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 ∈ K α = β/n β ∈ OK n ∈ Z Example.

K OK Q Z ( √ √ [ a] a ≡ 2, 3 mod 4 ( , square free) Z √ Q( a) a ∈ Z \{0, 1} 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 ∈ OK [x] is monic and f(α) = 0 for some α ∈ K, then α ∈ 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 Z√ √ √ √ 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 ∈ OK . The group of units is denoted ∗ . OK Example. ∗ K OK OK Q Z {±1} Q√(i) Z√[i] {±√1, ±i} n Q 2 Z[ 2] ±(1 + 2) : n ∈ 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] √ 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 ∈ a α = 17n n ∈ 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 a|b 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 ∈ OK (α) = (β) α = βu u ∈ OK 2. For ideals a, b then a|b if and only a ⊇ b. √ √ 3. To multiply ideals, just multiply their generators, e.g. , √ √ (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 ∈ OK \{0} β ∈ OK \{0} αβ ∈ Z \{0} Proof. Let n n−1 (with be the minimal polynomial of . Then n n−1 X + an−1X + ··· + a1X + a0 ai ∈ Z) α α + an−1α + , so take n−1 n−2 . ··· + α1α = −a0 ∈ 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 ∈ a \{0} β ∈ OK N = αβ ∈ Z \{0} N ∈ a [OK : a] ≤ [OK :(α)] ≤ [OK : [K:Q] (N)] = [OK : NOK ] = |N| (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 ∈ OK \{0} 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) = |det Tα| = [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 ∈ K such that a = λb. The ideal class group of K denoted ClK , is the set of classes {non-zero ideals}/ ∼. It is a group, the group structure coming from multiplication of ideals. The identity is {principal ideals} 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 {0, 1, 1+i, 2i, 2i+ . Let , then (= ) 1} p = (3) OK /p = F9 F3[i] √ h√ 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 ( √ h√ 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 ∈ 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 ψ : 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 ψ = ∩k pni = Q pni (by Lemma 1.8). i=1 i i=1 i So it remains to prove that ψ 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 ∈ p j and β ∈ 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 ψ 3 ψ(β) = (0, 0,..., 0, 1, 0,..., 0) j ψ

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 ∈ pn \ pn+1. Thus pn| (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 | (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) ∈ a

Proof. N(a) is zero in any abelian group of order N(a), in particular in OK /a.

1.6 Enlarging the eld

Example. Consider Q(i)/Q. Take primes in Q and factorise them in Q(i). • (2) = (1 + i)2

• (3) remains prime • (5) = (2 + i)(2 − i)

We only see those three properties/behaviour in Q(i), so we say • 2 ramies

• 3 is inert • 5 splits

Note that p (prime of Q(i)) contains p = charZ[i]/p, so p|(p). Thus factorising 2, 3, 5, 7,... will yield all the primes of Z[i]. Denition 1.20. Let be an extension of number elds and an ideal. Then the conorm of is the ideal L/K aCOK a aOL of OL. I.e., the ideal generated by the elements of a in OL. Equivalently, if a = (α1, . . . , αn) as an OK -ideal, then aOL = (α1, . . . , αn) as an OL-ideal.

Note. (aOL)(bOL) = (ab)OL (aOM ) = (aOL)OM when K ≤ L ≤ M

Warning: Sometimes write a for aOL.

Proposition 1.21. Let L/K be an extension of number elds and a ∈ OK a non-zero ideal. Then N(aOL) = N(a)[L:K].

4 Proof. If a = (α) is principal, then by Lemma 1.12, we get

N(aOL) = NL/Q(α) [L:K] = NK/Q (α) = |N(a)|[L:K]

k In general, as ClK is nite, a = (α) for some k ≥ 1. Hence

k k N(aOL) = N(a OL) = N(ak)[L:K] = N(a)k[L:K]

[L:K] Hence N(aOL) = N(a) .

Denition 1.22. A prime q of L lies above a prime of K if q|pOL. (Equivalently if q ⊇ p as sets) Lemma 1.23. Let L/K be a number eld. Every prime of L lies above a unique prime of K. In fact q lies above q ∩ OK = p.

Proof. q ∩ OK is a prime ideal of OK and is non-zero as it contains N(q). So q lies above p = q ∩ OK . 0 0 If q also lies above p then q ⊇ p + p = OK 3 {1}, which is a contradiction. Lemma 1.24. Suppose lies above (primes). Then is a eld extension of with q C OL p C OK OL/q OK /p φ : OK /p ,→ OL/q given by φ(x mod p) = x mod q.

Proof. φ is well-dene as q ⊇ p and is a ring homomorphism, so has no kernel as OK /p is a eld. Hence φ is an embedding OK /p ,→ OL/q

Denition 1.25. If q lies above p then its residue degree is fq/p = [OL/q : OK /p]. eq /p Its ramication degree is the exponent, , in the prime factorisations Qn i . eq/p pOL = i=1 qi Theorem 1.26. Let L/K be an extension of number elds, p a prime of K.

Qm ei Pm 1. If pOL decomposes as pOL = q (with q distinct and ei = eq /p). Then eq /pfq /p = [L : K]. i=1 i i i i=1 i i

2. If M/L is a further extension, r lies above q, which lies above p, then er/p = er/qeq/p and fr/p = fr/qfq/p. Proof. 1.

[L:K] N(p) = N(pOL) Y = N( qei ) i i Y = N(q )ei i i Y = N(p)fiei i P = N(p) eifi

2. Multiplicativity for e is trivial. For f just apply the Tower law

5 m Denition 1.27. Let be extensions of number elds, a prime of with Q ei ( distinct). L/K p K pOL = i=1 q q Then: i i

• p splits completely in L if m = [L : K], i.e., ei = fi = 1 • p splits in L if m > 1 • p is totally ramied if m = f = 1, e = [L : K]

We'll see that when L/K is Galois, then ej = ei and fi = fj for all i, j. Then we say p is ramied if e1 > 1 and unramied if e1 = 1. Pseudo-example Let , , and 2 F = Q(i)/Q OF = Z[i] f(X) = X + 1 Observe:

(2) = (i + 1)2 X2 + 1 = (X + 1)2 mod 2 (5) = (i + 2)(i − 2) X2 + 1 = (X + 2)(X − 2) mod 5 (3) prime X2 + 1 mod 3 is irreducible

Theorem 1.28 (Kummer - Dedekind). Let L/K be an extension of number elds. Suppose OK [α] ≤ OL has nite index N , for some α ∈ OL with minimal polynomial f(X) ∈ OK [X]. Let p be a prime of K not dividing N (equivalently ). charOK /p - N If m Y ei f(X) mod p = gi(X) i=1 where g are distinct irreducible, then i m Y pO = qei L i i=1 with , where satisfy . The are distinct primes of with q = pOL + gi(α)OL g (α) ∈ OK [α] gi(x) = gi(x) mod p q L i and . i i eq /p = ei fq /p = deg gi(X) i i Example. Let , , . Take , so and 4 3 2 . K = Q L = Q(ζ5) OL = Z[ζ5] α = ζ5 N = 1 f(X) = X + X + X + X + 1 Now f(X) mod 2 is irreducible, hence (2) is prime in OL. 4 5 f(X) = (X − 1) mod 5, hence (5) = (5, ζ5 − 1) 2 2 , hence 2 2 . f(X) = (X + 5X + 1)(X − 4X + 1) mod 19 (19) = (19, ζ5 + 5ζ5 + 1)(19, ζ5 − 4ζ5 + 1)

Q(ζ5) ·••

e=1,f=4 F16 e=4,f=1 F5 e=2,f=2 F192 Q (2) (5) (19) Proof of Theorem 1.28. Claim. 1: are primes with q fq /p = deg qi i i Set , , . Use the map to dene a map A = OK [α] F = OK /p p = charF x 7→ α A/pA + g (α)A ← O [x]/(f(x), p, g (x)) i K i ∼ = F[x]/(f(x), gi(x)) ∼ = F[x]/(gi(x)) a eld of degree over , as is irreducible fi = deg gi F gi

6 Now pick M ∈ such that NM ≡ 1 mod p and consider φ : A/pA + g (α)A → O /q dened by φ(x) Z i L i mod (pA + g (α)A) → x mod q , it is well dened as q ⊇ pA + g (α)A. It is surjective since: if x ∈ O then i i i i L Nx ∈ A and M(Nx) = MNx mod q = x mod q (since MN ≡ 1 mod q ). Now O /q is non-zero: otherwise i i i L i 1 ∈ pOL + gi(α)OL, so both p and MN ∈ pA + gi(α)A, hence 1 ∈ pA + gi(α)A a contradiction as this is a proper ideal of A. Therefore O /q is a eld (hence q is prime) with degree f over . L i i i F Claim. 2: q 6= q for i 6= j i j As gcd(g (x), g (x)) = 1 can nd λ(x), µ(x) ∈ O [x] such that λ(x)g (x) + µ(x)g (x) = 1 mod p. Then q + q i j K i j i j contains both p and λ(α)g (α) + µ(α)g (α) = 1 mod p, hence q + q = O i j i j L Claim. pO = Q qei L i i

Y Y qei = (pO + g (α)O ) i L i L i i

Y ei ⊆ pOL + gi(α) OL

= pOL

Q ei as gi(α) ≡ f(α) ≡ 0 mod p. But

m Y Y e f N( qei ) = | | i i i F i=1 i deg f(x) = |F| [L:K] = |F| = N(pOL)

n Xp −1 Example. Let K = , L = (ζ n ), p prime, O = [ζ n ] and α = ζ n . Then N = 1, f(X) = . Now Q Q p L Z p p Xpn−1 −1 pn−pn−1 , hence is totally ramied in . f(X) = (X − 1) mod p (p) Q(ζpn )/Q If is also prime, then, working , pn d pn , hence pn has no repeated q 6= p mod q gcd(X − 1, dx (X − 1)) = 1 X − 1 roots in . In particular, has no repeated factors, so all are , i.e., is unramied in . Fq f(X) mod q ei 1 q Q(ζpn )/Q

Remark. Can't always nd α such that OL = OK [α]. But by Primitive Element Theorem, there exists α such that [OL : OK [α]] < ∞, so can decompose almost all primes. Proposition 1.29. Let be a nite extension, with and minimal polynomial . L/Q α ∈ OL L = Q(α) f(X) ∈ Z[X] If has distinct roots in (equivalently ) then is coprime to (so Kummer - f(X) mod p Fp p - discf [OL : Z[α]) p Dedekind applies) Proof. Let be a -basis for so β1, . . . , βn Z OL     1 β1  α1  β2   .   .   .  = M  .   .   .  n−1 α βn for some with . M ∈ Matn×n(Z) |det M| = [OL : Z[α]]

7 Let F be a splitting eld for f(X). Write σ1, . . . σn for the embeddings of L,→ F and αi = σ(α) for the roots of f(x). Then

Y 2 p - disc(f) = (αi − αj) i

α1 α2 ··· αn 2 n−1 2 = α1 α2 αn ...... n−1 n−1 n−2 α1 α2 ··· αn   2 σ1(β1) ··· σn(β1) . .   . .  = det M  . .  σ1(βn) ··· σn(βn) = [OL : Z[α]] · B for some , hence . B ∈ OF \{0} p - [OL : Z[α]] Proposition 1.30. Let L/K be a nite extension of number elds, p a prime of K. Suppose L = K(α) for some α ∈ OL satisfying an Eisenstein minimal polynomial with respect to p, i.e.,

n n−1 f(X) = X + an−1X + ··· + a1X + a0 with and 2 . Then is totally ramied in p|(ai) p - (a0) p L/K Proof. Omitted

2 Decomposition of primes 2.1 Action of the Galois group

Let F/K be a Galois extension of number elds: • Gal(F/K) = Aut(F/K). • F/K is normal (is f(X) ∈ K[X] is irreducible and acquire a root in F then f splits completely) • |Gal(F/K)| = [F : K]

• H < Gal(F/K) → F H , Gal(F/L) ← K ≤ L ≤ F a 1-1 correspondence √ 3 Q(ζ3, 2)

C2 C3 √ 3 Q( 2) S3 Q(ζ3)

not Galois C2∼=S3/C3 Q

Lemma 2.1. Let g ∈ Gal(F/K):

1. α ∈ OL then gα ∈ OF

8 2. is an ideal, then ideal a ∈ OF g(a) C OF 3. Let be ideals, then , . a, b C OF g(a · b) = g(a)g(b) g(a + b) = g(a) + g(b) If q is a prime of F above p, a prime of K, then 4. g(q) is a prime of F above p (so Gal(F/K) permutes the primes above p)

5. eq/p = eg(q)/p and fq/p = fg(q)/p Proof. Clear Example. Let , , then and . K = Q F = Q(i) OF = Z[i] Gal(F/K) = C2 = {id, complex conjugation} Consider (1 + i), it is xed by Gal(F/K). (3) is also xed by Gal(F/K). But (2 + i) and (2 − i) are swapped by Gal(F/K). Theorem 2.2. Let F/K be a Galois extension of number elds, let p be a prime of K. Then Gal(F/K) act transitively on the primes of F above p. Proof. Let q , . . . , q be the primes above p. We need to show that there exists g ∈ Gal(F/K) such that g(q ) = q . 1 n 1 2 Pick x ∈ OF such that x ≡ 0 mod q1 but x 6≡ 0 mod qi for all i 6= 1. This is possible by the Chinese Remainder Theorem (Theorem 1.16). Then Q h(x) ∈ q ∩ O = p ⊆ q . So for some g, g(x) ≡ 0 mod q , hence h∈Gal(F/K) 1 K 2 2 x ≡ 0 mod g−1(q ). Therefore g−1(q ) = q by choice of x. I.e., q = g(q ). 2 2 1 2 1

Corollary 2.3. Let F/K be a Galois extension. If q , q lie above p, then eq /p = eq /p and fq /p = fq /p. 1 2 1 2 1 2

Hence we can write epand fp in the case of Galois extensions

Example. Suppose Gal(F/K) = S and p splits into q , q , q , q in F , with the usual action of S on 4 elements 4 1 2 3 4 4

F ···· H

K ·p

n o n o n o Say H = {id, (12)(34)} ∼ C , L = F H . H-orbits of q , . . . , q are q , q and q , q , so there exists 2 primes = 2 1 4 ! 2 3 4 r , r in L above p.(r splits into q and q in F and r splits into q and q ) 1 2 1 1 2 2 3 4 2.2 Decomposition Group Notation. If is prime of , write . p K Fp = OK /p Denition 2.4. Let F/K be a Galois extension of number elds, q a prime of F above p, a prime of K. The decomposition group Dq = Dq/p of q (over p) is Dq/p = StabGal(F/K)(q) Remark. xes so it acts on by . This gives a natural map g ∈ Dq q Fq x mod q 7→ g(x) mod q Dq → Gal(Fq/Fp)

9 Example. Let K = Q, F = Q(i). Let p = 3 and q = (3), complex conjugations xes q and acts as a + bi 3 . I.e., exactly as the frobenius automorphism 3 on . mod (3) 7→ a − bi mod 3 = (a + bi) mod 3 x → x Fq Theorem 2.5. Let F/K be a Galois extension of number elds, q a prime of F above p, a prime of K. Then the natural map is surjective. Dq → Gal(Fq/Fp) Proof. Pick such that . Let be its minimal polynomial and β ∈ Fq Fp(β) = Fq f(x) ∈ Fp[x] β = β1, β2, . . . , βn ∈ Fq be its roots (in as is Galois). Since determines so is suces to prove that there exists Fq Fq/Fp g(β) g ∈ Gal(Fq/Fp) g ∈ Dq with g(β) = g(β2). 0 0 Pick α ∈ OK with α ≡ β mod q and α ≡ 0 mod q for all other primes q above p (possible by CRT Theorem 1.16). Let F (X) ∈ OK [X] be its minimal polynomial over K, and α = α1, α2, . . . , αm ∈ OF be its roots (in F as F/K is Galois). F (X) mod q has β as a root, hence f(x) divides F (X) mod q, hence β2 also is a root of F (X) mod q. Without loss of generality α2 mod q = β2. Take g ∈ Gal(F/K) with g(α) = α2. Then g(α) 6= 0 mod q, hence g(q) = q so g ∈ Dq, and g(β) = β2

Corollary 2.6. Let K be a number eld, F/K the splitting eld of a monic irreducible f(x) ∈ OK [x], of degree n. Suppose for some prime of , with be distinct irreducible with p K f(x) mod p = g1(x)g2(x) . . . gk(x) gi(x) ∈ Fp[x] deg gi = di. Then Gal(F/K) ⊆ Sn contains an element of cycle type (d1, d2, . . . , dk)

Proof. Let α1, . . . , αn ∈ OK be the roots of f(x). Then βi = αi mod q (q any prime above p) are precisely the roots of in . Their generator of permutes the with cycle type . Hence its f(x) mod p Fq Gal(Fq/Fp) βi (d1, . . . , dk) lift to Dq ≤ Gal(F/K) has the claimed cycle type.

Denition 2.7. Let F/K be Galois, q a prime above p. The inertia subgroup Iq = Iq/p is the (normal) subgroup of that acts trivially on , i.e., . Dq Fq Iq = ker(Dq → Gal(Fq/Fp)) Note that as the amp is surjective ∼ . The latter group is cyclic and is generated by the Dq/Iq = Gal(Fq/Fp) frobenius map φ : x → x#Fp .

Denition 2.8. The (arithmetic) Frobenius element Frobq/p is the element of Dq/Iq that maps to φ.

Theorem 2.9. Let F/K be Galois extensions of number eld, q a prime of F above p, a prime of K. Then

1. Dq/p = eq/p · fq/p

2. Order of Frobq/p is fq/p

3. Iq/p = eq/p If K ≤ L ≤ F and s is above p, below q

4. Dq/s = Dq/p ∩ Gal(F/L)

5. Iq/s = Iq/p ∩ Gal(F/L) Proof.

1. Let n = #primes above p . Then

n · Dq/p = |Gal(F/K)| (orbit − stabiliser) = [F : K] X = eifi (Theorem 1.26)

= neq/pfq/p

2. order of fq/p = [Fq : Fp] = Gal(Fq/Fp) = Frobq/p

3. order of , hence Dq/p = Iq/p · Frobq/p Iq/p = eq/p 4. and 5. Just from denition.

Example. √ √ Q( 2, 3) = F

√ √ √ Q( 2) Q( 3) Q( 6)

Q Now 2 ramies in all three quadratic elds:

√ 2 • (2) = 2 • (x2 − 3) = (x + 1)2 mod 2 • (x2 − 6) = x2 mod 2 √ and use Kummer - Dedekind. Let in , hence , so , so contains for some . So q F e ≥ 2 Iq ≥ 2 Iq Gal(F/Q d ) d √ the prime above in is ramied, so and . 2 F/Q( d) eq = 2 · 2 = 4 Iq = C2 × C2 Example. Let , . Let be a prime, a prime of above . We know that is unramied, K = Q F = Q(ζn) p - n q F p p D E so and . Now p and hence p as i are distinct Iq/p = {id} Dq/p = Frobq/p Frobq/p(ζn) ≡ ζn mod q Frobq/p(ζn) = ζn ζn in . (Since n has distinct roots). In particular, and order of order of in Fq x − 1 mod p eq/p = 1 fq/p = Frobq/p = p (Z/nZ)∗. 2.3 Counting primes

Lemma 2.10. Let F/K be a Galois extension of number elds. 1. primes of K are in 1-1 correspondence with Gal(F/K)-orbits on primes of F via p ↔primes of F above p.

2. If q lies above p then gDq 7→ g(q) is a Gal(F/K) set isomorphism from {primes above p} to G/Dq

−1 −1 −1 3. Dg(q) = gDqg , Ig(q) = gIqg and Frobg(q)/p = gFrobq/pg . Proof. 1. is from transitivity of the Galois action while 2. and 3. are elementary check

Corollary 2.11. Let F/K be Galois, K ≤ L ≤ F , G = Gal(F/K),H = Gal(F/L). Then

{primes of L above p} ↔ H−orbits on primes of F above p ↔ H\G/Dq = {HgDq} via s 7→elements that sent q to some prime above s. Proposition 2.12. Let F/K be Galois extension of number elds, L = K(α) an intermediate eld, G = Gal(F/K) and H = Gal(F/L). Let X = {roots of min poly of α} =∼ {embeddings L,→ F } = H\G a G-set of size [L : F ]. 1:1 Then {primes of L above p} ↔ Dq-orbits on X with

11 • size Dq-orbits= es/p · fs/p

• size Iq-suborbits = es/p

• number Iq-suborbits= fs/p

Explicitly, s 7→Orbits of g−1(α) where g(q) lies above s. √ Example. Let , 5 , , Let be a prime of above K = Q F = Q(ζ5, 2) p = 73 q F p

√ 5 L = Q( 2)

Now p is unramied in F with residue degree 4 (use Kummer - Dedekind on L/ and (ζ )/ and that Gal(F/ ) √ √ Q Q 5 Q Q has no element of order ). Now permutes 5 5 transitively, , hence -orbits are 20 Gal(F/Q) 2, ζ 2,... eq/73 = 1 Iq √ trivial. hence , ∼ (generated by ). Without loss of generality xes 5 and permutes fq = 4 Dq = 4 Dq = C4 Frob Dq 2 the rest cyclically. Hence there are 2 primes in L above 73 with residue degree 1 and 4 and ramication degrees 1 and 1. Proof of Theorem 2.12. We have

{primes of L above p} ↔ H\G/Dq

↔ Dq−orbits on H\G =: X

−1 (Dq acts by d(Hg) = Hgd ). Size of Dq-orbits of

Dq −1 g (α) = Stab (g−1(α)) Dq

Dq = Stab −1 (α) gDq g

Dq = |gDg−1 ∩ H|

Dq =

Dg(q)/p

eq/p · fq/p = eq/s · fq/s

= es/p · fs/p

Similarly, size of Iq-orbits is es/p (same calculations as above, replacing Dq with Iq). And hence #Iq -suborbits es/p·fs/p is = fs/p. es/p

12 2.4 Representation of the decomposition group

Let F/K be a Galois Extensions of number elds. Let G = Gal(F/K), p a prime in OK , q a prime above p in OF . Let D = Dq/p and I = Iq/p, Frob = Frobq/p.

Denition 2.13. A representation V of D is unramied if I acts trivially on V , therefore V I = V (F/K is unramied if and only if all V are unramied) Notation. For a th -root of unity , dene the representation ∗ such that fq/p ζ Ψq/p,ζ = Ψζ : D → C = GL1(C) Ψζ (h) = 1 k ( ), . Thus k if acts as (#Fp) on . h ∈ I Ψζ (Frob) = ζ Ψζ (g) = ζ g x 7→ x Fq

I Lemma 2.14. If V is a irreducible representation of D then either V = 0 or V is unramied and V = Ψζ for some ζ with ζfq/p = 1.

I I I Proof. Since I C D it follows that V is a subrepresentation of V . Then either V = 0 or V = V .In the latter case, the action of D factors through D/I = hFrobi and hence V is 1-dimensional and V = Ψζ for some ζ. Notation. If q0 = g(q) is another prime above p for some g ∈ G and (ρ, V ) a representation of D, we write (ρg,V g) g −1 ∼ for the corresponding representation of Dq//p given by ρ (h) = ρ(ghg ). Clearly Dq/p = Dq0/p as groups. ∼ ∼ Example. Let G = S4, D = D8 and I = C4. Then there are |G/D|-prime above p. D8 = Dq/p. Others are the two other subgroup of S4 isomorphic to D8.

D8 e (1234) (13)(24) (12)(34) (13) 1 1 1 1 1 1 1 1 1 1 −1 1 The 2-dimensional irreducible representation of Dq0/p is ob- 2 1 −1 1 1 −1 3 1 −1 1 −1 1 ρ 2 0 −2 0 0 tained as ρ0(h) = ρ(g−1hg). Lemma 2.15.

1. If 0 another prime over then g q = g(q) p Ψq0/p,ζ = Ψq/p,ζ

2. If is an intermediate eld, is a prime below then f whence . In particular L Σ q ResDq/Σ Ψζ = Ψq/Σ,ζ f = fΣ/p fΣ/p ResDq/Σ Ψq/p,ζ = I ⇐⇒ ζ = 1 Proof.

−1 −1 −1 1. Follows from Dq0/p = gDg , Iq0/p = gIg and Frobq0/p = gFrobg

f 2. sends to and to f because acts as (#Fp) on . ResDq/Σ Ψq/p,ζ Iq/Σ 1 Frobq/Σ ζ Frobq/Σ x → x Fq

Proposition 2.16. Let K ⊆ L ⊆ F , H = Gal(F/L)

q − F − i

Σ G L −

p − K

13 Let {Σ } be the set of primes of L above p and pick q = g (q) above Σ for each i. For V a representation of H i i i i

g−1 M Dq /p i ResG IndG V =∼ Ind i ResH V . D H Dq /Σ Dq /Σ i i i i Σi In particular D G E X Ψζ , ResDIndH V = Ψ fΣ /p , ResDq ,Σ V . q /Σi,ρ i i i i Σi Proof. The main claim is precisely Mackey's formula for H,D ≤ G. The second claim then follows from

D g−1 E D E Ψζ , (IndResV ) i = Ψq /p,ζ , IndResV i Dq/p Dq /p i D E = ResΨq /p,ζ , ResV by Frobenius reciprocity i Dq /Σ i i

= Ψ fΣ /p , ResV q /Σi,ζ i i

th Corollary 2.17. Let ζ be a primitive n root of unity with n|fq/p. Then K ⊆ L ⊆ F with H = Gal(F/L). The D E number of primes of above G G L p = Ψζ , ResDIndJ I P fΣ/p Proof. By previous proposition, the Right Hand Side is Ψ f , = #Σ with ζ = 1 Σ above p q0/Σ0,ζ Σ/p I

3 L-series Aim:

1. If gcd(a, n) = 1 then there are innitely many primes p ≡ a mod n

2. If f(x) ∈ Z[x] monic and f(x) mod p has a root mod p for all p, then f(x) is reducible. Denition 3.1. An ordinary Dirichlet series is a series P∞ −s ( ). f(s) = s=1 ann an ∈ C, s ∈ C Convention: s = σ + it.

Convergence Property Lemma 3.2 (Abel's Lemma). PM PM−1 Pn PM n=N anbn = n=N ( k=N ak)(bn − bn+1) + k=N ak bM Proof. Elementary rearrangement (c.f., integration by part)

∞ Proposition 3.3. Let P −λns for an increasing sequence of the real number f(s) = n=1 ane λn → ∞ 1. If the partial sums PM are bounded, then the series converges locally uniformly of to an analytic N an Re(s) > 0 functions

2. If the series f(s) converges for s = s0, then it converges uniformly on Re(s) > Re(s0) to an analytic function.

Note. Dirichlet series are the case λn = log n

14 Proof. Note that 1. implies 2. by the change of variables 0 and 0 −λns0 . The new series converges s = s − s0 an = e an at and so must have PM 0 bounded 0 N an For 1. we will show uniform convergence on −A < arg(s) < A and Re(s) > δ. This will suce as the uniform limit of analytic functions is analytic and these regions cover Re(s) > 0. −λns Let > 0, nd N0 such that n > N0, e < in this domain. Now compute: for N1M ≥ N0

M n−1 n ! M ! X X X X −λns −λns −λn+1s −λM s ane = ak e − e + ak e n=N n=N k=N N M−1 X −λ s −λ s ≤ B e n − e n+! + B n=N where B is the bound for the partial sums. Observe:

β −αs −βs −xs e − e = s e dx ˆα β ≤ |s| e−xσdx ˆα |s| = (e−ασ − e−βσ) σ where σ = Re(s) for α > β. So

M M−1 X |s| X a e−λns ≤ B e−λnσ − e−λn+1σ + B n σ n=N N |s| ≤ B + B σ ≤ (BK + B) where s in our domain. σ ≤ K

P −λns Proposition 3.4. Let f(s) = ane for λn → ∞ an increasing sequence of positive reals. Suppose

• an ≥ 0 is real

• f(s) converges on Re(s) > R (∈ R) and hence is analytic • it has an analytic continuation to a neighbourhood of s = R Then f(s) converges on Re(s) > R − for some > 0.

Proof. Again, we can assume R = 0. As f is analytic on Re(s) > 0 and |s| < δ. Hence f is analytic on |s − 1| ≤ 1 + . The Taylor Series of f around s = 1 converges on all of |s − 1| ≤ 1 + . In particular ∞ ∞ P 1 k k (k) converges. For (k) P k −λn (ok since uni- f(−) = k=0 k! (−1) (1 + ) f (1) Re(s) > 0 f (s) = n=1 an(−λn) e

15 ∞ form convergence). Hence (k) P k −λn . Hence (−1)f (s) = n=1 anλne ∞ ∞ X 1 X f(−) = (1 + )k a λk e−λn k! n n k=0 n=1 X 1 = a λk e−λn e−λn (1 + )k as all terms positive n n k! k,n ∞ X −λn −λn(1+) = ane e n=1 ∞ X λn = ane n=1 is a convergent series for f converges at s = −. This implies the result. Theorem 3.5. 1. If are bounded, then P −s converges absolutely on to an analytic function. an n≥1 ann Re(s) > 1 2. If the partial sums PM are bounded then P −s converges absolutely on to an analytic N an ann Re(s) > 0 function. Proof. ∞ 1. an 1 where and P 1 does converge for real. Analytic comes from Proposition 3.3 ns ≤ k nσ σ = Re(s) n=1 nx x > 1 2. See Proposition 3.3

P −s P −s Remark. If ann and bnn converges on Re(s) > σ0 to the same function f(s), then an = bn for all n.

3.1 Dirichlet L-functions

Denition 3.6. Let N ≥ 1 be an integer ψ :(Z/NZ)∗ → C∗ a group homomorphism. Extend ψ to a function on ( ψ(n mod N) gcd(n, N) = 1 Z by ψ(n) = . Such a function is called a Dirichlet character modulo N. 0 else Its -series (or -function) is P∞ −s. L L LN (ψ, s) = n=1 ψ(n)n Lemma 3.7. Let ψ be a Dirichlet character modulo N. Then 1. ψ(a + N) = ψ(a) (so ψ is periodic) 2. ψ(ab) = ψ(a)ψ(b) (so ψ is strictly multiplicative) 3. The L-series for ψ converges absolutely on Re(s) > 1 and satises the Euler product Y 1 L (ψ, s) = . n 1 − ψ(p)p−s p prime

Proof. 1. and 2. are clear from the denition. 3. the L-series coecients an = ψ(n) are bounded, so absolute convergence follows from Theorem 3.5. For Re(s) > 1 X Y ψ(n)n−1 = (1 + ψ(p)p−s + ψ(p)2p−2s + ... ) (by 2. and abs conv) p prime Y 1 = (geometric series) 1 − ψ(p)p−s p prime

16 ∗ Remark. The case ψ(n) = 1 for all n ∈ (Z/NZ) gives the trivial Dirichlet character modulo N. In this case Y 1 L (ψ, s) = N 1 − p−s p prime Y = ζ(s) · (1 − p−s) p|N, prime where ζ(s) is the Riemann ζ-function. Example. Take , so ∗ ∼ , and , , and . Then N = 10 (Z/NZ) = {1, 3, 7, 9} = C4 ψ(1) = 1 ψ(3) = i ψ(7) = −i ψ(9) = −1 i i 1 1 i i 1 . Note that by the alternating series test (applied to the L10(ψ, s) = 1 + 3s − 7s − 11s + 11s + 13s − 17s − 19s + ... real part and imaginary part separately) implies convergence on s > 0 real. Theorem 3.8. Let N ≥ 1 and ψ :(Z/NZ)∗ → C∗.

1. If ψ is trivial then LN (ψ, s) has an analytic continuation to Re(s) > 0 except for a simple pole at s = 1

2. If ψ is non-trivial, then LN (ψ, s) is analytic on Re(s) > 0 Proof. 1. Follows from the previous remark and that ζ(s) has an analytic continuation to Re(s) > 0 with a simple pole at s = 1. PA+N−1 P PB 2. ψ(n) = ∗ ψ(n) · 1 = ψ(N) hψ, i = 0 as ψ 6= 1. So the partial sums ψ(n) are bounded n=A n∈(Z/NZ) I A and the result follows from Theorem 3.5ii)

Theorem 3.9. Let ψ be a non-trivial Dirichlet character modulo N. Then LN (ψ, 1) 6= 0. Proof. Let Y ζN (s) = LN (χ, s) ∗ ∗ χ:(Z/NZ) →C If then has an analytic continuation to , the pole form having been killed by LN (ψ, 1) = 0 ζN (s) Re(s) > 0 LN (I, s) the zero on LN (ψ, s). We'll show that this is not the case (and hence has a simple pole at s = 1) On Re(s) > 1, ζN (s) has the absolutely convergent expression Y Y 1 ζ (s) = N 1 − χ(p)p−s χ p-N Y Y 1 = 1 − χ(p)p−s p-N χ Y 1 = (1 − p−fps)φ(N)/fp p-N where is the Euler totient function and the order of in ∗. Hence φ fp p (Z/NZ) φ(N)/f Y −fps −2fps p ζN (s) = 1 + p + p + ... . p-N This is a Dirichlet series with positive real coecient so if it has an analytic continuation to Re(s) > 0, it must converge there by Proposition 3.4. But for s > 0 real it dominates

Y −φ(N)s −2φ(N)s (1 + p + p + ... ) = LN (I, φ(N)s) p-N which diverges for s = 1/φ(N).

17 3.2 Primes in Arithmetic Progression Strategy: P −s P P −s with . This is approximately P p≡a mod N p = χ λχ p χ(p)p λI 6= 0 χ λχ log LN (χ, s) ∼ 1 as . λI log s−1 → ∞ s → 1 Proposition 3.10. Let ψ be a Dirichlet character modulo N

n 1. The Dirichlet series P ψ(p) −ns converges on so it is an analytic function and denes p prime,n≥2 n p Re(s) > 1 (a branch of) log LN (ψ, s) there.

n 2. If is non-trivial then P ψ(p) −ns is bounded as . If then P 1 −ns 1 as . ψ p,n n p s → 1 ψ = I p,n n p ∼ log 1−s s → 1 Proof.

1. The series has bounded coecients so converges absolutely on Re(s) > 1 to an analytic function by Theorem 3.5 1. For a xed s with Re(s) > 1

X ψ(p)n X (ψ(p)p−s)2 p−ns = ψ(p)p−s + + ... n 2 p,n p X 1 x2 = log (branch with log(1 + x) = x − + ... ) 1 − ψ(p)p−s 2 p Y 1 = log (possibly a diﬀ branch) 1 − ψ(p)p−s p

= log LN (ψ, s)

n Hence P ψ(p) −ns gives an analytic branch of on . p,n n p log LN (ψ, s) Re(s) > 1 2. By Theorem 3.8, for , converges to a non-zero value as , so is bounded as ψ 6= I LN (ψ, s) s → 1 log LN (ψ, s) . For has a simple pole at (hence λ ) so 1 as . s → 1 LN (I, s) s = 1 ∼ s−1 log LN (I, s) ∼ log s−1 s → 1

Corollary 3.11. If is non-trivial then P −s is bounded as . If , then P −s ψ p prime ψ(p)p s → 1 ψ = I p prime ψ(p)p = P p−s ∼ log 1 as s → 1 and in particular tends to ∞ as s → 1. p-N s−1 n Proof. P −s P ψ(p) −ns so it suces to check that the last term is bounded on p ψ(p)p = log LN (ψ, s) − p,n≥2 n p Re(s) > 1.

n X ψ(p) −ns X 1 p ≤ n n n |ps| p,n≥ X 1 ≤ pn p prime,n≥2 X 1 = p(p − 1) p X 1 ≤ < ∞ k2 k≥1

Dirichlet's Theorem on primes in Arithmetic Progression. Let a, N be coprime integers. Then there are P 1 1 1 innitely many primes p ≡ a mod N. Moreover if P is the set of these primes then −s ∼ log a,N p∈Pa,N p φ(N) s−1 as s → 1.

18 Proof. The rst statement follows from the second as 1 as . Consider the (class-)function log s−1 → ∞ s → 1 ( 1 if n = a ∗ ∗ with . Then Ca,N :(Z/NZ) → C Ca,N (n) = 0 else

1 X hC , χi = C (n)χ(n) a,N φ(N) a,N ∗ n∈(Z/NZ) 1 = χ(a) φ(N) so P χ(a) by Character theory. Hence Ca,n = χ φ(N) χ X 1 X = C (p)p−s ps a,n p∈Pa,N p prime X χ(a) X χ(p) = ψ(N) ps p By Corollary 3.11, each term on RHS is bounded as s → 1 except for χ = I, and X (p) 1 X 1 I I = φ(N) ps φ(N) ps p-N 1 1 ∼ log φ(N) s − 1 as s → 1.

3.3 Artin L-functions Denition 3.12. Let F/K be a Galois extension of number elds and ρ a Gal(F/K)-represnetation. The Artin L-function of ρ is dened by the Euler product L(F/K, ρ, s) = L(ρ, s) Y 1 = −s Pp(ρ, N(p) ) p prime of K

Ip where the local polynomial of ρ at p, Pp(ρ, T ) is dened by Pp(ρ, T ) = det(1 − FrobpT |ρ ).(= det(I − MT ) where Ip M is the matrix by which Frobp acts on ρ ). Note that here Ip = Iq/p and Frobp = Frobq/p for some (Any) prime q above p. Example. Take , arbitrary, . Then for all . Hence Q 1 . • K = Q F ρ = I Pp(I, t) = det(1−1·t) = 1−t p L(I, s) = p 1−p−s = ζ(s) Similarly, for general and we get . Then Q 1 (The • K ρ = I Pp(I, t) = 1 − t∀p L(F/K, I, s) = p 1−N(p)−s = ζK (s) Dedekind ζ-function of K) , non-trivial -dimensional representation. Then • K = Q,F = Q(i) ρ : Gal(F/K) = C2 → {±1} 1  1 p = 2 ⇒ ρIp = 0  Pp(ρ, T ) = 1 − t p ≡ 1 mod 4 ⇒ Frobp = id  1 + t p ≡ 3 mod 4 ⇒ Frobp 6= id Hence Q 1 Q 1 Q 1 where ∗ is the non-trivial L(ρ, s) = p≡1 mod 4 1−p−s p≡3 mod 4 1+p−s = p 1−χ(p)p−s χ : Z → C Dirichlet character mod 4. That is L(ρ, s) = L4(χ, s) a Dirichlet L-function.

19 Lemma 3.13. The local polynomial Pp(ρ, T ) is independent of the choice of q above p and of the choice of Frobp.

Proof. For a xed q above p , independence of the choice of Frobp is clear. Since two choices dier by an element Ip Ip Ip σ ∈ Ip, which acts trivially on ρ . Hence det(1 − Frobpt|ρ ) = det(1 − σFrobpt|ρ ). 0 If q = g(q) is another prime above p, then the matrix of ρ(d) for d ∈ Gal(F/K) with respect to a basis {ei} is −1 −1 the same as that of ρ(gdg ) with respect to a basis {gei}. As d → gdg maps Dq/p to Dq0/p, Iq/p to Iq0/p and Frobq/p to Frobq0/p the result follows. Remark. If dim ρ = 1 then ( I 1 − ρ(Frobp)t if ρ = ρ Pp(ρ, t) = 1 if ρI = 0

Ip 2 n In general it is essentially the characteristic polynomials of Frobp on ρ : If Pp(ρ, t) = 1 + a1t + a”t + ··· + ant n n−1 then characteristic polynomials is t + a1t + ··· + an. 2 Remark. The polynomial Pp(ρ, T ) has the form 1 − (aT + bT + ... ) so (ignoring convergence questions) 1 = 1 + (aT + bT 2 + ... ) + (aT + bT 2 + ... )2 + ... Pp(ρ, T )

= 1 + apT + ap2 T + ... for some . Formally substituting this to the Euler product gives the Artin -series for . api ∈ C L ρ

Y −s −2s L(ρ, s) = (1 + apN(p) + aps N(p) + ... ) p

X −s = anN(n) n non−zero ideal of K for some . Note that grouping ideals of equal norm yields an expression for as an ordinary Dirichlet an ∈ C L(ρ, s) series. Lemma 3.14. The L-series expression for L(ρ, s) agrees with the Euler product on Re(s) > 1 where they converge absolutely to an analytic function. Proof. It suces to prove that Q −s (as in the previous remark) converges absolutely on p prime of K (1+apN(p) +... ) Re(s) > 1. This justies rearrangement of the terms, and the expression as an ordinary Dirichlet series for L(ρ, s) then proves analytically (Proposition 3.3). The polynomial factors over as Pp(ρ, T ) C Pp(ρ, T ) = (1 − λ1T )(1 − λ2T ) ... (1 − λkT ) for some k ≤ dim ρ with all |λi| = 1. So the coecients of 1 1 = Q Pp(ρ, T ) (1 − λiT ) 2 = 1 + apT + ap2 T + ... are bounded in absolute value by those of 1 = (1 + T + T 2 + ... )dim ρ (1 − T )dim ρ Hence

Y X −s Y 1 apn N(p) ≤ −s dim ρ p n p 1 − N(p) dim ρ Y 1 ≤ (pabove p) 1 − |p−s| p = ζ(σ)dimρ[K:Q] where σ = Re(s).

20 Proposition 3.15. Let F/K be a Galois extension of number elds ρ a Gal(F/K) representation. 1. If ρ0 is another Gal(F/K)-representation then L(ρ ⊕ ρ0, s) = L(ρ, s)L(ρ0, s)

00 N 2. If N C Gal(F/K) lies in ker(ρ) so that ρ comes from a representation ρ of Gal(F /K) = G/N then L(F/K, ρ, s) = L(F N /K, ρ00, s). 3. ( Artin Formalisation) If G 000 for a representation of then H 000 ρ = IndH ρ H ≤ G L(F/K, ρ, s) = L(F/F , ρ , s) Proof. It suces to check the statement prime-by-prime for the local polynomials

1. Clear (Note (ρ ⊕ ρ0)Ip = ρIp ⊕ ρ0Ip )

N 2. Straight from the denitions using: if G = Gal(F/K), N C G, q a prime above s in F , above p is K. Then Ds/p = Dq/pN/N, Is/p = Iq/pN/N, Frobq/pN = Frobs/p (proof, exercise)

3. This follows from Proposition 2.16: the second formula there show that the number of times (1 − ζT ) in p and in Q 000 Σi/p is the same. Pp(ρ, T ) Σ above p Ps(p ,T )

Example. Let , . ∼ ∗. Then K = Q F = Q(ζN ) G = Gal(F/K) = (Z/NZ)

ζF (s) = L(F/F, I, s) = L(F/Q, IndI, s) Y = L(χ, s)

χ∈Gb where Gb is the set of irreducible representations of G. For general F/Q would have

Y dim ρ ζF (s) = L(ρ, s) ρ∈Gb

3.4 Artin L-series for 1-dimesnional representation Lemma 3.16. Let and ∼ ∗ ∗. Then F = Q(ζN ) χ : Gal(F/Q) = (Z/NZ) → C Y 1 L(F/Q, χ, s) = LN (χ, s) · −1 Pp(χ, q ) p|N

Proof. We can compare the Euler factor prime by prime. For p|N we have equality as the one for LN (χ, s) is 1 (χ(p) = 0). p For , is unramied in so (so Ip ) and ∗. Thus p - N p Q(ζN ) Ip = {id} χ = χ Frobp(ζN ) = ζN ↔ p mod (Z/NZ)

Ip Pp(χ, T ) = det(1 − FrobpT |χ )

= 1 − χ(Frobp)T = 1 − χ(p)T as required

Remark. When N is minimal, the last term is 1 and L(χ, s) = LN (χ, s). Remark. By the Kroneckar-Weber theorem, every abelian extension of lies inside for some . So if Q Q(ζN ) N ρ : Gal(F/Q) → C∗ is a 1-dimensional representation then by Proposition 3.15 2. L(ρ, s) = L(ρ00, s) for some 00 ∗ which is then essentially a Dirichlet -function. ρ : Gal(Q(ζN )/Q) → C L

21 Theorem 3.17. (Hecke (1920)+ some Class Field Theory) Let F/K be a Galois extension of number elds and ψ : Gal(F/K) → C∗ a 1-dimensional representation. Then L(ψ, s) has an analytic continuation to C, except for a simple pole at s = 1 when ψ = I. Remark. When and this recovers Theorem 3.8 K = Q F = Q(ζN ) Proof. Way beyond the scope of this course

Corollary 3.18. If ψ 6= I then L(ψ, 1) 6= 0. Proof. By Proposition 3.15 2. we may assume F/K is abelian. Then by Proposition 3.15 1. and 3.

ζF (s) = L(F/K, IndI, s) Y = L(F/K, χ, s)

χ∈Gb Y = ζK (s) L(F/K, χ, s) χ6=I As both ζ-functions have a simple pole and the rest are analytic, this implies L(F/K, χ, 1) 6= 0.