Factoring polynomials over discrete valuation rings Adrien Poteaux - PowerPoint PPT Presentation

Introduction Definitions Irreducibility test Factorisation Conclusion Factoring polynomials over discrete valuation rings Adrien Poteaux ⋆ & Martin Weimann ✈ ⋆ : CFHP - CO2 - CRIStAL - Université de Lille ✈ : GAATI - Université de Polynésie Française 7 février 2019 Journées Nationales de Calcul Formel CIRM, Luminy adrien.poteaux@univ-lille.fr Factorisation over DVR 1 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion One example F = ( y α − x 2 ) 2 + x α ∈ A [ y ] with A = C [[ x ]] d = deg ( F ) = 2 α , δ = υ x ( Disc ( F )) = 2 α 2 − 4 α + 4. Assume α > 4 odd. Is F irreducible in C [[ x ]][ y ] ? adrien.poteaux@univ-lille.fr One example 1 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Using the Newton-Puiseux algorithm. F = ( y α − x 2 ) 2 + x α ∆ , slope − 2 4 α φ ∆ ( z ) = ( z − 1 ) 2 1 G ← F ( x α , x 2 ( y + 1 )) / x 4 α , 2 2 Hensel: G = A · B , α 2 α α 2 − 4 α 3 Recursive call with A A Polynomial size φ ∆ ( z ) = z + c α Θ( α 2 ) = Θ( δ ) F Θ( α 3 ) = Θ( d δ ) G B Θ( α 2 ) = Θ( δ ) A 2 2 α Answer: Yes complexity: Θ( d δ ) Answer in O ˜( δ ) ? adrien.poteaux@univ-lille.fr One example 2 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Another way ? α 2 − 4 α N ( A ) φ ∆ ( z ) = z + c α F = ( y α − x 2 ) 2 + x α 2 Writing ψ = y α − x 2 , we have F = ψ 2 + x α , Can we “guess” the second Newton polygon from ψ 2 + x α ? Can we “read” φ ∆ ? adrien.poteaux@univ-lille.fr One example 3 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Another way ? α 2 − 4 α N ( A ) φ ∆ ( z ) = z + c α F = ( y α − x 2 ) 2 + x α 2 Writing ψ = y α − x 2 , we have F = ψ 2 + x α , Can we “guess” the second Newton polygon from ψ 2 + x α ? Can we “read” φ ∆ ? Key ingredients: √ 2 ψ = F is an approximate root of F , F = ψ 2 + x α is the ψ -adic expansion of F . adrien.poteaux@univ-lille.fr One example 3 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Another way ? α 2 − 4 α N ( A ) φ ∆ ( z ) = z + c α F = ( y α − x 2 ) 2 + x α 2 Writing ψ = y α − x 2 , we have F = ψ 2 + x α , Can we “guess” the second Newton polygon from ψ 2 + x α ? Can we “read” φ ∆ ? Key ingredients: √ 2 ψ = F is an approximate root of F , F = ψ 2 + x α is the ψ -adic expansion of F . Questions: Why x α corresponds to α 2 − 4 α ? How to recover the correct characteristic polynomial ? adrien.poteaux@univ-lille.fr One example 3 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion This talk Context: A a discrete valuation ring (e.g. K (( x )) , Q p ), υ A valuation over A (e.g. υ x , υ p ), F ∈ A [ y ] (monic). Objective(s): 1 Irreducibility test in A [ y ] , 2 Factorisation of F in A [ y ] . 3 Case A = K [[ x ]] : Puiseux series of F ? Notations: d = deg ( F ) ; δ = υ A ( Disc ( F )) adrien.poteaux@univ-lille.fr Factorisation over DVR 4 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Approximate root of F ∈ A [ y ] monic [Ab10] Hyp: char ( A ) does not divide d , Let N ∈ N dividing d , Proposition There is an unique monic ψ ∈ A [ y ] such that: deg ( ψ ) = d / N , deg ( F − ψ N ) < d − d / N , √ N ❀ ψ = F is the N -th approximate root of F . √ F = y + a d − 1 d Example: ψ = is the d -th approximate root of F . d adrien.poteaux@univ-lille.fr Approximate roots of Abhyankar-Moh 5 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Valuations on A [ y ] Gauss valuation: F = � i a i y i , υ 0 ( F ) = min i υ A ( a i ) . Extended valuation: given ψ ∈ A [ y ] monic, m q ∈ Q : υ ψ = ( υ 0 , ψ, m q ) extends υ 0 . Defined by υ ψ ( ψ ) = m q , υ ψ ( y ) = m and υ ψ ( x ) = q , Expand F = � i a i ( y ) ψ i with deg ( a i ) < deg ( ψ ) , Generalised Newton polygon: N ψ ( F ) is the lower convex hull of ( i , υ ψ ( a i ψ i ) − υ ψ ( F )) i . adrien.poteaux@univ-lille.fr Extended valuation 6 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Improving the irreducibility test generalisation of the work of Abhyankhar to A [ y ] . link with the Newton–Puiseux algorithm for A = K (( x )) adrien.poteaux@univ-lille.fr Irreducibility in A [ y ] 7 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion We get the second Newton polygon ! F = ( y α − x 2 ) 2 + x α ∆ , slope − 2 4 α φ ∆ ( z ) = ( z − 1 ) 2 √ N ( F ) 2 With m = 2, q = α , ψ = F , we get: 2 α F = ψ 2 + x α . α 2 − 4 α υ ψ ( F ) = 4 α N ψ ( F ) υ ψ ( ψ 2 ) − υ ψ ( F ) = 0, υ ψ ( x α ) − υ ψ ( F ) = α 2 − 4 α . 2 Reminder: υ ψ ( x ) = α υ ψ ( y ) = 2 υ ψ ( ψ ) = 2 α adrien.poteaux@univ-lille.fr Irreducibility in A [ y ] 7 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Complexity ? √ N Computing F : O ( M ( d )) = O ˜( d ) op in A . F ∞ = y d F ( 1 / y ) the reciprocal polynomial of F , F ∞ ( 0 ) = 1 ❀ ∃ ! φ ∈ A [[ y ]] s.t. φ ( 0 ) = 1 and φ N = F ∞ , φ is the root of Z N − F ∞ = 0 ❀ Newton iteration ! d ψ is the reciprocal polynomial of ⌈ φ ⌉ N adrien.poteaux@univ-lille.fr Irreducibility in A [ y ] 8 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Complexity ? √ N Computing F : O ( M ( d )) = O ˜( d ) op in A . F ∞ = y d F ( 1 / y ) the reciprocal polynomial of F , F ∞ ( 0 ) = 1 ❀ ∃ ! φ ∈ A [[ y ]] s.t. φ ( 0 ) = 1 and φ N = F ∞ , φ is the root of Z N − F ∞ = 0 ❀ Newton iteration ! d ψ is the reciprocal polynomial of ⌈ φ ⌉ N ψ -adic expansion: O ( M ( d ) log ( N )) = O ˜( d ) op in A . N 2 + B ❀ O ( M ( d )) F = A ψ Recursive call on A and B . adrien.poteaux@univ-lille.fr Irreducibility in A [ y ] 8 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Complexity ? √ N Computing F : O ( M ( d )) = O ˜( d ) op in A . F ∞ = y d F ( 1 / y ) the reciprocal polynomial of F , F ∞ ( 0 ) = 1 ❀ ∃ ! φ ∈ A [[ y ]] s.t. φ ( 0 ) = 1 and φ N = F ∞ , φ is the root of Z N − F ∞ = 0 ❀ Newton iteration ! d ψ is the reciprocal polynomial of ⌈ φ ⌉ N ψ -adic expansion: O ( M ( d ) log ( N )) = O ˜( d ) op in A . N 2 + B ❀ O ( M ( d )) F = A ψ Recursive call on A and B . Truncation: n = 2 δ/ d . Total cost: δ plog ( d ) ! adrien.poteaux@univ-lille.fr Irreducibility in A [ y ] 8 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Miscellaneous More than one Newton–Puiseux recursive call ? Compute successive approximate roots ψ 0 , · · · , ψ k ψ − 1 = x Recursive augmented valuations υ k = ( υ k − 1 , ψ k , m k q k ) :  υ k ( ψ i ) = q k υ k − 1 ( ψ i ) − 1 ≤ i < k − 1   υ k ( ψ k − 1 ) = q k υ k − 1 ( ψ k − 1 ) + m k   υ k ( ψ k ) = q k υ k ( ψ k − 1 ) N k ( F ) via generalised ( ψ 0 , · · · , ψ k ) -adic expansions Compute the characteristic polynomials ? The coefficients of the ψ -adic expansions must be corrected , Compute some λ k ( ψ i ) ∈ K k (tower of fields). Make a single (univariate) irreducibility test ? Rely on dynamic evaluation. adrien.poteaux@univ-lille.fr Irreducibility in A [ y ] 9 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Hensel–Newton algorithm and extended valuations adrien.poteaux@univ-lille.fr Irreducibility in A [ y ] 10 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Slope factorisation [CaRoVa16] d � a i y i F ( y ) = i = 0 β a “break” of N ( F ) , A ( y ) β � a i y i , V 0 = 1, A 0 = i = 0 B ( y ) Newton iteration: A k + 1 = A k + ( V k F % A k ) β d B k + 1 = F � A k + 1 ( 2 V k − V 2 V k + 1 = k B k + 1 )% A k + 1 Factorisation up to precision n ❀ O ˜( n d ) adrien.poteaux@univ-lille.fr Newton–Hensel and extended valuations 10 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Hensel lemma works with extended valuations Lemma Assume B = ψ b + · · · and υ ( B ) = b υ ( ψ ) . Then υ ( A % B ) ≥ υ ( A ) , υ ( A � B ) ≥ υ ( A ) − υ ( B ) . adrien.poteaux@univ-lille.fr Newton–Hensel and extended valuations 11 / 14

Introduction Definitions Irreducibility test Factorisation Conclusion Hensel lemma works with extended valuations Lemma Assume B = ψ b + · · · and υ ( B ) = b υ ( ψ ) . Then υ ( A % B ) ≥ υ ( A ) , υ ( A � B ) ≥ υ ( A ) − υ ( B ) . Theorem Assume υ ( F − G H ) ≥ υ ( F ) + n and υ ( S G + T H − 1 ) ≥ n . Then ˜ G , ˜ H , ˜ S , ˜ T = HenselStep ( F , G , H , S , T ) satisfies: υ ( F − ˜ G ˜ H ) ≥ υ ( F ) + 2 n , υ ( ˜ S ˜ G + ˜ T ˜ H − 1 ) ≥ 2 n . adrien.poteaux@univ-lille.fr Newton–Hensel and extended valuations 11 / 14