- Date:27 Oct 2020
- Views:4
- Downloads:0
- Pages:157
- Size:1.08 MB

Transcription:

Thanks to Tom Church Rankeya Datta John Doyle Tyler Genao Ernest Guico. David Krumm and Todd Trimble for pointing out typos Thanks to Makoto. Suwama for contributing Exercise 1 50d which gives a counterexample to a. previously stated version of the exercise Thanks to Qun Li and Haiyang Wang. for contributing Exercise 4 28,Chapter 1 Normed Fields and Valuation Theory 5. 1 Absolute values and valuations 5,2 Completions 21. 3 Extending norms 37,4 The Degree in equality 51,5 Hensel s Lemmas 55. Chapter 2 Local Fields 59,1 Remedial number theory I Dedekind Kummer 59. 2 Unramified extensions 61, 3 Remedial Number Theory II Scho nemann Eisenstein 62.

4 Totally ramified extensions 63,5 Higher unit groups 64. 6 Locally compact fields 65,7 Squares in local fields 70. 8 Quadratic forms over local fields 71,9 Roots of unity in local fields 74. 10 N th power classes 76,11 Krasner s Lemma and applications 78. 12 Autoduality of locally compact fields 85,13 Structure theory of CDVFs 86.

Chapter 3 Adeles 95,1 Introducing the Adeles 95, 2 The Adelic Approach to Class Groups and Unit Groups 103. 3 Ray Class Groups and Ray Class Fields 113,Chapter 4 Complements and Applications 127. 1 Mahler series 127,2 Monsky s theorem 134,3 Linear groups over locally compact fields 140. 4 Cassels s embedding theorem 147,5 Finite matrix groups 150. Bibliography 155,Normed Fields and Valuation Theory.

1 Absolute values and valuations,1 1 Basic definitions. All rings are commutative with unity unless explicit mention is made otherwise. A norm on a field k is a map k R 0 satisfying,V1 x 0 x 0. V2 x y k xy x y,V3 x y k x y x y, Example 1 1 On any field k define 0 k R 0 by 0 7 0 x k 0 7 1. This is immediately seen to be a norm on k called the trivial norm In many. respects it functions as an exception in the general theory. Example 1 2 The standard norm on the complex numbers a bi a2 b2. The restriction of this to Q or to R will also be called standard. Example 1 3 The p adic norm on Q write b pn dc with gcd p cd 1 and. put ab p p n, It is straightforward to check that p is a norm on Q It will be more rewarding to. give a conceptual explanation Later we will see that we we can associate a norm. p to any nonzero prime ideal p in a Dedekind domain This hints that norms are. both plentiful and intimately related to classical algebraic number theory. Exercise 1 1 Let be a norm on the field k, a Show that the function d k k R by d x y x y is a metric.

b Reverse Triangle Inequality Show for all a b k a b a b 1. Exercise 1 2 Let R be a ring in which 1 6 0 Let R R 0 be a map. which satisfies V1 and V2 with k replaced by R above. a Show that 1 1, b Show that R is an integral domain hence has a field of fractions k. c Show that there is a unique extension of to k the fraction field of R satisfying. d Suppose that moreover R satisfies V3 Show that the extension of part b to k. satisfies V3 and hence defines a norm on k, 1This is the first of many points in the course where basic metric topology intervenes Because. students may be rusty on such things we prefer to err on the side of assuming too little rather. than too much familiarity with such topics,6 1 NORMED FIELDS AND VALUATION THEORY. e Conversely show that every integral domain admits a mapping satisfying. Exercise 1 3, a Let be a norm on k and let x k be a root of unity we have xn 1 for. some n Z Show x 1, b Show that for a field k the following are equivalent.

i Every nonzero element of k is a root of unity, ii The characteristic of k is p 0 and k Fp is algebraic. c If k Fp is algebraic show that the only norm on k is 0. In Chapter 2 we will see that the converse of Exercise 1 3c is also true a field that. is not algebraic over a finite field admits a nontrivial norm. Exercise 1 4 Let k be a normed field and let k k be a field. automorphism Define k R by x 7 x,a Show that is a norm on k. b Show that this defines a left action of Aut k on the set of all norms on k which. preserves equivalence, c Let d be a squarefree integer not equal to 0 or 1 Let k Q d viewed as a. subfield of C for specificity when d 0 we choose d to lie in the upper half. usual and let be the restriction of the standard valuation on C to k. Let d 7 d be the nontrivial automorphism of k Is Hint the. answer depends on d,1 2 Absolute values and the Artin constant. For technical reasons soon to be seen it is convenient to also entertain the fol. lowing slightly weaker version of V3 for C R 0 let V3C be the statement. V3C x k x 1 x 1 C, A mapping k R 0 satisfying V1 V2 and V3C for some C will be.

called an absolute value, For an absolute value on a field k we define the Artin constant Ck to be. the infimum of all C R 0 such that satisfies V3C,Exercise 1 5 Let be an absolute value on k. a Show that satisfies V3C for some C then C 1, b Let Ck be the Artin constant Show that satisfies V3Ck. c Compute Ck for the standard norm on C and the p adic norms on Q. Lemma 1 4 Let k be a field and an absolute value and C 1 Then. the following are equivalent,i x k x 1 x 1 C,ii x y k x y C max x y. Proof Assume i and let x y k Without loss of generality we may assume. that 0 x y Then xy 1 so xy 1 C Multiplying through by y gives. x y C y C max x y,1 ABSOLUTE VALUES AND VALUATIONS 7.

Now assume ii and let x k be such that x 1 Then,x 1 C max x 1 C max x 1 C. Lemma 1 5 Let k be a field and an absolute value with Artin constant C. Then is a norm iff C 2, Proof Let be a norm on k and let x k be such that x 1. x 1 x 1 x 1 1 1 2, Suppose C 2 Let x y k Without loss of generality we may assume. that 0 x y Then xy 1 so 1 xy C 2 Multiplying through by y. we get x y 2 y 2 max x y Applying this reasoning inductively we get. that for any x1 x2n k with 0 x1 x2n we have,x1 x2n 2n max xi. Let r be an integer such that n 2 2n Then,1 x1 xn x1 xn 0 0 2r max xi 2n max xi.

Applying this with x1 xn 1 gives that n 2n Moreover by replacing. the max by a sum we get the following weakened version of 17. x1 xn 2n xi, Finally let x y k be such that 0 x y Then for all n Z. X n i n i X n,x y x y 2 n 1 x i y n i,4 n 1 x i y n i 4 n 1 x y n. Taking nth roots and the limit as n gives x y x y,Why absolute values and not just norms. Lemma 1 6 Let k R 0 be an absolute value with Artin constant C. k R 0 x 7 x, a The map is an absolute value with Artin constant C. b If is a norm need not be a norm,Exercise 1 6 Prove Lemma 1 6.

This is the point of absolute values the set of such things is closed under the op. eration of raising to a power whereas the set of norms need not be. Moreover Lemma 1 6 suggests a dichotomy for absolute values We say an ab. solute value is non Archimedean if the Artin constant is equal to 1 the smallest. possible value Conversely if the Artin constant is greater than one we say that. the norm is Archimedean,8 1 NORMED FIELDS AND VALUATION THEORY. For example on k Q the p adic norm p is non Archimedean whereas the. standard absolute value is Archimedean with Artin constant 2. Exercise 1 7 Let be an absolute value on l and let k be a subfield of l. a Show that the restriction of to k is an absolute value on k. b If is a norm on l then the restriction to k is a norm on k. 1 3 Equivalence of absolute values, Two absolute values 1 2 on a field k are equivalent if there exists R 0. such that 2,1 When convenient we write this as 1 2. By a place on a field k we mean an equivalence class of absolute values 2 It. is easy to check that this is indeed an equivalence relation on the set of absolute. values on a field k Moreover immediately from Lemma 1 6 we get. Corollary 1 7 Each absolute value on a field is equivalent to a norm. Theorem 1 8 Let 1 2 be two nontrivial absolute values on a field k The. following are equivalent,i There exists R 0 such that 1 2. ii x k x 1 1 x 2 1,iii x k x 1 1 x 2 1,iv x k all of the following hold.

x 1 1 x 2 1,x 1 1 x 2 1,x 1 1 x 2 1, Remark This may seem like a strange way to organize the equivalences but it will. be seen to be helpful in the proof which we give following Ws Thm 1 1 4. Proof We shall show i ii iii iv i That i,ii and in fact all the other properties is clear. ii iii let x k be such that x 1 1 We must show that x 2 1 Since. 1 is nontrivial there exists a k with 0 a 1 1 and then by ii we have. 0 a 2 1 Then for all n Z xn a 1 1 so xn a 2 1 so x 2 a 2 n. Taking n to infinity gives x 2 1 We may apply the same argument to x 1. getting x 2 1, iii iv Choose c k such that 0 c 2 1 Then for sufficiently large n. x 1 1 x n1 c 1 1 1 x n2 c 2 1 x 2 1, So far we have shown iii ii As in the proof of ii iii we have. x 1 1 x 2 1 Moreover,x 1 1 1 1 2 1 x 2 1, 2Warning in more advanced valuation theory one has the notion of a K place of a field k.

a related but distinct concept In these notes we shall always use place in the sense just defined. 1 ABSOLUTE VALUES AND VALUATIONS 9,This establishes iv. iv i Fix a k such that a 1 1 Then a 2 1 so,We will show that 2. 1 For this let x k and put for i 1 2, It suffices to show 1 2 Let r q be a rational number with q 0 Then. r 1 p log a 1 q log x 1,ap 1 xq 1 p,p log a 2 q log x 2 2. Exercise 1 8 Let be an absolute value on a field k Show that. is Archimedean resp non Archimedean iff every equivalent absolute value is. Archimedean resp non Archimedean,1 4 Artin Whaples Approximation Theorem.

Theorem 1 9 Artin Whaples Let k be a field and 1 n be inequivalent. nontrivial norms on k Then for any x1 xn k and any 0 there exists. x k such that,i 1 i n x xi i,Proof Our proof closely follows A 1 4. Step 1 We establish the following special case there exists a k such that a 1 1. a i 1 for 1 i n, Proof We go by induction on n First suppose n 2 Then since 1 and 2 are. inequivalent and nontrivial by Theorem 1 8 there exist b c k such that b 1 1. b 2 1 c 1 1 c2 1 Put a cb, Now suppose the result holds for any n 1 norms so that there exists b k. with b 1 1 and b i 1 for 1 i n 1 Using the n 2 case there is c k. such that c 1 1 and c n 1, Case 1 b n 1 Consider the sequence ar cbr Then for all r Z we have. ar 1 1 while ar n 1 For sufficiently large r ar i 1 for all 2 i n so we. may take a ar,Case 2 b n 1 This time for r Z we put.

Then for i 1 and i n,br 1 br i c i,lim ar c i lim c i r. r r 1 b i r 1 br i,so for sufficiently large r we have. ar 1 c 1 1 and ar n c n 1,10 1 NORMED FIELDS AND VALUATION THEORY. On the other hand for 1 i n,ar i c i bi r 1, Therefore we may take a ar for sufficiently large r. Step 2 We claim that for any 0 there exists a k such that a 1 1 and. a i for 1 i n, Proof If b is such that b 1 1 and b i 1 for 1 i n then the computations.

of Step 1 show that we may take ar 1 b r for sufficiently large r. Step 3 Fix 0 By Step 2 for each 1 i n there exists ai k such that. a i 1 and for all j 6 i ai j Put A maxi j xi j Take. x a1 x1 an xn,x xi i ai xi xi i aj xj i A n 1 A nA. Thus taking nA does the job, Remark Theorem 1 9 also goes by the name weak approximation By any name. it is the most important elementary result in valuation theory playing a role highly. analogous to that of the Chinese Remainder Theorem in commutative algebra On. other hand when both apply the Chinese Remainder Theorem is subtly stronger. in a way that we will attempt to clarify at little later on. 1 5 Archimedean absolute values, In the land of Archimedean absolute values there is one theorem to rule them. all It is as follows, Theorem 1 10 Big Ostrowski Theorem Let k be a fi eld and an Archimedean. absolute value on k Then there exists a constant R 0 and an embedding. k C such that for all x k x x,In other words up to equivalence.

every Archimedean absolute value arises by embedding k into the complex numbers. and restricting the standard norm, Theorem 1 10 is a deep result every known proof from first principles takes several. pages It immediately implies all of the other results in this section and conversely. these results and more are used in its proof Indeed to prove Big Ostrowski it. is convenient to use aspects of the theory of completions so the proof is deferred. until Chapter 2, For a field k let Z 1 be the additive subgroup generated by 1 Recall that if. k has characteristic 0 then Z 1 is isomorphic to the integers whereas if k has. characteristic p 0 Z 1, Proposition 1 11 Let be an absolute value on a field k The following are. equivalent,i is non Archimedean,ii Z 1 is bounded, 3Thus either way Z 1 is a subring of k often called the prime subring. 1 ABSOLUTE VALUES AND VALUATIONS 11, Proof Since both conditions i and ii are unaffected by changing the abso.

lute value within its equivalence class we may and shall assume that is a norm. The implication i ii follows from the remark preceding the statement. of Proposition 1 11 Now suppose ii for specificity suppose that there exists. M 0 such that n 1 M for all n Z Let a b k and n Z Then. X n i n i X,a b n a b n ab M a i b n i M n 1 max a b n. Taking nth roots of both sides and letting n approach infinity gives the result. Corollary 1 12 An absolute value on a field of positive characteristic is non. Archimedean, Lemma 1 13 Ostrowski Lemma Every Archimedean absolute value on Q is. equivalent to the standard Archimedean norm, Proof Let be an Archimedean absolute value on Q Once again we may. Chapter 1 Normed Fields and Valuation Theory 5 1 Absolute values and valuations 5 2 Completions 21 3 Extending norms 37 4 The Degree in equality 51 5 Hensel s Lemmas 55 Chapter 2 Local Fields 59 1 Remedial number theory I Dedekind Kummer 59 2 Unrami ed extensions 61 3 Remedial Number Theory II Sch onemann Eisenstein 62 4 Totally

Recent Views:

- Livecode 8 0 1 release notes 5 19 16
- Quicker maths kopykitab
- Maintenance safety manual gitlab
- Nilai pengembaraan pada cerita rakyat roro kuning di desa
- El concilio de nicea la construcci n del hereje en el
- Mm caster camber plates 1994 04 mmcc9994
- System pro m library e abb com
- Long 2610 tractor parts manual jensales tractor manuals
- Manual service disconnect
- The washoe county child advocacy center quarterly newsletter