Character sums for primitive root densities Peter Stevenhagen - PowerPoint PPT Presentation

Character sums for primitive root densities Peter Stevenhagen Geocrypt, Bastia June 28, 2011

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 . q prime

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 √ a ) = Split Q ( X q − a ) , q K q = Q ( ζ q , of degree [ K q : Q ] = q ( q − 1 ) . 1 For fixed q , a fraction q ( q − 1 ) of all primes p is eliminated. Now “take the limit” over all primes q .

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 √ n K n = Q ( ζ 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....

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 F p [ X ] ) or being proved around that time by Weil. For number fields, decent remainder terms in density theorems only exist under GRH. 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 .

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)

An entanglement problem We want p that do not split completely in any of the fields √ q K q = Q ( ζ q , a ) . The quadratic field K 2 = Q ( √ a ) is abelian, hence cyclotomic. For K 2 = Q ( √ a ) of prime discriminant ± ℓ we simply have K 2 ⊂ Q ( ζ ℓ ) ⊂ K ℓ . The condition at q = ℓ can be left out, and we gain a factor 1 1 + ℓ 2 − ℓ − 1 .

Correcting the density The density (under GRH) is the inclusion-exclusion-value ∞ � µ ( n ) δ ( a ) = [ K n : Q ] , n = 1 which can be different from the naive value � � � 1 1 − . [ K q : Q ] q The only possible dependency between conditions occurs when K 2 = Q ( √ a ) is quadratic of odd discriminant d , leading to an inclusion K 2 ⊂ K d = (compositum of K q with q | d ) .

The corrected density Theorem. For a ∈ Q ∗ different from ± 1 we have � � � 1 δ ( a ) = E · 1 − , [ K q : Q ] q with E = 1 if d = disc ( Q ( √ a )) is even, and � 1 E = 1 − µ ( | d | ) [ K q : Q ] − 1 q | d 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!

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: � a 1 � ⊂ � a 2 � ⊂ F ∗ p ◮ higher-rank Artin: F ∗ p = � a 1 , a 2 , . . . , a r � ◮ multiple primitive roots: F ∗ p = � a 1 � = � a 2 � = . . . = � a r � . ◮ same order: � a 1 � = � a 2 � = . . . = � a r � ⊂ F ∗ p .

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).

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 � √ a 1 , √ a 2 , . . . , √ a r ) qk qk qk F q ∞ = Q ( ζ q k , k gotten by adjoining to Q all q -power roots of a finite number of elements a 1 , a 2 , . . . , a r ∈ Q . One wants to impose a splitting condition on p in the fields F q ∞ for all primes q < p .

Frobenius conditions (2) Example: we have q k || [ F p : � a � ] if and only if √ a ) , but not in Q ( ζ q k + 1 , √ a ) . qk qk + 1 p splits completely in Q ( ζ q k , In many Artin conjectures, the condition on p does not ‘factor’ via a finite degree subfield of F q ∞ . Just think of the condition implying � 2 � = � 3 � ⊂ F ∗ p .

Radical Galois groups The need arises to understand the radical Galois groups Gal ( Q (Γ 1 / n ) / Q ) for extensions generated by the radicals Γ 1 / n = { x ∈ Q : x n ∈ Γ } of order n ∈ Z > 0 of finitely generated subgroups Γ ⊂ Q ∗ . These are division points of the multiplicative group.

Galois representations In the case of elliptic curves, for torsion points, one views Galois groups as subgroups of generic groups GL 2 ( Z / n Z ) = Aut ( E [ n ]) of 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!

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 ∗ : x n ∈ Γ for some n ∈ Z > 1 } . Γ ∞ = { x ∈ Q Consider the injection of topological groups G = Gal ( Q (Γ ∞ ) / Q ) − → A = Aut Γ ∞ ∩ Q (Γ ∞ ) . Which group automorphisms come from field automorphisms?

A = Aut Γ ∞ ∩ Q (Γ ∞ ) We may assume Γ ∞ ∩ Q ∗ > 0 = Γ , and write Γ = � b 1 � × � b 2 � × . . . � b r � with b i > 0 and r the rank of Γ . This yields Γ ∞ = b Q 1 × b Q 2 × . . . b Q r × µ ∞ , ∗ . with µ ∞ the group of roots of unity in Q

A = Aut Γ ∞ ∩ Q (Γ ∞ ) A group automorphsm σ of Γ ∞ = b Q 1 × b Q 2 × . . . b Q r × µ ∞ , fixing µ ∞ and all b i is determined by the elements � � ∞ σ ( b 1 / n ) i ∈ � µ = lim ← n µ n , b 1 / n i n = 1 µ of the multiplicative group. in the Tate module � It is a homomorphism Γ = � b 1 � × � b 2 � × . . . � b r � − → � µ.

A = Aut Γ ∞ ∩ Q (Γ ∞ ) The group A fits in a split exact sequence 1 − → Hom (Γ , � µ ) − → A − → Aut ( µ ∞ ) − → 1 . µ is free of rank 1 as a � As � Z -module, we find A = Hom (Γ , � µ ) ⋊ Aut ( µ ∞ ) µ r ⋊ Aut ( µ ∞ ) ∼ = � Z r ⋊ � ∼ = � Z ∗ � � � � ∼ Z r q ⋊ Z ∗ = = A q . q q prime q prime

G versus A Q (Γ ∞ ) contains Q ( µ ∞ ) = Q a b , which has Galois group Gal ( Q a b / Q ) = Aut ( µ ∞ ) = � Z ∗ . We obtain a commutative diagram Gal ( Q (Γ ∞ ) / Q a b ) Gal ( Q a b / Q ) 1 − → − → G − → − → 1 ↓ ↓ ↓ 1 − → Hom (Γ , � µ ) − → A − → Aut ( µ ∞ ) − → 1

G versus A Over Q a b , all radical extensions become Kummer extensions, and, in principle, we know their Galois groups. Gal ( Q (Γ ∞ ) / Q a b ) Gal ( Q a b / Q ) 1 − → − → G − → − → 1 ↓ ↓ ↓ 1 − → Hom (Γ , � µ ) − → A − → Aut ( µ ∞ ) − → 1. Key lemma : Γ ∞ ∩ Q a b = Γ 1 / 2 ∞ .

Cutting out G using characters An element σ ∈ A is a field automorphism if and only if its √ b ∈ Γ ∞ ∩ Q a b its compatible with action on the square roots its cyclotomic action: √ ψ b ( σ ) := σ ( b ) √ = χ Q ( √ b ) ( σ ) , b with Z ∗ − → Aut ( µ ∞ ) = � √ χ Q ( b ) : A − → {± 1 } √ the Dirichlet character corresponding to Q ( b ) .

Cutting out G using characters An element σ ∈ A is in G if and only if it is in the kernel of the characters √ χ b := ψ b · χ Q ( b ) : A − → {± 1 } ( b ∈ Γ) . We have a perfect pairing A / G × Γ / Γ 2 − → {± 1 } ( α, b ) �− → χ b ( α ) of elementary abelian 2-groups.

Artin density problems In Artin problems, we compute the density of primes p such that for all q < p , its Frobenius in G q = Gal ( Q (Γ 1 / q ∞ ) / Q ) ⊂ A q is contained in some subset S q ⊂ A q . Example: in the original conjecture, take Γ = � a � and → Aut (Γ 1 / q )] . S q = A q \ ker [ A q −