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

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Introduction Definitions Irreducibility test Factorisation Conclusion Factoring polynomials over discrete valuation rings Adrien Poteaux & Martin Weimann : CFHP - CO2 - CRIStAL - Universit de Lille : GAATI - Universit de

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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

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Introduction Definitions Irreducibility test Factorisation Conclusion

One example

F = (yα − x2)2 + xα ∈ A[y] with A = C[[x]] d = deg(F) = 2 α, δ = υx(Disc(F)) = 2 α2 − 4 α + 4. Assume α > 4 odd. Is F irreducible in C[[x]][y] ?

adrien.poteaux@univ-lille.fr One example 1 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

Using the Newton-Puiseux algorithm.

F = (yα − x2)2 + xα

1 G ← F(xα, x2(y + 1))/x4 α, 2 Hensel: G = A · B, 3 Recursive call with A

Polynomial size F Θ(α2) = Θ(δ) G Θ(α3) = Θ(d δ) A Θ(α2) = Θ(δ)

4 ∆, slope − 2

α

φ∆(z) = (z − 1)2 2 α 2 α B 2 α α2 − 4 α A φ∆(z) = z + cα 2

Answer: Yes complexity: Θ(d δ) Answer in O˜(δ) ?

adrien.poteaux@univ-lille.fr One example 2 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

Another way ?

F = (yα − x2)2 + xα α2 − 4 α N(A) φ∆(z) = z + cα 2 Writing ψ = yα − x2, we have F = ψ2 + xα,

Can we “guess” the second Newton polygon from ψ2 + xα ? Can we “read” φ∆ ?

adrien.poteaux@univ-lille.fr One example 3 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

Another way ?

F = (yα − x2)2 + xα α2 − 4 α N(A) φ∆(z) = z + cα 2 Writing ψ = yα − x2, we have F = ψ2 + xα,

Can we “guess” the second Newton polygon from ψ2 + xα ? Can we “read” φ∆ ?

Key ingredients:

ψ =

2

√ F is an approximate root of F, F = ψ2 + xα is the ψ-adic expansion of F.

adrien.poteaux@univ-lille.fr One example 3 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

Another way ?

F = (yα − x2)2 + xα α2 − 4 α N(A) φ∆(z) = z + cα 2 Writing ψ = yα − x2, we have F = ψ2 + xα,

Can we “guess” the second Newton polygon from ψ2 + xα ? Can we “read” φ∆ ?

Key ingredients:

ψ =

2

√ F is an approximate root of F, F = ψ2 + xα is the ψ-adic expansion of F.

Questions:

Why xα corresponds to α2 − 4 α ? How to recover the correct characteristic polynomial ?

adrien.poteaux@univ-lille.fr One example 3 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

This talk

Context: A a discrete valuation ring (e.g. K((x)), Qp), υA valuation over A (e.g. υx, υp), 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 = υA(Disc(F))

adrien.poteaux@univ-lille.fr Factorisation over DVR 4 / 14

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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 − ψN) < d − d/N, ❀ ψ =

N

√ F is the N-th approximate root of F. Example: ψ =

d

√ F = y + ad−1 d is the d-th approximate root of F.

adrien.poteaux@univ-lille.fr Approximate roots of Abhyankar-Moh 5 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

Valuations on A[y]

Gauss valuation:

F =

i ai y i,

υ0(F) = mini υA(ai).

Extended valuation: given ψ ∈ A[y] monic, m

q ∈ Q:

υψ = (υ0, ψ, m

q ) extends υ0.

Defined by υψ(ψ) = m q, υψ(y) = m and υψ(x) = q, Expand F =

i ai(y) ψi with deg(ai) < deg(ψ),

Generalised Newton polygon: Nψ(F) is the lower convex hull of (i, υψ(ai ψi) − υψ(F))i.

adrien.poteaux@univ-lille.fr Extended valuation 6 / 14

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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

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Introduction Definitions Irreducibility test Factorisation Conclusion

We get the second Newton polygon !

F = (yα − x2)2 + xα With m = 2, q = α, ψ =

2

√ F, we get: F = ψ2 + xα. υψ(F) = 4 α υψ(ψ2) − υψ(F) = 0, υψ(xα) − υψ(F) = α2 − 4 α.

4 ∆, slope − 2

α

N(F) φ∆(z) = (z − 1)2 2 α α2 − 4 α Nψ(F) 2

Reminder: υψ(x) = α υψ(y) = 2 υψ(ψ) = 2 α

adrien.poteaux@univ-lille.fr Irreducibility in A[y] 7 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

Complexity ?

Computing

N

√ 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 φN = F∞, φ is the root of Z N − F∞ = 0 ❀ Newton iteration ! ψ is the reciprocal polynomial of ⌈φ⌉

d N

adrien.poteaux@univ-lille.fr Irreducibility in A[y] 8 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

Complexity ?

Computing

N

√ 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 φN = F∞, φ is the root of Z N − F∞ = 0 ❀ Newton iteration ! ψ is the reciprocal polynomial of ⌈φ⌉

d N

ψ-adic expansion: O(M(d) log(N)) = O˜(d) op in A.

F = A ψ

N 2 + B ❀ O(M(d))

Recursive call on A and B.

adrien.poteaux@univ-lille.fr Irreducibility in A[y] 8 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

Complexity ?

Computing

N

√ 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 φN = F∞, φ is the root of Z N − F∞ = 0 ❀ Newton iteration ! ψ is the reciprocal polynomial of ⌈φ⌉

d N

ψ-adic expansion: O(M(d) log(N)) = O˜(d) op in A.

F = A ψ

N 2 + B ❀ O(M(d))

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

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Introduction Definitions Irreducibility test Factorisation Conclusion

Miscellaneous

More than one Newton–Puiseux recursive call ?

Compute successive approximate roots ψ0, · · · , ψk ψ−1 = x Recursive augmented valuations υk = (υk−1, ψk, mk

qk ):

     υk(ψi) = qkυk−1(ψi) − 1 ≤ i < k − 1 υk(ψk−1) = qkυk−1(ψk−1) + mk υk(ψk) = qkυk(ψk−1) Nk(F) via generalised (ψ0, · · · , ψk)-adic expansions

Compute the characteristic polynomials ?

The coefficients of the ψ-adic expansions must be corrected, Compute some λk(ψi) ∈ Kk (tower of fields).

Make a single (univariate) irreducibility test ?

Rely on dynamic evaluation.

adrien.poteaux@univ-lille.fr Irreducibility in A[y] 9 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

Hensel–Newton algorithm and extended valuations

adrien.poteaux@univ-lille.fr Irreducibility in A[y] 10 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

Slope factorisation [CaRoVa16]

F(y) =

d

  • i=0

ai yi β a “break” of N(F), A0 =

β

  • i=0

aiyi, V0 = 1, Newton iteration: Ak+1 = Ak + (Vk F%Ak) Bk+1 = F Ak+1 Vk+1 = (2 Vk − V 2

k Bk+1)%Ak+1

B(y) A(y) d β

Factorisation up to precision n ❀ O˜(n d)

adrien.poteaux@univ-lille.fr Newton–Hensel and extended valuations 10 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

Hensel lemma works with extended valuations

Lemma

Assume B = ψb + · · · 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

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Hensel lemma works with extended valuations

Lemma

Assume B = ψb + · · · 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

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Good initialisation ?

F = ψ3 + y2 x3 ψ + x6 y with ψ = y3 − x2 υψ(x) = 3, υψ(y) = 2, υψ(ψ) = 6. Extend υψ with the lower edge: υ(x) = 6, υ(y) = 4, υ(ψ) = 13

2 Nψ(F) slope − 1

2

3 1 1

G0 =

26

  • ψ2 + y2 x3, H0 =

13

  • ψ

= ⇒

39

  • F

− G0H0 =

40

  • x6 y

adrien.poteaux@univ-lille.fr Newton–Hensel and extended valuations 12 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

Good initialisation ?

F = ψ3 + y2 x3 ψ + x6 y with ψ = y3 − x2 υψ(x) = 3, υψ(y) = 2, υψ(ψ) = 6. Extend υψ with the lower edge: υ(x) = 6, υ(y) = 4, υ(ψ) = 13

2 Nψ(F) slope − 1

2

3 1 1

G0 =

26

  • ψ2 + y2 x3, H0 =

13

  • ψ

= ⇒

39

  • F

− G0H0 =

40

  • x6 y

With s0 = 1 and t0 = − T, we have s0 (T 2 + 1) + t0 T = 1, S0 = x−5y

−26

, T0 = −x−5y ψ

  • −13

= ⇒ S0G0 + T0H0 − 1 = x−2ψ

1

adrien.poteaux@univ-lille.fr Newton–Hensel and extended valuations 12 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

State of the art (sketch)

Abhyankar-Moh [Ab06]: approximate roots, Mac Lane, Abhyankar [Ma362,Ab90,Ru14]: extended valuations, Montes et al [Mo99,GuMoNa11&12,BaNaSt13,GuNaPa12] O˜(d2 + dδ2), Caruso et al [CaRoVa16]: slope factorisation,

Case A = K[[x]]:

Sasaki et al [KaSa99,AlAtMa17]: Extended Hensel Construction

at least O˜(d2 (δ + d2)),

Puiseux [PoRy15,PoWe]: Newton–Puiseux algorithm O˜(d δ).

adrien.poteaux@univ-lille.fr Factorisation over DVR 13 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

Conclusion

Irreducibility test in A[y] in O˜(δ), ← improved by a factor d ! “direct” factorisation in A[y]: O˜(ρ n d), ← was O˜(n d2) Sage prototype,

“Bivariate” computations above the residue field of A (no field extension).

Puiseux series ?

N1 = d/2: ψ1 = ψ2

0 + X m1S1(X)2

❀ S1(X) is an approximate root (❀ Newton iteration !) q1 > 2 ? Solving some linear system ?

Example: if S1(x) = x

1 3 P1(x) + x 2 3 P2(x),

ψ1 = ψ3

0 − 3 x P1 P2 ψ0 − x P3 1 − x2 P3 2

adrien.poteaux@univ-lille.fr Factorisation over DVR 14 / 14

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Bibliographie

adrien.poteaux@univ-lille.fr bibliography 14 / 14

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Introduction Definitions Irreducibility test Factorisation Conclusion

  • S. Abhyankar.

Irreducibility criterion for germs of analytic functions of two complex variables.

  • Adv. Mathematics, 35:190–257, 1989.
  • S. Abhyankar.

Algebraic Geometry for Scientists and Engineers, volume 35 of Mathematical surveys and monographs.

  • Amer. Math. Soc., 1990.
  • S. Abhyankar.

Lectures on Algebra. Number vol. 1 in Lectures on Algebra. World Scientific, 2006.

  • P. Alvandi, M. Ataei, and M. Moreno Maza.

On the extended hensel construction and its application to the computation of limit points. In ISSAC ’17, pages 13–20. J.-D. Bauch, E. Nart, and H. Stainsby.

adrien.poteaux@univ-lille.fr bibliography 14 / 14

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Complexity of the OM factorizations of polynomials over local fields. LMS Journal of Computation and Mathematics, 16:139–171, 2013.

  • X. Caruso, D. Roe, and T. Vaccon.

Division and slope factorization of p-adic polynomials. In ISSAC ’16, pages 159–166.

  • J. v. z. Gathen and J. Gerhard.

Modern Computer Algebra. Cambridge University Press, New York, NY, USA, 3rd edition, 2013.

  • J. Guàrdia, J. Montes, and E. Nart.

Higher Newton polygons in the computation of discriminants and prime ideal decomposition in number fields.

  • J. Théor. Nombres Bordx., 23(3):667–696, 2011.
  • J. Guàrdia, J. Montes, and E. Nart.

adrien.poteaux@univ-lille.fr bibliography 14 / 14

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Newton polygons of higher order in algebraic number theory. Transsactions of the American Mathematical Society, 364:361–416, 2012.

  • J. Guàrdia, E. Nart, and S. Pauli.

Single-factor lifting and factorization of polynomials over local fields. Journal of Symbolic Computation, 47(11):1318 – 1346, 2012.

  • F. Kako and T. Sasaki.

Solving multivariate algebraic equations by Hensel construction. Japan J. of Industrial and Applied Math., 16:257–285, 1999.

  • S. MacLane.

A construction for absolute values in polynomial rings.

  • Trans. Amer. Math. Soc., 40(3):363–395, 1936.
  • S. Mac Lane.

adrien.poteaux@univ-lille.fr bibliography 14 / 14

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A construction for prime ideals as absolute values of an algebraic field. Duke Math. J., 2(3):492–510, 1936.

  • J. Montes Peral.

Polígonos de newton de orden superior y aplicaciones aritméticas. PhD thesis, Universitat de Barcelona, 1999.

  • A. Poteaux and M. Rybowicz.

Improving complexity bounds for the computation of puiseux series over finite fields. ISSAC ’15, pages 299–306

  • A. Poteaux and M. Weimann.

Computing Puiseux series: a fast divide and conquer algorithm. arXiv:1708.09067, pages 1–27, 2017.

  • J. Rüth.

Models of curves and valuations.

adrien.poteaux@univ-lille.fr bibliography 14 / 14

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PhD thesis, Universität Ulm, 2014.

adrien.poteaux@univ-lille.fr bibliography 14 / 14