.||2) respectively Then Ly = Lạ

and that || ||) and || \\o are equivalent.

Proof Thank to Theorem 1.1.6, we know that, respectively, Ly, Le are Cy Ny, C2/ No,

where C¡.C¿ are all the Cauchy sequences in (K.{\1) (K.||2) and Vị N2 are all the

null Cauchy sequences in (IK, |{)) , (IK, | |2), respectively Because | |) = | |§ (c > 0), we

obtained that Cy = C2, Ny = N»2 (with the same operation) and that L; = Ly

Moreover, the dense embedding from (K, | |1)} , (K, | |2) to Ly, La are the same, say j For

every z € lU, there exists a sequences (z„) C K such that lim j{2,,) = + (with respect to

both equivalent valuations | |¡ | |2), then

lzzi'¡ = lim ||j(z„)||¡ = lim zaÍ¡ = lim|za 5 = lim ||7(za)|(§ = |lz |5.

So || |Í¡ = || l|š and that they are equivalent "

We have considered a valued field and its completion From the very first

defini-tion of a valuadefini-tion, the triangle inequality can be strengthened in a way that results in

many interesting properties In next subsection, we shall investigate the strong triangle

inequality

1.2 Non-archimedean valuations

In this subsection, whenever we write ø| with ø € N, it means (n1x| where 1z is the unit

of the valued field (K, | |).

Definition 1.2.1 A valuation on a field KK is archimedean if |2| > 1, non-archimedean if

|2| < 1.

Example 1.2.1 All trivial valuations are non-archimedean

Example 1.2.2 So far, we have encountered some valued fields, but with non-trivial

valued fields, we only met Q.R,C and their valuations are all archimedean Now, weshall construct a valued field with non-trivial non-archimedean valuation

Let IK be a field and K(X) be the field of the rational functions over the field KK, i.e

K(X) = {/.ø' | f.g KỊX].ø 40}.

For a fixed real number p > 1, we define the rule || : K(X) + B by

0 iff=0

1 _ E ,

We now show that || is a non-archimedean valuation on K(X),

First, for two elements ƒg~' and ƒ'g~} of K(X) such that ƒg ! = f'g + We have

ƒ =0 Ƒƒ =0 On the other hand, if { # 0, we have fq' = {'¢ so that deg{ f} —deg(g) =

deg{ {") — deg(g') Therefore | | is well-defined.

It is easy to see that for fg-* € K(X), we have |fg-| = 0 and | fg-*| = 0 if and only if

fg" =0.

It is also clear that \fg7*.f'g’—| = |fgU || fq’ | for all fg), fg’! e K(X).

Take fg 1, fg’) € K(X) ƒƒfƑ'd + f3) = 0, it is true that |ƒg Ì + fig | <

fo" + | Ƒø~}\ And if f{'(f¢ + fg) # 0, we have deg( fg’ + f'g) < max(deg( fg’) +

deg(f’g)) It follows that | fg-' + f'g' = |(fg + f'e)(ạg))Ì| = p*ẽUz+f s)=dee(ag) <

pmaxideg(Sy’}+-deg({"9))- deg(42') < max(øe(/)-dee(2), peel) -đe£(2)) < Jfø-!|+|/'ø#-1.

Finally, we have |3L = 1 ifchar() # 2 and |2| = 0 if char(K) = 2 Therefore || is a

non-archimedean valuation on K(X),

With this example, we may acknowledge that |fg~! + f’g’"| < max(|fø-}|.| a1).

And that is presented in the following results

Theorem 1.2.2 Let || be a non-archimedean valuation on a field Then

i) |n| < Lforalln e Ñ

ti) |r + yÌ < max (Iz| \y|) for all x.y © KK (the strong triangle inequality).

Proof i) Forn € Ñ and n > 2, we write n in the base of 2 as follows

With this, we obtained that ifn < 2Š, then |n| < s

For an arbitrary k € N*, we have n* < 2**, so that [n*| < ks.

Take & — th root of both side, we have

Let k > œ, we obtain |n| < 1.

ii} For 2,y € K and rn € N*, apply 7), we have

l(z+u)"I=I3ˆ (Q7 <5” (2) ey *| < 3 ` IzF||y"”#| < (n+1) max((z],

It follows that

# + | < lim(n 4 1 max(ilz!, ||} = max(iz|, |y!).

With this theorem, we obtain other definitions of non-archimedean valuation as acorollary.

Corollary 1.2.3 For a valuation on a field , the three following conditions are

equiv-alent

i) The valuation satisfies the strong triangle inequality

ii) The valuation is non-archimedean

iii) The set {1, 2,3, } is bounded

Proof i) = ii) It is clear that (2| < max({1], (1|) = 1

it} => dé} It is obviously true that | is an upper bound of the set {1, 2,3, }

+ +y| < lim(n + 1l MỸ max(|+| iu|) = max{'z', |ự ).

An interesting property of archimedean valuation is that if (IK,||) is a

non-archimedean valued field, then | í* is also a non-non-archimedean valuation on K for all s > 0.Meanwhile for an archimedean valuation, it is not generally true for the case s > 1.

Beside, with two equivalent valuation || and | |’, we have |2| < 1 if and only if |2\’ < 1,

so that two equivalent valuations are both archimedean or non-archimedean.

Now, we shall present one of some important results of the strong triangle inequality

as the following proposition

Proposition 1.2.4 Let (K.||) be a non-archimedean valued field For every pair of

elements x,y € IK, we have

i) If |x| # \y|, then |x + y| = max (|2| |y]).

ii) If |x + y| > |x|, then |x + y| = ly.

Proof i) Assume that |z| < |y|, we shall show that |z + | = ở

Indeed, we have |z + | < max (|z| lw|} = |y| On the other hand, we have |y! = {a + =

a| < max (|x + y|, |z|), but |y] > |z|, so that |y| < Ja + y| It follows that |x + g| = |yl.

ii} Suppose that + + y > x), by strong triangle inequality, we obtain that

jy] < max (lz + yl, |z|) = |z +

ø-1.3 The residue class field, the value group and series expansions of

ele-ments of non-archimedean valued fields

A non-archimedean valued field may have lots of amazing properties that an archimedean

may not, it is all presented in this subsection FOR THE LAST OF THIS SUBSECTION,

THE VALUATION | | OF THE FIELD KK IS NON-TRIVIAL, NON-ARCHIMEDEAN

Definition 1.3.1 The closed unit disc and the open unit dise of K, respectively, are the

sets

Bo(1) = {x € K | |x| < 1} and Bo(1-) = {z e K | || < 1}:

For the case of archimedean valuations, it is easy to see that 7Z2¿(1) is not a subring

of the field KK because |1 + 1| > 1 But with non-archimedean valued field as we are

considering, the situation is much better

Proposition 1.3.2 Bg(1) is a subring of K and Ba(1—} is a maximal ideal of Ba(1)

Proof First, we show that 8o(1) is a subring of the field K Indeed, for every x.y €

Boll), we have

ja — y| < max (|e, |y|) < Land |xy| = |z||u| < 1.

Now, let «,v € bọ(1—) and z € Bof1), it is clear that

Ju — t| < max (Iu| e|} < 1 and |uz| = |u||z| < 1.

So that Bol )3 Ba(1)

Moreover, every element of /2o(1) \ /Za(1~} is multiplicative invertible in Bo(1), then

u(1~} is a maximal ideal of Bp{1} a

By this proposition, we see that By(1}/ Ba(1~) shall be a field

Definition 1.3.3 The residue class field of IK is the field defined by

We shall investigate the connection between the residue class field and the value

group of a valued field and its completion, the field itself but with another equivalent

valuations

Proposition 1.3.5 Let U be the completion of \K, then their value groups are the same

and their residue class fields are isomorphic

Proof For an arbitrary element x € L*, there exist a sequence of elements in Ï<*, says (z„), that is convergent to x There is ng such that |x — z„„í < |z|/2 it follows that

|z„„ — z| < |x| and it implies that |z,,,| = (za¿ — «+2! = (2í, In other words, we have

|L*| = |K'!.

Let &,! be the residue class fields of K,L, respectively It is easy to see that the map

Ti kla+Bo(l ) z+ Ba(1—}], where Bof{1~) and Bp{1~) respectively are the open

unit discs of K, L, is an injection of & into / So we may write k = z(k) < l

On the other hand, for «+ Bp{17) € J, ifa+ Bo(1~) = 0+ Bọ(1~), then 2+ Bạ(1~} € 7{k)

If |x| = 1, then by the same approach as above, there exist yy € IX such that || = 1 and

jy — z| < 1, therefore z + Bo(1~) = y+ Øạ(1~) € r(É) It implies that r is an isomorphic

Proposition 1.3.6 The value groups of a field with respect to different equivalent

valua-tions are isomorphic, the residue class fields of a field with respect to different equivalent

valuations are the same

Proof Let || and | |’ be two equivalent valuations

First, there exists e > 0 such that |) = | |, and this is the isomorphic between two valued

group Second, thanks to Theorem 1.1.9, we have 8o{1) and Øụ(1} are the same with

respect to both valuations, so that the residue class fields are also the same "

Let G is an arbitrary subgroup of the multiplicative group (0, +2).

Suppose that | is an accumulation point of G, we shall show that G is dense in (0 +00)

Indeed, by assuming the contrary, there exist 0 < r < x such that (+ —7,2+rj)NG =

Because | is an accumulation point of G, there exist (z„} C G \ {1} such that 2, > 1,

we can choose z„ > 1 for all x For every «,, there uniquely exists & € Z such that

ak<cx-—re<atr < zỀ*Ì, sothat x, — aktlok > (z + r)/(x — r) > 1 forall x This

contradiction shows that G is dense in (0, +00)

Now, suppose that | is notan accumulation point of G, we shall show that G = {m" |ne 5}

for some € {0,+00) Indeed, if the set {x EG\a< 1} does not have maximal

ele-ment, then there exists a sequence (z„) of it such that z„ + sup {z € Ở | z < 1} and that

(#ø„/#a+L) 1S convergent to 1 or equivalently, ] is an accumulation point of G Therefore

there exists 7 € G such that z = max {x €G|z< 1} Let y be any element in G, there uniquely exists & € Z such that x݆! < y < +*, then x < yx *® < 1 By the definition of

x, it follows that yx—* = x or y = x**! and that G = {x" | n € Z} This results come to

another definition

Definition 1.3.7 The valuation on K is discrete if Ì is not an accumulation point of |K*|.Otherwise, the valuation is dense

The term discrete/dense above can be understood as the set |K*| is discrete/dense in

(0,+00) Now, we shall answer the question we set before We fixed an element 7 € K

such that 0 < || < 1 Then we can easily check that the set J = {x €K| |2| < |z } is an

ideal of Bo(1) and 2ø(1)/./ is a ring, let R be a set of all representations in Bo{1) modulo

J, ie RC Ball) satisfying

(i) Forry,r2 € Rand ry, # ra, we have |x| < |ry — rai < 1

(it) For every s 6 Bo(1), there exists » € R such that |s — r| < |x|

Beside, the element r in (¿} is unique More clearer, if z € R satisfying |s — z| < |x|,

then |r — z| < max {|r — s|.|s — |) < |x|, it implies that + = z With this z and R above,

we have the following theorem

Theorem 1.3.8 Let (Kí ') be a valued field and x, R be defined as above Then for

cách + & lk, there exists a unique two-sided sequence a_—\ da, dị of elements in Rsuch that a_, = 0 for large n and

Proof For an element x € IK First we prove the existence of the two-sided sequences

satisfying the theorem Let » € § be the minimal natural number such that zz”| < 1.Then there exist a_,, € R such that (+x” — a_,| < (2| Therefore we may write zz” asfollows

ax” =(t_n +717,

where (z¡' < 1 Using the same reasoning with 2, we shall find a_,,,; € 2 such that

|rì =ø_„¿:: < 'xÍ and we may write

Using the same reasoning, we shall get a; = }, for all i € Z That means every element

x in IK has a unique extension of the form (+)

In addition, if K is complete, with any two-sided sequence 2_¡.ø¿,øị such that

a_, = 0 for large ø Set r„ = $0" aja’, it is clear that (x,,) is Cauchy and that

I=—%

tlm —? an ayn in K,

Note 1.3.1 For a discrete valuation ' | on K, there exists z € K such that

|x| = max {|z| | x € Bo(17)}.

Let this element + be as the element z in the theorem above, then we have J = u(1~}

and that # are similar to the residue class field ¿ Now suppose that & is finite, observe

a sequence {x,,} C By{1} We have

%

) F=:

In = (ny7T.

;=

Because # is finite, so is R, there is a subsequence #1), z;, such that the terms of 7°

of them are the same and equal to ag € ?#? By using the same reasoning for the sequence

(ay) but with the term zÌ, there exists a subsequence sa), 222, of (z„) such that the

terms x" z! of them are the same and equal to øo + a¡z with ap, a, € R Continuing this

reasoning, for every k, we get a subsequence z¿„„ such that the terms 7", ,7*~* of them

are the same Take the diagonal sequence (z¿¿), it obviously is a subsequence of {x,,}and convergent to ag + a,x + +» © Pa(1) It follows that Bo(1) is compact Moreover, let

+ € Bo(1), for every y € K such that |„— z| < 1/2, we have |y| < max(|z— z!, |+|} < 1 and

that € By(1), that means Bo{1) is open That means Bo{1) is a compact neighbourhood

of 0 and z + Bp{1} is a compact neighbourhood of x This lead to our next corollary.

Corollary 1.3.9 4 non-archimedean valued field is locally compact iff its residue classfield is finite and its valuation is discrete

Proof (<=) can be obtained by the very previous note We show that (=>) is true There

exists an clement d, with 0 < |d| < 1 anda compact set U of IK such that {x | |z| < |a|} Cc

U Then dBp{1) = {a |+z[ < (4|} € U, it follows that (1) is compact and that Bp{1)}

is compact

If & is infinite, there exist a sequence (z„} such that |x,,| = 1 for all n and +„ = z„| = 1for all n z m Observe that (z„}) C Bo(1) but |za — #„ | = 1 for all n 4 m, equivalently,

(z„) doesn’t have any convergent subsequence It contradicts the compactness of Z¿(1).

If || is not discrete, then there exists (z„} C Øa(1~) such that |x| < (z„+¡| and (z„( > 1

By compactness, there exists a subsequence („) of (z„) such that y, — y € Boll).

It is clear that || = 1 There exists m > 0 such that |y,, — | < 1/2, it follows that

lym | = max (|ym — y!, |v!) = 1, this contradicts the fact that y,, € Bo(17) .

We have got a brief understanding about a non-archimedran field, in next section,

we shall dive into the context of p—adic numbers ficlds, they looks quite unnatural at

first side but they have more interesting properties than the archimedean valued fields

that we usually encounter,

2 The fields Q,, C,,

The field Q of rational numbers together with the absolute valuation, of course, is a

archimedean valued field But we may replace the absolute valuation with some

archimedean valuations, the results may be a little different In this section, let p be a

fixed prime number

2.1 The field Q,,

Definition 2.1.1 For an integer n, the order of nm modulo p is defined by

onal tai) too ifn=0,

ord, (rn) =

, m if pTM|n but pTM*! tn.

It is easy to see that for all n,m Z 0, we have ord,{nm) = ord,(njord,(m) and ord,(n +

ra) > min(ord,(n), ord,{m)}

For s positive real number ø > 1, we define a map ||: @ > E by

ord, (b)—ord,(a) otherwise,

0 ifa = 0.

|a/b| = ue , for all a,b € Z.b # 0.

p

Theorem 2.1.2 The map || defined above is a non-archimeadean valuation on Q.

Proof First, we prove that | | is well-defined Indeed, for a/b = c/d € Qwitha,b,c.d€ 5

and fd # 0 Ifa = e = 0, then |œ¿b' = 0 = fe/d| On the other hand, if ac 4 0, we

have ad = be and that ord,(a) + ord,{d}) = ord,{c) + ordp{b) = ordy{b} — ordy{a) =

ord,{d) = ord,{c), it follows that |a/b| = 'c/4|

It is clear that |x| > 0 for all 2 € Q and |x| = 0 HỶ z = 0,

For a/b, c/d € Q (a,b.e,d € Zand cd # 0) If ac = 0, then it is clear that |a/b.c/d!

œ/bl|c/d| and la¿b + ¢/d| < la/b| + lefd| If ac 4 0 We have

a/b.cjd = prrde{bd)-ordy{ac) = prrdr(t}-ords(a) ordp(d}—ordp(c) = |a/®|Ìc/4|.

|a/b h c/4| = g d4) ordy{ad-+be) < p?ar(04)- min(ord, (œd)-+ord; (be})

< praxtord,(b} —ord,(a),ord,(d)—ord,(c})

= max(|a/b), |c/d|) < |a/b| + |e/d

Therefore || is a valuation on Q, moreover, the above inequality implies that || is

non-archimedean 2

Note 2.1.1 For every ø > 0, we get a valuation but all the these valuations are all

equiv-alent For convenient reason, we shall choose » = p, denote the valuation by | |, and call

it the p—adic valuation on Q

For two different primes p,q, we have |g), = g7! < 1 but |g/p = 1, it follows that

||, and ||, are two inequivalent valuations on Q Note that the absolute valuation is anarchimedean valuation on Q That means | |, and the absolute valuation are also inequiv-alent Next, we will show that any non-trivial valuation on @ must be equivalent to the

absolute valuation or some p—adic valuation

Theorem 2.1.3 Every non-trivial valuation on Q must be equivalent to the absolute

valuation or some p—adic valuation.

Proof, Let || be a non-trivial valuation on Q

First, we consider the case || is archimedean Then, we have | < 2| < 1| + 1| = 2

There exists c > 0 such that (2| = 2° For any n € N, œ > 2, we may write n in the base

of 2 as follows:

n= do, ay.2 ofe sss be ,—12° V

where a; € {0,1} for all i = 0,s — 1 and a;_¡ # 0 It follows that

Therefore |n| — n° It follows that || is equivalent to the absolute valuation on Q

Finally, let || be a non-archimedean valuation Because || is non-trivial, there exists

rn & § such that |n| < 1, let p be the minimal positive integer such that |p| < 1 If pis notprime, then p = ah with p > a,b > 1, then |p| = |ø||b' = 1, a contradiction Therefore p is

a prime For any ¿ € Z such that p { g, we write g = ap +r with a,r € Zand 0 < r < p,then [ap| < |r| and this implies |ạ| = |r| = 1 That means for every n € ©, we have

|n| = Ip|®Í"), By setting e = logyy)-1(p), we get | |, = | Í° and that | | is equivalent to the

p-adic valuation on Q a

By this theorem, to investigate all the non-archimedean valuation on Q, we onlyhave to investigate the p—adic valuation For a p—adic valuation on Q, it is clear that the

sequence {p"),, does not have any subsequence that is convergent With this remark, we

shall complete the field Q with respect to the p—adic valuation as other valued fields

But first, let investigate the value group and the residue class field of (Q ||,).

Because ord,(n) € Ñ for all nonzero integer n and |p"|, = p~" ¥n € 5, we have |Q*), =

{p~" In € Z} and the valuation ||, is discrete Recall the unit closed disc and the unit

open disc of are By{1} and Ba{l~} For every a/b € By{1}, we may assume that p ‡ 6.There exist 7 € {0,1, ,p — 1} such that a = 6) (mod p), so that |ø/b— j|p„ = |[a—bj) /blp <

p~} < 1 Furthermore, we have |i = 7| = 1 for all i # 7, ¿, j € {0,1, ,p = 1} It follows

that k = {i+ Bọ(1 )|O<i<p-1}2F,.

Now, we shall give that name of the completion of (2, | ») and show that this

com-pletion is a real extension of Q later

Theorem and Definition 2.1.4 The valued field (Q,||p) has the completion, denoted

by Q,

Remark 2.1.1 Thanks to Proposition 1.3.5, the value group of Q,, and Q are the sameand the valuation on @, is discrete Their residue class fields are isomorphic, they areboth finite Recall R = {0.1 p = 1} and x = pare #,z in Theorem 1.3.8, we may

write Q, as

x

Q, = » aipÌ (*) | a; € {0,1, ,p — 1} and a_, = 0 for large n

t==%

i.e every clement in Q, has a uniquely extension of the form (*) and vice versus

More-over, if n is the smallest integer such that a, # 0, then |2 _ _ ap"|) = p7”.

If Q, is countable, then we can write Q, — {z¡ | i ¢ N} where x; = i a;jp)

Con-sider the sequence (a;) of {0,1, p — 1} defined by

0 ifay #0

a= PY

1 ifay =0

Clearly, the clement $>*,, a;p’ of, is different from every z„, a contradiction So that

Q, is uncountable and Q, #4 Q That means (@Q.||,,} is not complete and Q, is a real

7

completion of @.

From the very remark above, we obtain another result as follows

Theorem 2.1.5 @,, is locally compact and separable Q is a subfield of Q, and a

count-able, dense subset of Q).

The unit closed dise of Q, is the set

©

Bụ(1) = {z <$; | |z|p = 1} = Se ap! | a; € {0,1, ,p— 1}

i—)

We denote By(1) by @, and call it the ring of p—adic integers Note that for every integer

n, we have |r|, = p-°'’*(") < 1, Then we may consider Z as a subring of the ring Z,.

Furthermore, because @, is locally compact, we shall get that /29(1) is a open compactneighbourhood of 0 So that z + 2, is an open compact neighbourhood of « for every

xe Q,.

For every « € (Q„, we have |x*| = ||? = p-*" for some integer n, therefore the

equation «* = p has no root on Q, That means Q, is not algebraically closed Look

back at our archimedean fields Q < R < © f3 is the completion of Q and C is algebraicclosure of K, also the completion of B We may acknowledge that the evaluation on IR

and © are the extensions of the absolute valuation on @ For the non-archimedean valued

field Q, we point out its completion, that is Q, and the valuation on Q,, is an extension

of | |, Now, before we get the algebraic closure of Q,, let make it sure that there exists

a valuation on this algebraic closure that extend | |,

2.2 The algebraic closure of a non-archimedean valued field

First, let introduce some aspect of calculus: norm, normed spaces,

IN THIS SUBSECTION, K IS A COMPLETE VALUED FIELD WITH RESPECT TO

NON-TRIVIAL, NON-ARCHIMEDEAN VALUATION

Definition 2.2.1 Let J2 be a vector space over the field IK A norm on F is a map q :

E — IK such that for z.„ € FE and À &€ IK, we have

i) gfx) > Oand g(r) = 0 z = 0

ii) g{Ax) = |Alg(x)

HH) g{2 + y) < max(2(2),

Note 2.2.1 iii) in the definition above may be replaced by di’) g(2 +) < g(x) + q() for

a general case But in this subsection, we shall consider 72 as an extension field of K, so

that it shall be more convenient to use 22), Beside, we can easily show that g{2) = g(—z}

for all z € #

We often denote a norm by | || instead of gy A norm (vector) space over ÏŠ is a

pair (7 | ||) simply written 2, where ƒ is a vector space over K and || || is a norm on

E E is a Banach space over K if F is complete with respect to the induced metric

(x,y) => l|z — y|| Fora valued field (L, ||) containing K as a sub-valued field, we can

consider |L is a vector space over K and so on, IL is a norm space over K

Two norms |||); and || [lạ on a vector space & over KK are called equivalent if they

induce the same topology on /Z We have the following result

Theorem 2.2.2 7ivo norms || ||) and | ||» on a vector space E over K are equivalent iffthere exists ¢,, 62 > 0 such that c¡ | ||) < || lo < cel} a

Proof (=>) If there doesn’t exist c2, then there exists a sequence (z„) of such that

— > œ forall rn > 1 Equivalently, we have

Let z € KK* such that |x| < 1 For each rn > 1, there uniquely exists k„ such that (z#*| <

|lrallz < |x| Set un = aay for every œ Then 1 < |lws|ls < |=~}{ for all x Now,

we have ||a|lạ < 1/nlluall2 < 1/a|z7}| — 0 So that y, 3 0 with respect to || ||» but (y„}

is obviously not convergent to 0 with respect to |Í |(¡

With the similar reasoning, we shall point out the existence of ¢)

(<=) Let V be an open set of (£, || ||) and a € V There exist 6 > 0 such that

{z € B|lla-ai<s}cv.

Then, we have

{x € B| lla - z||s < cd} c {x € E| |la—x) < 5} C V.

Thar means V is also open in (£, | ||2) With the similar reasoning, we obtain that | ||;

and |Í |/2 are equivalent 2

The last result we shall use is presented as follows

Theorem 2.2.3 All norms on a finite dimensional vector space E over KK are equivalentand l2 is a Banach space with respect to each norms

Proof We prove the theorem by induction on n = dimy 2 For n = 1, it is obvious as

the norm on £ must be in the form a|| where 2 is a positive constant Assume that the

statement is true for all finite dimensional norm space whose dimension is less than 1.Let dimy E = ø and e), - ,e„ 1S a base of E.

|IAzll„ = max; |Aa¿| = |A|llzllx

| + yllac = max; fa; + đ;Í < max(max; |a;|, max; (đ;!) < max(||zÍ|¬‹

ÍÍw||¬‹}-Therefore || ||, is a norm Moreover, for any Cauchy sequence (z/„) C & where z¿„, =

327 emits The sequence (œ„¿) is also Cauchy in K and it is convergent to a; € IK.

Then the sequence (+„) is convergent to 5Д aye; That means (£, || ||.) is Banach.

Now, let || || be a norm on £ It suffices to prove that || || and | ||, are equivalent

For every x = 57>" _, aye; € E, we have

n

xl] = lỀ 2 «¡si < max |aillleill < Cllzlboo,

¿1

where C' = max, |Íe,|Í On the other hand, let D is a subspace of F that is generated by

€1,++* ,€n-1 Note that |] | and || || are norms on 2 By induction hypothesis, there

exists e such that ||2|| > e||zi|„„ and D is complete (so that it is a close subspace of #) Itfollows that

e:= inf {len — yl] |y € D} > 0,

or else, there exists a sequence of D that is convergent to e„ ¢ D

Set e’ = min(ellen ||}, 1) and CY = min(ce’,e’|le, ||) For z = el aje; € E, we shall

prove that Í|z|| > C”l|z||x

If a„ = 0, then it is true that |||! > <flla„e„l| Ifa, 4 0, we have

|zl| = lan||len + œ2 yll > |anle > e'lanen |:

IIz|\ = e’ max{||y||, ||ec„eøl|) = ef max(e||wl|~ |e» |Í(ea ||)

= max(ce’ |lw|̬ e Í|ea | œn|)} > c max(||¥|| 20, |@n!)

= C' max aj! = C'|#||+

t

Comeback to our question, can we extend the p—adic valuation on Q,, to its algebraic

closure? Fortunately, the answer is yes

Theorem 2.2.4 (Krull’s existence theorem) Let % be a subfield of a field L, let || be a

non-archimedean valuation on TK Then there exists a (non-archimedean) valuation on

lL that extend | |

Proof Let Sy be the set of all the valued subfield # of L containing K as a subfield,

together with its valuation ||; that extends || It is true that Sy # 0 as K € Sx We

define a relation <s on S by: E <s F if F is a subfield of F and ||, extends ||, for

ELF € Sx.

It is casy to check that <¢ is a partial order on Sy For a totally ordered subset T of Sự,

we show that = UpeyF is an upper bound of T in Sy as follows

For z, € U and z € U \ {0} There exists £, F,G € 7 such that z € E,y € Ƒ,z € G.

We may assume that G = max(E£, F,G), then x,y,z € Gand z — y,2z7!' € GC U So

that U is a field

Consider the map | |y on T defined by: +; = z|g if a € EB € 7 (if there are #, F such

that z € HO F, then /zizg = |x|; as one valuation shall be an extension of the other, i.e

||u is well-defined) Using the similar reasoning, we obtain that | '¿; is a valuation on UL

that extends || and | 'g for all # e T

By Zorn’s lemma, Sy has a maximal element, says L, if L # L, let a € L \ L, we prove

that there exists a valuation on (a) that extends ||; and that is a contradiction So,

simply, it suffices to prove the theorem in the case [L = K{a)

First, we consider the case a is not algebraic over K, then K(a} > IKCY) the field of

rational functions over K For f = ag + aja +++: a,x" € KIX), set

[fl] = max |a;).

It is clear that || || : K[X] > E is a map that extends | |

It is easy to see that || f|| => 0 for all £ € IK{X] and || || = 0 iff f = 0

For ƒ = a9 + - - - + an2",g = by + - +b," € IK[X], we have

If + gl] = max a; + ¿| < max(|a¡| |b¿|) = maxi] fl], llgll).

Ilfgll < max q;|Íb;| < max |a| max |b;| = |L/llllull.

Beside, let r,s be the minimal integers such that @, = max, (+; and b¿ = max; |b,|

r+s

Observe the term +7!“ in fg, we get S2; `2 aibpss—i For i < r we have |aib-+5-i| <

res 9

drbe 45a] < |ab;| and for¿ > r we have |a¿b;;¿- ¿| < |aibs| < Jaybs|, so that | 52, “4 aibyys—s|

(arb„|„ it implies || ƒø|| =

||lIllgll-For f,9, ƒ, ø,h,& € K[X] such that ø, Z, & # 0 and f/9 = f'/¢’, we have

$9 = ## = lUllløll " = II/IIllv II.

So that || f/g|| := |Lfllløl|? is well-defined and it is a map K(X) > R.

Moreover, we casily have

l//ø|| z 0 and || //ø|| = 0 uf f/g = 0

Iƒ/ø.h/R| = l|ƒ/øl|llh/Š |:

| //ø + h/R|| < max(I|ƒ&|\ |løh|l)/I|øk|| = max(̓/ø|| |/È|).

Therefore || || is a valuation on K(X) = K{a) that extend ||

For the last part of the proof, let L = K(a) where a is algebraic over K Set n = [L - K|and e¡ c„ is a base of L over the field K As in the proof of Theorem 2.2.3, we haveanorm || || on L defined by

nn

lx||„ = max|ai| (x = 3 aye; € L).

i=1

For + = $7, avei,¥ = S07.) Biei € L, we have

|zwil < max [loi 3;ere;lloo < max |a||f;|l|ee;ll < Cllxlllull.

where Œ := max; ; |Ìe¿e;|| Set || || = Cll] ||0, for z, € Land A € K, it is clear that

|z|| = 0 and ||z|' = 0 IÝ+ =

ll+z + yl] < max(|Iz | ll+w|| = IIzllllsll:

|w||)-We define

v:LoR,ae—- Jim /||z°||

We shall prove the following properties of v For all x,y € L and À e KK

(a) v(x) = lim t/|Jz"|| = inf,«w /|lz°|:

Let x y be fixed elements of IL, & ¢ N* and A e K

Set a = infew /||z°Í Let ¢ > 0, there is n € N such that ||z"|| < (a + z)", For each

rm > rn, we write m = gn +» where g.r € Ñ and 0 <r <n = 1 So that we get

Wa" |) < |x? lal” < (a + z)"#||z|lT < (a + z)"{(œ +e)" Ya)”.

With Mo = maxyer<n—i{(a + €)74 Vall"), we have mạ, ,x &/|z”"|| < limpoola +

©) WM = a+c Therefore limmoc 4/||z"|| = a Note that a < lim ‡/||z"|| < lim ‡/||z"|

and we get (a).

It is clear that

< u(x} < lim VIz*||< < lim Vllzl| = = |Iz||.

v(Ar) = lim ÿ/J(Az)*|| = |À| im /ÍJz^|| = |A|e(2).

s(zy) = lim $/|(zw)*|[ < lim §/|Iz*|lly"|| = e(z)e(y).

v(1) = lim $/4|1"|| = lim 1 = 1.

s(z*) = lim i/o] = lim ( "V({ Viel) = v(x)".

and we get (b)-(f).

To prove (g), we observe that

n : :

an aye yy! < „hy?

Ke +a" 2 ()› yf] < max |ja'y"““].

D<i<n

For i = 0, we have j(+'y°~!|| = ||y"|[ For 0 <i < v2, we have

le'v" "I < IIy"IIllz/wll' < lly"|I.max(1 |z/w||Y^).

With i > vú, let s be the smallest integer such that s > vú We writen = ¿ = ms +1

and i = qos + rạ with 9), ga, rị,ra € Nand 0 < rị,?z2 < s—1 < s/n, we have

Wary FI] < IIz*Ifllw°ll#llz"'»"*|I < max(||z*|l#'?*, lu? |4) max(1, |Iz|IY^|||IX”).

Note that (rị + 72)/s < 2, so that so that n/s — 2 < ứi + g < n/s and we obtain

Viet < max(z*I⁄,lle"I⁄*~8/", py I2, fy 2) max, fay Y).

With n — 2c (so is s => 00), from all above observe, we get v(x + y) < max{v(x), v{y))

Set /‡ is the set of all norm on L over K that satisfies all the properties (b)-(g) On R, we

define the relation <g by: eị <p vy Im¡(£) < w{x) Ve € L for rị,tạ € R It is easy to

see that (R, <,) is a partially ordered set.

For a totally ordered subset T of R, we show that T has a lower bound in 2 We define

t:z = inf{s(z) |v € T} For x,y € Land 4 € K,we have

0 < u(x) < |z| => 0 < tz) <

||z||-t(Ax) = inf {v(A2) | v € T} = [Alinf {e(x) | v € T} = |Als(z)

t(1) = inf {v(1) |v € T} = inf{1 | ve 7} = 1.

te!) = int {o(2*) | v eT} = int ofa)" |v e T} = inf {o(2) | ve TH = tk

Now, we show that inf{e(œ)e(w) lục T} = inf { v(x) Juve T} inf {u(y) Juve rh.

For every + € 7, we have w(x) > inf {v(x) | v € T} and v(y) > inf {v(y) | e € 7} so that

inf{o(z)e(w) jve T} > inf { v(x) lục T} inf{e(w) lục }.

With a > inf { v(x) [ve T} inf { v(y) jve T}, there exists ¥), € 7 such that vy (2)eofy) <

a We can assume that 1) <p 2% so that øị(z)mj(} = vifa)jety) < a, it follows that

inf {v(x)u(y) |e € T} <a.

Therefore inf {ø(z)(y) |» € 7} = inf { v(x) | v € T} inf{s(w) | € T} And we get

tay} = inf{e(zw) luc 7) < inf{e(z)e(w) Jve T} = t(z)H).