Galois action on the homology of Fermat curves Rachel Pries - PowerPoint PPT Presentation

Galois action on the homology of Fermat curves Rachel Pries Colorado State University pries@math.colostate.edu KR 2 V : joint work with R. Davis, V. Stojanoska, K.Wickelgren Thanks to Joe! Silvermania, Brown University August 12, 2015 Pries (CSU) Galois action on Fermat curves 1 / 26

Abstract: Fix p odd prime. Let K = Q ( ζ p ) . Let X be the Fermat curve x p + y p = z p . Anderson studied the action of the absolute Galois group G K on a relative homology group of X (Duke, 1987). He proved that the action factors through Q = Gal ( L / K ) where L is the the splitting field of 1 − ( 1 − x p ) p . Using this, he obtained results about the field of definition of points on a generalized Jacobian of X . We build upon Anderson’s work: for p satisfying Vandiver’s conjecture, we compute Q and find explicit formula for the action of q ∈ Q on the relative homology. Using this, we obtain information about maps between several Galois cohomology groups which arise in connection with obstructions to rational points. This is joint work with R. Davis, V. Stojanoska, and K. Wickelgren. Pries (CSU) Galois action on Fermat curves 2 / 26

Background on Fermat curve Let p be an odd prime. Let ζ be a p th root of unity. Let X be the Fermat curve x p + y p = z p , having genus g = ( p − 1 )( p − 2 ) . 2 Let U = X − Z where Z is closed subscheme of p points where z = 0. Let Y ⊂ X be closed subscheme of 2 p points where xy = 0. Pries (CSU) Galois action on Fermat curves 3 / 26

Background on Fermat curve Let p be an odd prime. Let ζ be a p th root of unity. Let X be the Fermat curve x p + y p = z p , having genus g = ( p − 1 )( p − 2 ) . 2 Let U = X − Z where Z is closed subscheme of p points where z = 0. Let Y ⊂ X be closed subscheme of 2 p points where xy = 0. (this is not a talk about) Fermat’s Last Theorem: X ( Q ) = Z ∪ Y . Pries (CSU) Galois action on Fermat curves 3 / 26

Other results about rational points Theorem - Greenberg (paraphrased) Let p ≥ 5 and let L 0 be the field generated over K = Q ( ζ ) by the p th roots of the real cyclotomic units of K . Then L 0 is the field generated by the points of order p on Jac ( X ) . Theorem - Anderson For p an odd prime, let L be the splitting field of 1 − ( 1 − x p ) p . Let S be the generalized Jacobian of X with conductor ∞ . Let b = “( 0 , 1 ) − ( 1 , 0 )” , a Q -rational point of S . Then L is the number field generated by the p th roots of b in S ( Q ) . Similar results obtained by Ihara and Coleman. Pries (CSU) Galois action on Fermat curves 4 / 26

´ Etale homology groups (coefficients in Z / p ) There is an action of µ p × µ p on X : x p + y p = z p (stabilizing U and Y ) given by ( ζ i , ζ j ) · [ x , y , z ] = [ ζ i x , ζ j y , z ] . Let Λ 1 = ( Z / p )[ µ p × µ p ] , generators ε 0 and ε 1 . The Jacobian (and other (co)homology groups) are Λ 1 -modules and also modules for G K , the absolute Galois group of K = Q ( ζ ) . Consider the homology group H 1 ( U ) , dimension ( p − 1 ) 2 and its quotient H 1 ( X ) , dimension 2 g = ( p − 1 )( p − 2 ) and the relative homology group M = H 1 ( U , Y ) , dimension p 2 . Consider the class β ∈ M of the path (singular 1-simplex) √ √ p p β : [ 0 , 1 ] → U ( C ) given by t �→ ( t , 1 − t ) (real p th roots). Pries (CSU) Galois action on Fermat curves 5 / 26

Galois action on relative homology Recall β ∈ H 1 ( U , Y ) chosen singular 1-simplex and Λ 1 = ( Z / p )[ µ p × µ p ] . Theorem - Anderson M = H 1 ( U , Y ) is a free Λ 1 -module of rank 1 with generator β . Let K = Q ( ζ ) . The action of σ ∈ G K on M is determined by its action on β . For p an odd prime, let L be the splitting field of 1 − ( 1 − x p ) p . Theorem - Anderson Then σ ∈ G K acts trivially on M = H 1 ( U , Y ) if and only if σ fixes L . Pries (CSU) Galois action on Fermat curves 6 / 26

More on the Galois action on relative homology The G K -action on H 1 ( U , Y ) factors through Q = Gal ( L / K ) . For q ∈ Q , write q β = B q β for some unit B q ∈ Λ 1 . Let ε 0 , ε 1 generate µ p × µ p . Recall Λ 1 = ( Z / p )[ µ p × µ p ] . 0 ε j Write B q = ∑ 0 ≤ i , j < p b i , j ε i 1 (view as p × p matrix). Anderson: (i) B q is a symmetric unit ( b i , j = b j , i ). (ii) B q − 1 is in the augmentation ideal ( 1 − ε 0 )( 1 − ε 1 )Λ 1 . (rows and columns of B q − 1 sum to zero mod p ). Observation: Identify Λ 1 with H 1 ( U , Y ) , then B q − 1 ∈ H 1 ( U ) . (iii) (Cliff note version) There are maps Λ ∗ d ′ 1 ) sym d ′′ → (Λ ∗ → Λ ∗ 2 and B q ∈ Ker ( d ′′ ) . 0 There is Γ q ∈ Λ sh 0 , unique up to Ker ( d ′ ) sh , s.t. ( d ′ ) sh (Γ q ) = B q . The logarithmic derivative of Γ q in Ω(Λ sh 0 ) is represented by the class of ( q − 1 ) ◦ [ 0 , 1 ] in H 1 ( A 1 − µ ∗ p ) . In theory, this determines the action of G K on H 1 ( U , Y ) completely. Pries (CSU) Galois action on Fermat curves 7 / 26

Our program: for all odd primes p Make Anderson’s work more explicit, (1) Determine Q = Gal ( L / K ) and (2) Determine formula for B q in order to compute Galois cohomology groups of Fermat curves which arise in connection with obstructions to rational points. (3) Main target: X ( K ) → H 1 ( G K , M ) (with restricted ramification) Quotient of target: kernel of the differential d 2 : H 1 ( N , M ) Q → H 2 ( Q , M ) when N = G L (with restricted ramification). Main result so far: for all odd p , have bounds on Ker ( d 2 ) (4) lower bound arising from (mod p ) Heisenberg extensions of K . (5) upper bound arising from Q -invariant local units of O L . Application: If p = 3, then 12 ≤ dim ( H 1 ( G K , M )) ≤ 22. Pries (CSU) Galois action on Fermat curves 8 / 26

(1) The Galois group Q of L / K Vandiver’s Conjecture (first conjectured by Kummer in 1849) The prime p does not divide the class number h + of K + = Q ( ζ p + ζ − 1 p ) . True for all p < 163 million (Buhler/Harvey) and for all regular primes. Let E be the units in O K and E + = E ∩ K + . Let C = V ∩ E be the cyclotomic units where V ⊂ K ∗ is generated by p : 1 ≤ i ≤ p − 1 } . Let C + = C ∩ O ∗ {± ζ p , 1 − ζ i K + . Then h + is the index of C + in E + . If Vandiver’s conjecture is true for p , then E / E p is generated by C . Pries (CSU) Galois action on Fermat curves 9 / 26

(1) The Galois group Q of L / K Let K = Q ( ζ p ) . Let r = ( p − 1 ) / 2. Let L be splitting field of 1 − ( 1 − x p ) p . Proposition: KR 2 V If Vandiver’s Conjecture is true for the prime p , then the Galois group Q of L / K is an elementary abelian p -group of rank r + 1. � 1 − ζ i Proof: L = K ( p p : 1 ≤ i ≤ p − 1 ) so Q is elem. abel. p -group and L / K ramified only over � 1 − ζ p � . Note rank ≤ r + 1 because L / K generated by p th roots of elements in subgroup B ⊂ K ∗ / ( K ∗ ) p generated by ζ p and 1 − ζ i p for 1 ≤ i ≤ r . Then B = � 1 − ζ p , B ′ � where B ′ ⊂ K ∗ / ( K ∗ ) p is generated by the cyclotomic units C . By Vandiver hypothesis, B ′ = E / E p . By Dirichlet’s unit theorem, E ≃ Z r − 1 × µ p so B ′ has rank r . Pries (CSU) Galois action on Fermat curves 10 / 26

(2) The Galois action The action of G K on M = H 1 ( U , Y ) factors through Q = Gal ( L / K ) . If q ∈ Q , then action determined by q · β = B q β for some B q ∈ M . Fixed isomorphism Q ≃ ( Z / p ) r + 1 with q �→ ( c 0 ,..., c r ) . i = 1 c i and F a root of F p − F + c = 0. Let c = ∑ p − 1 Let γ ( ε ) = ∑ p − 1 ) ε i − ∑ p − 1 i where ε p = 1. i = 1 ( c i + c − F c i i i = 1 Let Λ 0 = Z / p [ µ p ] and y = ε − 1 nilpotent variable since y p = 0. For f ∈ y Λ 0 , define E ( f ) = ∑ p − 1 i = 0 f i / i ! . Theorem: KR 2 V The action of q ∈ Q on H 1 ( U , Y ) is determined explicitly by: B q = E ( γ q ( ε 0 )) E ( γ q ( ε 1 )) E ( − γ q ( ε 0 ε 1 )) . Pries (CSU) Galois action on Fermat curves 11 / 26

(2) Example when p = 3 � If p = 3, then L = K ( ζ 9 , 3 1 − ζ − 1 ) and Q = � σ , τ � (commuting elements of order 3) σ acts by multiplication by ζ on ζ 9 and � 3 1 − ζ − 1 . τ acts by multiplication by ζ on M = Z / 3 [ µ 3 × µ 3 ] generated by ε 0 and ε 1 Our formula simplifies to:   0 − 1 1 B σ − 1 = − ( ε 0 + ε 1 )( 1 − ε 0 )( 1 − ε 1 ) = − 1 − 1 − 1   1 − 1 0 and  0 1 − 1  B τ − 1 = ( ε 0 + ε 1 ) − ( ε 2 0 + ε 0 ε 1 + ε 2 1 )+ ε 2 0 ε 2 1 = 1 − 1 0   − 1 0 1 Pries (CSU) Galois action on Fermat curves 12 / 26

(2) Example when p = 5 When p = 5, then Q = � σ , τ 1 , τ 2 � ≃ ( Z / 5 ) 3 . B σ = 2 ε 4 0 ε 3 1 + ε 4 0 ε 2 1 + 2 ε 4 0 ε 1 + 2 ε 3 0 ε 4 1 + ε 3 0 ε 3 1 + ε 3 0 ε 2 1 + ε 3 0 ε 1 + ε 2 0 ε 4 1 + ε 2 0 ε 3 1 + ε 2 0 ε 2 1 + 2 ε 2 0 ε 1 + 2 ε 0 ε 4 1 + ε 0 ε 3 1 + 2 ε 0 ε 2 1 + ε 0 ε 1 + 4 ε 0 + 4 ε 1 + 2. B τ 1 = 2 ε 4 0 ε 4 1 + 4 ε 4 0 ε 3 1 + 4 ε 4 0 ε 1 + 4 ε 3 0 ε 4 1 + 3 ε 3 0 ε 3 1 + 3 ε 3 0 + 3 ε 2 0 ε 2 1 + 4 ε 2 0 ε 1 + 3 ε 2 0 + 4 ε 0 ε 4 1 + 4 ε 0 ε 2 1 + 2 ε 0 + 3 ε 3 1 + 3 ε 2 1 + 2 ε 1 + 3 B τ 2 = 2 ε 4 0 ε 4 1 + ε 4 0 ε 2 1 + 2 ε 4 0 + 2 ε 3 0 ε 3 1 + ε 3 0 ε 2 1 + ε 3 0 ε 1 + ε 3 0 + ε 2 0 ε 4 1 + ε 2 0 ε 3 1 + ε 2 0 ε 2 1 + 2 ε 2 0 + ε 0 ε 3 1 + 2 ε 0 ε 1 + 2 ε 0 + 2 ε 4 1 + ε 3 1 + 2 ε 2 1 + 2 ε 1 + 4. Pries (CSU) Galois action on Fermat curves 13 / 26