introduction to cyclotomic fields - l. washington (springer, 1982) ww

[...]... to be rather easy The cyclotomic reciprocity law has the same advantages As an example for the theorem, let p = 2 and n = 23 Then 2 11 1 (mod 23), so f = 11 Therefore 2 splits into two factors in 0 ( 23 ) But we already know that 2 splits as Y - in C1( \/ -2 3), where Y -= (2, (1 + \. /- 23) 7 2) (see Exercise 1.4) Going from 0(\/ -2 3) to 0( 23 ), Y and must remain prime Therefore (2) = (2, (1 + ,/ -. .. (mod p) Also x + C- l y = C - re, a (mod 15) = c - rg l a (mod p) since 5 — a and p =- 15 We obtain - - r(x + 07) = C r(x + C - 1 Y) (mod p) or x + cy _ 2rx _ 2r- 'y _ = 0 (mod p) (*) If 1, , c2r , v2r - 1 are distinct, then (since p > 5) Lemma 1.9 says that p ç divides x and y, which is contrary to our original assumptions Therefore, they are not distinct Since 1 0 C and C2r o c 2r- 1, we have three... Theorem 87 93 100 102 107 CHAPTER 7 Iwasawa's Construction of p-adic L-functions 7.1 Group rings and power series 7.2 p-adic L-functions 7.3 Applications 7.4 Function fields 7.5 p = 0 113 113 117 125 128 130 CHAPTER 8 Cyclotomic Units 8.1 Cyclotomic units 8.2 Proof of the p-adic class number formula 8.3 Units of0( p) and Vandiver's conjecture 8.4 p-adic expansions 143 143 151 153 160 CHAPTER 9 The Second... theory Therefore Y - Y il = Gal(K/L) L = Gal(L/CP)" lt follows that we have a one-one correspondence between subgroups of X and subfields of K given by Gal(K/L) I 4- L Y- fixed field of 1 7-1 - 23 3 Dirichlet Characters This gives us a one-one correspondence between all groups of Dirichlet characters and subfields of cyclotomic fields, since any two groups may be regarded as subgroups of some larger group... 02 (X) Similarly, 04 (X) is used to obtain primes of the form 4n + 1 Now we turn our attention to the splitting of primes in cyclotomic fields First we need the following useful result Lemma 2.12 Suppose p,i'n and let Y be a prime of() lying above p Then the nth roots of unity are distinct mod Y PROOF The result follows immediately from the equation n- 1 n= n 0 - w J-0 0 14 2 Basic Results Note that... completes the proof of the lemma CI Remark The units of Lemma 1.3 are called cyclotomic units and will be of great importance in later chapters Lemma 1.4 The ideal (1 — ) is a prime ideal of and (1 — OP -1 = (p) Therefore p is totally ramified in Q(C) PROOF Since Xi' + XP - 2 + • • + X + 1 = Fr= 1 1- (X — Ç), we let X 1 to obtain p = - 0 From Lemma 1.3, we have the equality of ideals (1 — = (1 — C) Therefore... cases: (1) 1 =- Cr We have from (*) that x + cy — x — c - 'y 0 (mod p), so, cy V - 'y 0 (mod p ) Lemma 1.9 implies that y =- 0 (mod p ) , contradiction (2) 1 _, 2r- 1 or, equivalently, C = c, 2r Equation (*) becomes — Y (x — y) Lemma 1.9 implies x — y - 0 (mod p), which contradicts the choice of x and y made at the beginning of the proof (3) c2r - 1 • Equation (*) becomes x— 0 (mod p), so x = =- 0 (mod... equal to - = (unit)(power of 1 — 1 .