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Seminar Talk: Galois Representations Seminar Talk: Galois Representations Denition Let K be a eld and K a xed algebraic closure of K. Then we call the group GK := Gal(K=K) the absolute Galois group of K. 0 Note that for a dierent algebraic closure K , the corresponding group 0 Gal(K =K) is isomorphic to GK . As the talk will treat dierent objects dended via limits, we recall the denition of an inductive limit: Denition Let I be a small category (i. e. objects and morphisms are sets), C be an arbritary category, F : I !C a functor and write Xi for the image of i 2 I under F . Then the projective limit of F (if it exists) is an object lim Xi − i2I together with morphisms (ρj : lim Xi ! Xj)j2I − i2I so that for every object T 2 Ob(C) the map ρj◦ : Hom(T; lim Xi) −! ffj : T ! Xj j F (φ) ◦ fi = fj for all (φ : i ! j) 2 Ig − i2I g 7−! ρj ◦ g is an isomorphism. We can write the absolute Galois group as a limit: Gal(K=K) = lim Gal(L=K) − L=K where galop is the opposite category of nite galois extensions of K, i.e. I = K Ob(I) = fL a eld j L=K Galois and nite g; f1g if L0 ⊆ L Mor (L; L0) = : I ; otherwise Writing the Group as a projective limit endows it with a topology compatible with the group structure, called the Krull topology, so that it has the structure of a topological group, i. e. composition and the inverse map are continuous). The open subgroups are of the form Gal(K=L), where L=K is a nite Galois extension. As GK is a fairly complicated group, we have to develop some tools to study it. One important approach is to use Galois representations. 1 Denition A Galois representation (of dimension n) is a continuous homo- morphism ρ : Gal(K=K) −! GLn(F ); where F is a topological eld (i. e. (F; +) and (F n f0g; ·) are topological groups. The topology on GLn(F ) is induced by the topology F by considering GLn(F ) as the complement of the open subset ∼ n2 fM j det(M) = 0g ⊆ Matn×n(F ) = F , 2 where F n has the canonical topology. Denition ρ is unramied at a prime p of K if it factors as ρ Gal(K=K) GLn(F ) res Gal(L=K) for some Galois extension L=K (innite extensions are allowed). Recall the denition of the Frobenius automorphism: For a Galois extension L=K, p a prime of K taht is unramied in L, P a prime in L above p, we dene FrobP jp as Np FrobP jp(x) ≡ x modP for all x 2 OL and we can easily see that two Frobenius automorphisms for dierent primes P; P 0 above p are conjugate in Gal(L=K), so that we get a well-dened conjugacy class [ρ(Frobp]. If ρ is unramied, the class [ρ(Frobp] ⊆ GLn(F ) is well-dened. This notion allows us to formulate a more general statement as question B: Question C Given a Galios representation ρ : GK ! GLn(F ), is there a rule to determine the conjugacy class of ρ(Frobp) for p an unramied prime of K? Denition If the eld F is equal to the complex numbers equipped with the discrete topology, a Galois representation GL is called Artin ρ : GK ! n(C) representation. Note that images of Artin representations are nite, as ρ−1(id) is open. Therefore they factor through representations of nite groups Gal(L=K), where L=K is a nite Galois extension. Finding an answer for Question C would also answer Question B, als for L=K nite and Galois, we can nd an Artin representation ρ with Ker(ρ) = Gal(K=L), so that p splits in L if and only if [ρ(Frobp] is trivial. Example If , we have GL ∗, so an Artin representation of n = 1 1(C) = C dimension one is just a continuous homomorphism ∗. By the ρ : GK ! C Kronecker-Weber Theorem (every number eld is contained in a cyclotomic eld) we know that there ist some so that factors through Gal ∼ m ρ Q(ζm)=Q) = (Z=mZ)∗. Thus we get a corresponence 2 ∗ fone-dimensional artin representationsg ! fDirlichet Charactersgχ :(Z=mZ) ρ 7−! χ with χ(p) = ρ(Frobp): p 4 Example For L = Q(i; 2), the splitting eld of the polynomial x4 − 2 oder , we know that the Galois group is isomorphis to the dihedral group that Q D8 is generated by the two elements r and s, with p p r( 4 2)) = i 4 2); r(i) = i; p p s( 4 2)) = 4 2); s(i) = −i: We have r4 = 1; s2 = 1; srs−1 = r−1. We get a two-dimensional Artin represen- tation Gal ρ GL (Q=Q) 2(C) res Gal ∼ (L=Q) = D8 0 1 0 1 via ρ(r) = ; ρ(s) = . −1 0 1 0 Fact We have 8 if 2 2 > 1 p = a 64b > 2 if 2 2 odd <> r p = a + 16b ; b [ρ(Frobp] = rs if p ≡ 3(8) ; > if > r p ≡ 5(8) :> s if p ≡ 7(8) so that 8 2 if p = a264b2 < 2 2 tr(ρ(Frobp) = −2 if p = a + 16b ; b odd : : 0 otherwise Example If we take some prime p ≡ 1(8), we know by the quartic reciproctiy law that 2 is a quadratic, but not biquadratic residue mod p, so we get the following splitting behaviour: p 4 Q(i; 2) p1 p2 p Q(i; 2) p1 p2 Q(i) p1 p2 Q p 3 The fact stated above can be proven by using the quadratic reciprocity law. Recall The set of rational primes p so that a polynomial f 2 Q[x] irredu- cible splits completely has a density that is (#G)−1;G = Gal(Q[x]=(f)=Q). If we dene G0 := fσ 2 G j σ has no sulutions mod pg has a density equal to #G0=#G. For , prime, we denote by the number of zeros of in f 2 Z[x] p Np(f) f Fp (counted without multiplicity). n Example We want to compute Np(f) for f(x) = x − x − 1. In the case n = 2, the discriminant of f is 5, so we can conclude N5(f) = 1. If p 6= 5, we have 2 if 5 is a square mod p 2 if p ≡ ±1 mod p N (f) = = p 0 otherwise 0 if p ≡ ±2 mod p p because for p 6= 2, the roots of f in Fp are (1 ± p)=2. The second equality is just quadratic reciprocity. In the case n = 3, we write E = Q[x]=(f), L a Galois closure of E. The Galois group is Gal . We know the discirminant of is , so that (L=Q) =pS3 f −23 L contains the eld K = Q( −23) and L is a cubic extension of K. If p 6= 23, let Frobp be the Frobenius substitution of p in S3. As we know there is a unique quadratic character associated to , we know sgn Frob , K=Q ( p) = " where q . This implies that Frob is a transposition if p , so "(q) = 23 p 23 = −1 we have Np(f) = 1 in this case. If p , Frob has either order or , so or . 23 = 1 p 1 3 Np(f) = 1 3 To distinguish these cases, we decompose p = p1p2 2 K. By the coresondence between ideal classes and binary quadratic forms we know that p1 is principial if and only if we can write p = a2 + ab + 6b2 with a; b 2 Z. In the other case we can write p = 2a2 + ab + 3b, and as a result we get 8 3 if p = a2 + ab + 6b2 < 2 Np(f) = 0 if p = 2a + ab + 3b : : if p 1 23 = −1 Denition If F is the eld of the p−adic numbers, equipped with the p-adic topology, a Galois representation GL is called p-adic Galois ρ : GK ! n(Qp) representatnion. Recall The p-adic integers are dened as the projective limit lim =pm ; − Z Z m2N where the limit runs over all positive integers and we have a morphism from 0 Z=pmZ to Z=pm Z if and only if m ≤ m0, and that morphism is just the inclusion. Alternatively, they can be dened as the completion of Z with respect to the p-adic absolute value maxfk : pk j ng if n 6= 0 j n j := : p 0 if n = 0 4 We then dene the p-adic numbers to be the fraction eld of the p-adic integers. On those we dene the -adic norm: For any there is a p x 2 Qp n f0g unique so that n a , so that does not divide and , and we dene n x = p b p a b p−n if x 6= 0 j x j := : p 0 if x = 0 Remark Every nonzero p-adic integer a can be written uniquely as n−1 ∗ a = a0 + a1p + ::: + an−1p 2 Zp with ai < p for all i. Example The p-adic cyclotomic character is the one-dimensional representa- tion Gal ρcycl ∗ ∼ GL (Q=Q) Qp = 1(Qp) res Gal ∼ ∗ (Qp=Q) = Zp n−1 given by ρcycl(σ) = a = a0 + a1p + ::: + an−1p for n−1 a0+a1p+:::+an−1p σ(ζpn ) = ζpn We know that the only prime ramied in is .
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