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 ∈ I under F . Then the projective limit of F (if it exists) is an object

lim Xi ←− i∈I together with morphisms

(ρj : lim Xi → Xj)j∈I ←− i∈I so that for every object T ∈ Ob(C) the map

ρj◦ : Hom(T, lim Xi) −→ {fj : T → Xj | F (φ) ◦ fi = fj for all (φ : i → j) ∈ I} ←− i∈I

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) = {L a eld | L/K Galois and nite }, {1} 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 homomorphism

ρ : Gal(K/K) −→ GLn(F ), where F is a topological eld (i. e. (F, +) and (F \{0}, ·) 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 {M | det(M) = 0} ⊆ 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 |p as

Np FrobP |p(x) ≡ x modP for all x ∈ 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 ∗ {one-dimensional artin representations} ←→ {Dirlichet Characters}χ :(Z/mZ) ρ 7−→ χ with χ(p) = ρ(Frobp). √ 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 √ √ r( 4 2)) = i 4 2), r(i) = i, √ √ s( 4 2)) = 4 2), s(i) = −i. We have r4 = 1, s2 = 1, srs−1 = r−1. We get a two-dimensional Artin representation

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  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  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: √ 4 Q(i, 2) p1 p2

√ 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 ∈ Q[x] irredu- cible splits completely has a density that is (#G)−1,G = Gal(Q[x]/(f)/Q). If we dene G0 := {σ ∈ G | σ has no sulutions mod p} has a density equal to #G0/#G. For , prime, we denote by the number of zeros of in f ∈ 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 √ 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) =√S3 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) = ε where q . This implies that Frob is a transposition if p , so ε(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 ∈ 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 ∈ Z. In the other case we can write p = 2a2 + ab + 3b, and as a result we get

 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 m∈N 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

max{k : pk | n} if n 6= 0 | n | := . 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 ∈ Qp \{0} 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 | x | := . p 0 if x = 0

Remark Every nonzero p-adic integer a can be written uniquely as

n−1 ∗ a = a0 + a1p + ... + an−1p ∈ Zp with ai < p for all i.

Example The p-adic cyclotomic character is the one-dimensional representation

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 . For prime, we Qp/Q p l 6= p have Frob l l(ζpn ) = ζpn and therefore ρcycl(Frobp = l.

Artin representations coming from geometry Let f(x) = x3 + ax + b ∈ Q[x], E the elliptic curve given by y2 = f(x), i.e.

2 3 2 3 E = ProjQ[X,Y,Z]/(Y Z − X − aXZ − bZ ) E is an abelian scheme with the well-known group law on E(L) for any eld extension L/K. The multiplication by m ∈ Z, denoted by [m] is a group homomorphism. We dene the m-torsion points to be

E[m] = {P ∈ E(Q | [m](P ) = O} = Ker([m]), where O = (0 : 1 : 0) is the point at innity. E[m] is a subgroup of E(Q as it is the kernel of a homomorpism. We have E =∼ (Z/mZ)2 as groups because E[m] is a free Z/MZ-module of rank two. Why is this the case? We know that over C, every elliptic curve can be written as

(E[m], +) =∼ (C/L, +),

5 where L ⊆ C is a lattice, i. e. a free abelian group of rank two, so that C/L is compact. For a class of complex numbers [z] corresponding to a point P ∈ E we have [m]P = [m][z] = [mz], because 1 Ker( /L → /L) = {z ∈ | [m]z ∈ L} = ( L)/L =∼ L/mL =∼ 2/m 2 =∼ ( /m )2, C C C m Z Z Z Z because L =∼ Z2. This implies E[m](C) = E[m](Q) =∼ (Z/mZ)2 because all points on the curve satisfy an algebraic equation. n Consider the action of GK on E[p ]:

n n σ GQ × E[p ] −→ E[p ](σ, P ) 7−→ P := (σ(x), σ(y)) for P = (x, y)O 7−→ O

This action is well-dened because (σx)3 = a(σx) + b = σ(x3 + ax + b) = σ(y2) = σ(y)2. This action gives us a Galois representation GL and ρm : GK → 2(Z/mZ) we have a chain

[p] ... E[pr] E[pr+1] ...

We dene the Tate module to be the projective limit

lim (E[pm], [p]). ←− m∈N It is a free module over

lim /m = p. ←− Z Z Z m∈N We write m ∼ 2 TpE := lim E[p ] = p ←− Z m∈N The Tate module acts on and we obtain a Galois representation GQ

Gal ρ GL (Q/K) 2(Qp) act

Aut ∼ GL (TpE) = 2(Zp) This ρ is unramied at all primes l not dividing p∆, ∆ = disc(f(x)). The caracteristic polynomial of ρFrobl is

2 det(xId − ρ(Frobl)) = x − (l + 1 + Nl)x + l,

where Nl = #E(Fl).