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A Second Course in Algebraic Number Theory

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A Second Course in Algebraic Number Theory A second course in Algebraic Number Theory Vlad Dockchitser Prerequisites: • Galois Theory • Representation Theory Overview: ∗ 1. Number Fields (Review, K; OK ; O ; ClK ; etc) 2. Decomposition of primes (how primes behave in eld extensions and what does Galois's do) 3. L-series (Dirichlet's Theorem on primes in arithmetic progression, Artin L-functions, Cheboterev's density theorem) 1 Number Fields 1.1 Rings of integers Denition 1.1. A number eld is a nite extension of Q Denition 1.2. An algebraic integer α is an algebraic number that satises a monic polynomial with integer coecients Denition 1.3. Let K be a number eld. It's ring of integer OK consists of the elements of K which are algebraic integers Proposition 1.4. 1. OK is a (Noetherian) Ring 2. , i.e., ∼ [K:Q] as an abelian group rkZ OK = [K : Q] OK = Z 3. Each can be written as with and α 2 K α = β=n β 2 OK n 2 Z Example. K OK Q Z ( p p [ a] a ≡ 2; 3 mod 4 ( , square free) Z p Q( a) a 2 Z n f0; 1g a 1+ a Z[ 2 ] a ≡ 1 mod 4 where is a primitive th root of unity Q(ζn) ζn n Z[ζn] Proposition 1.5. 1. OK is the maximal subring of K which is nitely generated as an abelian group 2. O`K is integrally closed - if f 2 OK [x] is monic and f(α) = 0 for some α 2 K, then α 2 OK . Example (Of Factorisation). Z is UFD. When factorisation can only get dierent orders of factors and dierent signs. The latter come from the units in . ±1 Zp p p p may not even be a UFD, e.g., , , . OK K = Q( −5) OK = Z[ −5] 6 = 2 · 3 = (1 + −5)(1 − −5) To x this one works with ideals. 1 1.2 Units −1 Denition 1.6. A unit in a number eld K is an element α of OK with α 2 OK . The group of units is denoted ∗ . OK Example. ∗ K OK OK Q Z {±1g Qp(i) Zp[i] {±p1; ±ig n Q 2 Z[ 2] ±(1 + 2) : n 2 Z Dirichlet's Unit Theorem. Let be a number eld. Then ∗ is nitely generated. More precisely K OK ∗ r1+r2−1 OK = ∆ × Z where is the (nite) group of roots of unity in , is the number of distinct embeddings of into , the ∆ K r1 K R r2 number of pairs of complex conjugates into with image not in . (Hence ) K C R r1 + 2r2 = [K : Q] p Corollary. The only number elds with nitely many units are Q and Q( −D) for D > 0 integer. 1.3 Ideals Example. , , all multiples of . So if and only if for some . K = Q OK = Z a = (17) = 17 α 2 a α = 17n n 2 Z Multiplication: (3) · (17) = (51). Unique factorisation of ideals. Let Kbe a number eld. Every non-zero ideals of OK admits a factorisation into prime ideals. This factorisation is unique up to order. Denition 1.7. Let be two ideals. Then divides (written ) if for some ideal . (Equivalently a; bCOK a b ajb a·c = b c if in the prime factorisation, n1 nk m1 mk we have for all ) a = p1 ··· pk ; b = p1 : : : pk ni ≤ mi i Remark. 1. For , if and only if for some ∗ . α; β 2 OK (α) = (β) α = βu u 2 OK 2. For ideals a; b then ajb if and only a ⊇ b. p p 3. To multiply ideals, just multiply their generators, e.g. , p p (2) · (3) = (6) (2; 1 + −5) · (2; 1 − −5) = (4; 2 + 2 −5; 2 − 2 −5; 6) = (2) 4. To add ideals, combine their generators, e.g., (2) + (3) = (2; 3) = (1) = OK . Lemma 1.8. , Q ni , Q mi with and prime ideals. Then a; b C OK a = i pi b = i pi ni; mi ≥ 0 pi 1. Q max(ni;mi) (lowest common multiple) a \ b = i pi 2. Q min(ni;mi) (greatest common divisor) a + b = i pi Lemma 1.9. Let . Then there exists such that . α 2 OK n f0g β 2 OK n f0g αβ 2 Z n f0g Proof. Let n n−1 (with be the minimal polynomial of . Then n n−1 X + an−1X + ··· + a1X + a0 ai 2 Z) α α + an−1α + , so take n−1 n−2 . ··· + α1α = −a0 2 Z β = α + an−1α + ··· + a1 Corollary 1.10. If is a non-zero ideal, then is nite. a C OK [OK : a] Proof. Pick and with . Then and α 2 a n f0g β 2 OK N = αβ 2 Z n f0g N 2 a [OK : a] ≤ [OK :(α)] ≤ [OK : [K:Q] (N)] = [OK : NOK ] = jNj (By Proposition 1.4) Denition 1.11. The norm of a non-zero ideal is . a C OK N(a) = [OK : a] 2 Lemma 1.12. Let , then α 2 OK n f0g NK=Q(α) = N((α)) Proof. Let be a -basis for and for the -linear map . Then v1; : : : ; vn Z OK Tα : K ! K Q Tα(v) = αv NK=Q(α) = jdet Tαj = [hv1; : : : ; vni : hαv1; : : : ; αvni] = [OK :(α)] = N((α)) 1.4 Ideal Class Group Let K be a number eld. We can dene an equivalence relation on non-zero ideals of OK by a ∼ b if and only ∗ if there exists λ 2 K such that a = λb. The ideal class group of K denoted ClK , is the set of classes fnon-zero idealsg= ∼. It is a group, the group structure coming from multiplication of ideals. The identity is fprincipal idealsg and OK . Note that PID if and only if ClK = 1 if and only if UFD Theorem 1.13. ClK is nite 1.5 Primes and Modular Arithmetic Denition 1.14. A prime p in a number eld K is a non-zero prime ideal in OK Its residue eld is . OK =p = Fp Its residue characteristic, p, is the characteristic of OK =p. Its (absolute) residue degree is . fp = OK =p : Fp Lemma 1.15. The residue eld of a prime is indeed a nite eld. Proof. Let p be a prime, then OK =p is an integral domain. Furthermore OK =p = OK : p = N(p) which is nite by Corollary 1.10. Hence OK =p is a eld. Note. The size of the residue eld is N(p) Example. Let then . Let , then the residue eld . K = Q OK = Z p = (17) OK =p = Z=(17) = F17 Let then . Let , then and its representatives can be K = Q(i) OK = Z[i] p = (2+i) OK =p = F5 f0; 1; 1+i; 2i; 2i+ . Let , then (= ) 1g p = (3) OK =p = F9 F3[i] p hp i Let where , so . Let be a prime with residue characteristic , then K = Q( d) d ≡ 2; 3 mod 4 OK = Z d p p ( p hp i is generated by and the image of . Thus Fp if d a square mod p OK =p Fp d OK =p = Fp d = Fp2 else Notation. If is a non-zero ideal, we say that if . a C OK x ≡ y mod a x − y 2 a Theorem 1.16 (Chinese Remainder Theorem). Let K be a number eld and p ; : : : ; p be distinct primes. Then 1 k O = pn1 : : : pnk !O =pn1 × · · · × O =pnk K 1 k K 1 K k via x mod pn1 : : : pnk 7! x mod pn1 ; : : : ; x mod pnk is a ring isomorphism. 1 k 1 k Proof. Let : O !O =pn1 ×· · ·×O =pnk by x 7! x mod pn1 ; : : : ; x mod pnk . This is a ring homomorphism K K 1 K k 1 k k with ker = \k pni = Q pni (by Lemma 1.8). i=1 i i=1 i So it remains to prove that is surjective, so that we can apply the rst isomorphism theorem. By Lemma 1.8, n n p j + Q pni = O , so there α 2 p j and β 2 Q pni such that α + β = 1, now β ≡ 0 mod pni for all i 6= j and j i6=j i K j i6=j i i nj . So . This is true for all , hence is surjective. β ≡ 1 mod pj im 3 (β) = (0; 0;:::; 0; 1; 0;:::; 0) j 3 ni Remark. CRT implies that we can solve any system of congruences, i.e., x ≡ ai mod p for 1 ≤ i ≤ k. (This is called the Weak Approximation Theorem) i Lemma 1.17. Let be prime p C OK n n 1. OK =p = N(p) n n+1 2. p =p = OK =p as OK -module Proof. 2. implies 1: as n 2 n n+1 OK =p = OK =p · p=p ··· p =p = N(p)n 2.) By unique factorisation pn 6= pn+1, so pick π 2 pn n pn+1. Thus pnj (π) and pn+1 - (π). So (π) = pn · a with . So dene n+1 by n+1, an -map. Note that n+1 p - a φ : OK !OK =p φ(x) = πx mod p OK ker φ = x : p j (πx) = p ∼ n+1 n+1 n n+1 = n n+1 and im φ = (π) + p mod p = p mod p (by Lemma 1.8). Hence OK =p ! p =p . Corollary 1.18. N(ab) = N(a)N(b). Proof. Use Theorem 1.16 and Lemma 1.17. Lemma 1.19. N(a) 2 a Proof.
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