SLIDE 1
Character sums for primitive root densities
Peter Stevenhagen Geocrypt, Bastia June 28, 2011
SLIDE 2 Artin’s primitive root conjecture
We call a ∈ Z a primitive root modulo a prime p if we have F∗
p = a mod p.
Suppose a ∈ Z is an integer that is not an exact power. Artin’s conjecture (1927): The set of primes p for which a is a primitive root modulo p is infinite, with natural density
(1 − 1 q(q − 1)) ≈ .3739558.
SLIDE 3 Heuristic derivation
The index [F∗
p : a mod p]
is divisible by a prime q if and only if p splits completely in the number field Kq = Q(ζq,
q
√ a) = SplitQ(X q − a),
- f degree [Kq : Q] = q(q − 1).
For fixed q, a fraction
1 q(q−1) of all primes p is eliminated.
Now “take the limit” over all primes q.
SLIDE 4 Why is this still a conjecture?
Artin’s conjecture has not been proved for a single value of a. Imposing the condition for finitely many q is fine: use Chebotarev for primes with prescribed splitting in Kn = Q(ζn,
n
√ a), with n the product of the q’s involved. Dealing with infinitely many q is much harder. Throwing away one prime at the time (0%), we may be left in the end with the empty set....
SLIDE 5 History
Erdös tried in vain (1935) to combine infinitely many q. Hasse and Bilharz proved function field analogues of Artin’s conjecture (late 1930s). In this setting, the Riemann hypothesis was obvious (for Fp[X])
- r being proved around that time by Weil.
For number fields, decent remainder terms in density theorems
Under GRH, one can handle infinitely many q. This is non-trivial: Hooley (1967), Cooke-Weinberger (1975). Unconditionally, we still do not know whether (say) 2 is a primitive root modulo infinitely many primes p.
SLIDE 6
Dependence of conditions
The version of the conjecture as originally stated by Artin is actually wrong - but this went unnoticed for 30 years. It was discovered numerically in 1957 by Derrick and Emma Lehmer, who used a computer to extend tables of Kraitchik. Artin had consulted these to see if his conjecture made sense. He now realized conditions at different q may be dependent: “I was careless but the machine caught up with me.”
(Letter to Emma Lehmer, January 1958)
SLIDE 7 An entanglement problem
We want p that do not split completely in any of the fields Kq = Q(ζq,
q
√ a). The quadratic field K2 = Q(√a) is abelian, hence cyclotomic. For K2 = Q(√a) of prime discriminant ±ℓ we simply have K2 ⊂ Q(ζℓ) ⊂ Kℓ. The condition at q = ℓ can be left out, and we gain a factor 1 + 1 ℓ2 − ℓ − 1.
SLIDE 8 Correcting the density
The density (under GRH) is the inclusion-exclusion-value δ(a) =
∞
µ(n) [Kn : Q], which can be different from the naive value
1 [Kq : Q]
The only possible dependency between conditions occurs when K2 = Q(√a) is quadratic of odd discriminant d, leading to an inclusion K2 ⊂ Kd = (compositum of Kq with q|d).
SLIDE 9 The corrected density
- Theorem. For a ∈ Q∗ different from ±1 we have
δ(a) = E ·
1 [Kq : Q]
with E = 1 if d = disc(Q(√a)) is even, and E = 1 − µ(|d|)
1 [Kq : Q] − 1 if d is odd. The correction factor E for odd d comes out of a calculation found in Hooley (1967). Note the nice multiplicative structure!
SLIDE 10
Lots of Artin conjectures
After Hooleys proof under GRH, people have considered many variants of the conjecture. There are various classical extensions:
◮ primes in arithmetic progressions with given primitive root; ◮ near-primitive roots: [F∗ p : a] = t ∈ Z>0; ◮ two-variable Artin: a1 ⊂ a2 ⊂ F∗ p ◮ higher-rank Artin: F∗ p = a1, a2, . . . , ar ◮ multiple primitive roots: F∗ p = a1 = a2 = . . . = ar. ◮ same order: a1 = a2 = . . . = ar ⊂ F∗ p.
SLIDE 11
Lots of Artin conjectures (2)
The published calculations tend to get very messy, but the correction factors coming out of the lengthy calculations often exhibit some ‘nice’ structure. This talk: conceptual way to arrive at these structures. It leads to an algorithm to compute the correction factors that is so much simpler that it can be applied in many more cases. The results are joint work with Hendrik Lenstra (Leiden) and Pieter Moree (Bonn).
SLIDE 12 Frobenius conditions
In all cases, the condition on p we want to be satisfied is a splitting condition in some number field F∞ of infinite degree, i.e., on the Frobenius of p in Gal(F∞/Q). F∞ is a compositum of infinite degree number fields Fq∞ =
Q(ζqk,
qk
√a1,
qk
√a2, . . . ,
qk
√ar) gotten by adjoining to Q all q-power roots of a finite number of elements a1, a2, . . . , ar ∈ Q. One wants to impose a splitting condition on p in the fields Fq∞ for all primes q < p.
SLIDE 13 Frobenius conditions (2)
Example: we have qk||[Fp : a] if and only if p splits completely in Q(ζqk,
qk
√a), but not in Q(ζqk+1,
qk+1
√a). In many Artin conjectures, the condition on p does not ‘factor’ via a finite degree subfield of Fq∞. Just think of the condition implying 2 = 3 ⊂ F∗
p.
SLIDE 14 Radical Galois groups
The need arises to understand the radical Galois groups Gal(Q(Γ1/n)/Q) for extensions generated by the radicals Γ1/n = {x ∈ Q : xn ∈ Γ}
- f order n ∈ Z>0 of finitely generated subgroups Γ ⊂ Q∗.
These are division points of the multiplicative group.
SLIDE 15 Galois representations
In the case of elliptic curves, for torsion points, one views Galois groups as subgroups of generic groups GL2(Z/nZ) = Aut(E[n])
- f automorphisms of the groups involved.
These generic groups satisfy a Chinese remainder theorem with respect to n that the actual Galois groups may not share. This point of view is also very fruitful for our Galois groups!
SLIDE 16
Describing G = Gal(Q(Γ∞)/Q)
Let Γ ⊂ Q∗ be a finitely generated subgroup We need the Galois group of F∞ = Q(Γ∞) over Q, with Γ∞ = {x ∈ Q
∗ : xn ∈ Γ for some n ∈ Z>1}.
Consider the injection of topological groups G = Gal(Q(Γ∞)/Q) − → A = AutΓ∞∩Q(Γ∞). Which group automorphisms come from field automorphisms?
SLIDE 17
A = AutΓ∞∩Q(Γ∞)
We may assume Γ∞ ∩ Q∗
>0 = Γ, and write
Γ = b1 × b2 × . . . br with bi > 0 and r the rank of Γ. This yields Γ∞ = bQ
1 × bQ 2 × . . . bQ r × µ∞,
with µ∞ the group of roots of unity in Q
∗.
SLIDE 18 A = AutΓ∞∩Q(Γ∞)
A group automorphsm σ of Γ∞ = bQ
1 × bQ 2 × . . . bQ r × µ∞,
fixing µ∞ and all bi is determined by the elements
i
) b1/n
i
∞
n=1
∈ µ = lim
←n µn,
in the Tate module µ of the multiplicative group. It is a homomorphism Γ = b1 × b2 × . . . br − → µ.
SLIDE 19 A = AutΓ∞∩Q(Γ∞)
The group A fits in a split exact sequence 1 − → Hom(Γ, µ) − → A − → Aut(µ∞) − → 1. As µ is free of rank 1 as a Z-module, we find A = Hom(Γ, µ) ⋊ Aut(µ∞) ∼ = µr ⋊ Aut(µ∞) ∼ = Zr ⋊ Z∗ ∼ =
q ⋊ Z∗ q
Aq.
SLIDE 20
G versus A
Q(Γ∞) contains Q(µ∞) = Qab, which has Galois group Gal(Qab/Q) = Aut(µ∞) = Z∗. We obtain a commutative diagram 1 − → Gal(Q(Γ∞)/Qab) − → G − → Gal(Qab/Q) − → 1 ↓ ↓ ↓ 1 − → Hom(Γ, µ) − → A − → Aut(µ∞) − → 1
SLIDE 21
G versus A
Over Qab, all radical extensions become Kummer extensions, and, in principle, we know their Galois groups. 1 − → Gal(Q(Γ∞)/Qab) − → G − → Gal(Qab/Q) − → 1 ↓ ↓ ↓ 1 − → Hom(Γ, µ) − → A − → Aut(µ∞) − → 1. Key lemma: Γ∞ ∩ Qab = Γ1/2
∞ .
SLIDE 22
Cutting out G using characters
An element σ ∈ A is a field automorphism if and only if its action on the square roots √ b ∈ Γ∞ ∩ Qab its compatible with its cyclotomic action: ψb(σ) := σ( √ b) √ b = χQ(
√ b)(σ),
with χQ(
√ b) : A −
→ Aut(µ∞) = Z∗ − → {±1} the Dirichlet character corresponding to Q( √ b).
SLIDE 23 Cutting out G using characters
An element σ ∈ A is in G if and only if it is in the kernel of the characters χb := ψb · χQ(
√ b) : A −
→ {±1} (b ∈ Γ). We have a perfect pairing A/G × Γ/Γ2 − → {±1} (α, b) − → χb(α)
- f elementary abelian 2-groups.
SLIDE 24
Artin density problems
In Artin problems, we compute the density of primes p such that for all q < p, its Frobenius in Gq = Gal(Q(Γ1/q∞)/Q) ⊂ Aq is contained in some subset Sq ⊂ Aq. Example: in the original conjecture, take Γ = a and Sq = Aq \ ker[Aq − → Aut(Γ1/q)].
SLIDE 25 Artin density problems
Artin’s error: the density of such p is not the relative density ν(S) ν(A) =
νq(Sq) νq(Aq)
q Sq in A. Here ν = q νq is the normalized Haar
measure on the profinite group A =
q Aq.
Under GRH, it is ν(S ∩ G) ν(G) . Problem: usually G
q Gq due to entanglement of radical
extensions.
SLIDE 26 Main theorem
ν(G ∩ S) ν(G) = E · ν(S) ν(A), for an entanglement correction factor E given by E =
Eb,q, with Eb,q = 1 νq(Sq)
χb,qdνq the average value on Sq of the q-primary component χb,q of χb =
q χb,q.
SLIDE 27 Proof
The characteristic function of G in A is 1G = 2−r
χb. Compute ν(G ∩ S) by integrating 1G over S with respect to ν: ν(G ∩ S) ν(G) = ν(S) ν(A) ·
ν(S)
χbdν
Over S =
q Sq it all decomposes into q-primary parts...
SLIDE 28 Example: back to Artin
Kq = Q(ζq,
q
√a).
- Theorem. For a ∈ Q∗ different from ±1 we have
δ(a) = E ·
1 [Kq : Q]
with E = 1 if d = disc(Q(√a)) is even, and E = 1 − µ(|d|)
1 [Kq : Q] − 1 if d is odd.
SLIDE 29 Proof of Artin (under GRH)
The set of ‘good’ Frobenius classes Sq = Aq \ ker[ϕq : Aq − → Aut(a1/q)] is a set-theoretic difference of groups of measure ν(Sq) = 1 − 1 [Kq : Q]. We have ν(S) ν(A) =
1 [Kq : Q]
SLIDE 30 Proof of Artin (under GRH)
The average of any character χ over Sq = Aq \ ker ϕq equals Eq = 1 ν(Sq)
χ dνq −
χq dνq
It has value 1 if χ is trivial, 0 if χ is non-trivial on ker ϕq, and −νq(ker ϕq) νq(Sq) = −1 [Kq : Q] − 1
SLIDE 31 Proof of Artin (under GRH)
In this rank-1 case the correction factor reads 1 +
Eq, with Eq the average of χa,q over Sq. Eq = 1 at q ∤ 2 · disc(Q(√a). Eq = −1/([Kq : Q] − 1) at odd primes q|disc(Q(√a). E2 = 0 (no correction!) for disc(Q(√a) even. E2 = −1 for disc(Q(√a) odd.
SLIDE 32
Impact of this strategy
Can be applied to all variants over Q that we mentioned. Yields very efficient proofs of older results. Provides extensions to cases excluded so far. Provides immediate non-vanishing conditions in the rank-1 case. Extends (non-trivially) to Artin type problems over arbitrary number fields.
SLIDE 33
Questions?
