Success and challenges in determining the rational points on curves ANTS X, San Diego, July 13, 2012 N ILS B RUIN R ATIONAL POINTS ON CURVES
Diophantine equations Example problems: Find the solutions x , y ∈ Q to x 2 + y 2 = 1 x 2 + y 2 = − 1 x 2 + y 2 = 5 x 2 + y 2 = 3 3 x 3 + 4 y 3 = 5 x 6 + 8 x 5 + 22 x 4 + 22 x 3 + 5 x 2 + 6 x + 1 = y 2 x 6 + x 2 + 1 = y 2 x 6 + 6 x 5 − 15 x 4 + 20 x 3 + 15 x 2 + 30 x − 17 = y 2 ( x 3 − x 2 − 2 x + 1 ) y 7 − ( x 3 − 2 x 2 − x + 1 ) = 0 x 4 + y 4 + x 2 y + 2 xy − y 2 + 1 = 0 x 2 y 2 − xy 3 − x 3 − 2 x 2 + y 2 − x + y = 0 Note: All of these ask for the rational points on curves. N ILS B RUIN R ATIONAL POINTS ON CURVES
Central Questions Definition: A curve C over Q is nice if it is: smooth , projective , absolutely irreducible . Typical example: Smooth plane projective curve: C : X 4 + Y 4 + X 2 YZ + 2 XYZ 2 − Y 2 Z 2 + Z 4 = 0 Decision problem: Given a nice curve C over Q , decide if C ( Q ) = / 0 . Determination problem: Given a nice curve C over Q , find a useful description of C ( Q ) . For curves of genus > 1 : List the finite set C ( Q ) . N ILS B RUIN R ATIONAL POINTS ON CURVES
Outline 1. Outline of a procedure to tackle the decision problem 2. Highlight challenges in executing the procedure 3. Finite Descent as a tool to face these challenges 4. Results for smooth plane quartics N ILS B RUIN R ATIONAL POINTS ON CURVES
Local obstructions Adelic points: → C ( A ) : = C ( R ) × ∏ C ( Q ) ֒ C ( Q p ) p Global-Local principle: C ( Q ) � = / 0 implies C ( A ) � = / 0 Happy fact: Deciding if C ( A ) = / 0 is decidable. Local-Global principle fails: C ( A ) � = / does not imply C ( Q ) � = / 0 , 0 Examples: 3 X 3 + 4 Y 3 + 5 Z 3 = 0 X 4 + Y 4 + X 2 YZ + 2 XYZ 2 − Y 2 Z 2 + Z 4 = 0 N ILS B RUIN R ATIONAL POINTS ON CURVES
Better information Alternative approach: Embed curve C in another variety with a sparser set of rational points, e.g., an Abelian variety J . Theorem (Mordell-Weil): J ( Q ) is a finitely generated abelian group: × Z r J ( Q ) ≃ J ( Q ) tors � �� � finite Principal homogeneous space: C ⊂ Pic 1 C under J = Pic 0 C . Pic 1 Pic 1 C ( Q ) � = / 0 if and only if C ≃ J Challenge : Decide if Pic 1 0 or find d ∈ Pic 1 C ( Q ) = / C ( Q ) . If Pic 1 C ( Q ) = / 0 then C ( Q ) = / 0 . Otherwise ι d : C ֒ → J . Challenge : Compute J ( Q ) ≃ J ( Q ) tors × Z r , in particular r . N ILS B RUIN R ATIONAL POINTS ON CURVES
� � � Mordell-Weil group combined with adelic information Assume: ◮ We have d ∈ Pic 1 C ( Q ) . ◮ We have generators for J ( Q ) . Commutative diagram: ι C ( Q ) J ( Q ) ρ ˜ ρ ˜ ι � J ( A ) • C ( A ) (Watch the Poonen • which modifies the J ( R ) factor) Conjecture: Writing C ( Q ) ⊂ C ( A ) for the topological closure, ? C ( Q ) = ι ( C ( A )) ∩ ˜ ρ ( J ( Q )) N ILS B RUIN R ATIONAL POINTS ON CURVES
� � � The Mordell-Weil sieve (see [Scharaschkin, B-Elkies (ANTS V), Flynn, B.-Stoll]) ι C ( Q ) J ( Q ) / BJ ( Q ) ρ S ρ S ι S � ∏ p ∈ S J ( F p ) / B · im ( ρ p ) ∏ p ∈ S C ( F p ) ◮ Let S be a finite set of primes ; B a positive integer � ◮ Let Λ p = ker ( ρ p : J ( Q ) → J ( F p )) and Λ S : = Λ p p ∈ S J ( Q ) ◮ C ( Q ) → V S , B : = im ( ι S ) ∩ im ( ρ S ) ⊂ Λ S + BJ ( Q ) Heuristic (Poonen): For appropriate S , B , the set V S , B consists only of cosets containing a point from C ( Q ) . N ILS B RUIN R ATIONAL POINTS ON CURVES
Decision procedure INPUT: A nice curve C of genus g > 0 . OUTPUT: P ∈ C ( Q ) or Unsolvable if C ( Q ) = / 0 . Execute in parallel: 0. Try candidates for P ∈ C ( Q ) and return P if one is found. Information from V S , B (step 5) helps. and 1. If C ( A ) = / 0 return Unsolvable 2. Determine d ∈ Pic 1 C ( Q ) or return Unsolvable if Pic 1 C ( Q ) = / 0 . 3. Determine J ( Q ) . 4. Choose reasonable values for S , B . 5. Mordell-Weil sieving: If V S , B = / 0 return Unsolvable . 6. Increase S , B ; go to 5. N ILS B RUIN R ATIONAL POINTS ON CURVES
How well does this work? Test case (B.-Stoll): Consider genus 2 curves admitting a model C : y 2 = f 6 x 6 + f 5 x 5 + ··· + f 0 with f i ∈ {− 3 ,..., 3 } Success: We were able to decide for all of them! All curves 196 171 100.00 % Curves with rational points 137 490 70.09 % Curves without rational points 58 681 29.91 % Curves with C ( A ) � = / 0 166 768 85.01 % Curves with C ( A ) � = / 0 and C ( Q ) = / 29 278 14.92 % 0 Curves that need BSD conjecture 42 0.02 % Disclosure: We only really needed MW-sieving for 1445 of these curves ( 27786 of these curves have a non-trivial 2 -cover obstruction to having rational points) N ILS B RUIN R ATIONAL POINTS ON CURVES
How to deal with rational points (see [Chabauty, Coleman, Flynn]) Problem: If P ∈ C ( Q ) then V S , B is never empty. Idea (Chabauty): Construct a p -adic analytic function Φ p on C ( Q p ) that vanishes on C ( Q ) . Restriction : Construction only works if rk J ( Q ) = r < g . Sketch of procedure: 1. Use MW-Sieving to find S , B and P i ∈ C ( Q ) such that V S , B = { P 1 ,..., P n } + Λ S + BJ ( Q ) 2. Find prime p with BJ ( Q ) ⊂ Λ p such that P i �≡ P j ( mod p ) for any i � = j 3. For each P i , use Φ p to show that there are no other rational points Q with Q ≡ P i ( mod p ) N ILS B RUIN R ATIONAL POINTS ON CURVES
Computational Challenges No guarantee that either procedure will terminate, i.e.: ◮ We only have a heuristic that MW-sieving converges to a sharp result. ◮ We have no guarantee we can always find a p such that Φ p does not have inconvenient extraneous p -adic zeros. Bigger problem: we cannot guarantee we can get started: For decision procedure: ◮ Decide if Pic 1 0 or find d ∈ Pic 1 C ( Q ) = / C ( Q ) . ◮ Determine the r in J ( Q ) ≃ J ( Q ) tors × Z r ◮ Find generators for J ( Q ) For determination procedure: ◮ What to do if r ≥ g ? (See [Wetherell, B.; future: Kim, Balakrishnan?]) N ILS B RUIN R ATIONAL POINTS ON CURVES
� � � n-descent Multiplication-by- n : n 0 → J [ n ] → J → J → 0 Taking galois cohomology: 0 → J ( Q ) γ → H 1 ( Q , J [ n ]) → H 1 ( Q , J ) nJ ( Q ) Approximate image locally: J ( Q ) γ H 1 ( Q , J [ n ]) nJ ( Q ) ∏ ρ p J ( Q p ) ∏ γ p � ∏ p H 1 ( Q p , J [ n ]) ∏ p nJ ( Q p ) Sel n ( J / Q ) : = { δ ∈ H 1 ( Q , J [ n ]) : ρ p ( δ ) ∈ im γ p for all p } N ILS B RUIN R ATIONAL POINTS ON CURVES
Computational considerations Explicit descent computations: We need to work with γ : J ( k ) nJ ( k ) → H 1 ( k , J [ n ]) for k = Q , R , Q p ◮ How do we represent J ( k ) ? ◮ How do we represent H 1 ( k , J [ n ]) ? ◮ How do we compute γ ? Representing J ( k ) : Pic 0 ( C / k ) ⊂ J ( k ) ; equality if C ( A ) � = / 0 . Use divisors on the curve. N ILS B RUIN R ATIONAL POINTS ON CURVES
Representing H 1 ( k , J [ n ]) Problem: We only know how to efficiently represent H 1 ( k , M ) for a very limited class of Galois modules. Twisted power: Let M be a Galois module and ∆ = Spec L = { θ 1 ,..., θ m } a Galois set. Define M ∆ : = M θ 1 ⊕···⊕ M θ m Hilbert 90: H 1 ( k , µ ∆ n ) = L × / L × n . Let J [ n ] = Spec ( L ) . Consider 0 → J [ n ] → ( µ n ) J [ n ] → R ∨ → 0 Cohomology: H 1 ( k , J [ n ]) → L × / L × n . N ILS B RUIN R ATIONAL POINTS ON CURVES
� � � Computations using descent setups (see [Cassels, Schaefer, Poonen-Shaefer, B.-Poonen-Stoll]) Writing L p = L ⊗ Q p J ( Q ) L × γ ˜ nJ ( Q ) L × n � L × J ( Q p ) γ p ˜ p L × n nJ ( Q p ) p ◮ Map ˜ γ is induced by a function f ∈ k ( C ) ⊗ L . ◮ Images of ˜ γ p are computable. ◮ For most p , this image lands in “unramified” part ◮ Image of ˜ γ is generated by S -units. γ ( J ) = { δ ∈ L × / L × n : ρ p ( δ ) ∈ im ˜ Sel ˜ γ p for all p } N ILS B RUIN R ATIONAL POINTS ON CURVES
Application to two challenges Bounding Ranks: � Z � r nJ ( Q ) = J ( Q ) tors J ( Q ) × nJ ( Q ) tors n Z So bounding the size of im γ bounds r (hopefully sharply). Embedding curve in J : [ Pic 1 C ] ∈ H 1 ( Q , J [ 2 g − 2 ]) There exists d ∈ Pic 1 C ( Q ) if and only if [ Pic 1 C ] ∈ im γ . Bonus: Map ˜ γ can be evaluated immediately on C . γ ( C ) = { δ ∈ L × / L × n : ρ p ( δ ) ∈ ˜ Sel ˜ γ p ( C ( Q p )) for all p } N ILS B RUIN R ATIONAL POINTS ON CURVES
� � � Example: Smooth plane quartics (B.-Poonen-Stoll) Let C be a smooth plane quartic. ◮ Set ∆ = Spec ( L ) of 28 bitangents ◮ Even weight vectors E ⊂ ( Z / 2 Z ) ∆ : µ ∆ 2 � J [ 2 ] � E ∨ � R ∨ � 0 0 ◮ Cohomology: J ( k ) L × γ � ˜ 2 J ( k ) L × 2 k × � � γ � J [ 2 ]( k ) � E ∨ ( k ) � R ∨ ( k ) � H 1 ( J [ 2 ]) � H 1 ( E ∨ ) 0 N ILS B RUIN R ATIONAL POINTS ON CURVES
� � We need all of computational algebraic number theory ... J ( k ) L × γ � ˜ 2 J ( k ) L × 2 k × � � γ � J [ 2 ]( k ) � E ∨ ( k ) � R ∨ ( k ) � H 1 ( J [ 2 ]) � H 1 ( E ∨ ) 0 ◮ ˜ γ consists of evaluation at the “generic” bitangent. ◮ We need the ring of integers of L and S -units in L . ◮ J [ 2 ]( k ) , R ∨ ( k ) , E ∨ ( k ) follow from identifying Gal ( L / k ) ⊂ Sp 6 ( F 2 ) . N ILS B RUIN R ATIONAL POINTS ON CURVES
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