Division and Slope Factorization of p-Adic Polynomials · Division and Slope Factorization of p-Adic Polynomials Introduction Why should one work with p-adic numbers ? p-adic methods

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Division and Slope Factorization of p-Adic Polynomials

Division and Slope Factorization of p-AdicPolynomials

Xavier Caruso, David Roe Tristan Vaccon

Univ.Rennes 1, Univ. Pittsburgh, 立教大学

July 22nd, 2016

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Division and Slope Factorization of p-Adic PolynomialsIntroduction

1 Division and Differential Precisionp-Adic PrecisionStudy of the divisionModular Multiplication

2 Newton PolygonsBasicsEuclidean divisionPrecision: Return on Modular Multiplication

3 Slope factorizationA Newton schemeApplying differential precision

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Why should one work with p-adic numbers ?

p-adic methodsWorking in Qp instead of Q, one can handle more efficiently thecoefficients growth ;

e.g. Dixon’s method (used in F4), Polynomial factorization viaHensel’s lemma.

p-adic algorithmsGoing from Z/pZ to Zp and then back to Z/pZ enables morecomputation, e.g. the algorithms of Bostan et al. and Lercier et al.using p-adic differential equations ;Kedlaya’s and Lauder’s counting-point algorithms via p-adiccohomology ;

My personal (long-term) motivationComputing (some) moduli spaces of p-adic Galois representations.

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p-adic methodsWorking in Qp instead of Q, one can handle more efficiently thecoefficients growth ;e.g. Dixon’s method (used in F4), Polynomial factorization viaHensel’s lemma.



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p-adic algorithmsGoing from Z/pZ to Zp and then back to Z/pZ enables morecomputation,

e.g. the algorithms of Bostan et al. and Lercier et al.using p-adic differential equations ;Kedlaya’s and Lauder’s counting-point algorithms via p-adiccohomology ;


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p-adic algorithmsGoing from Z/pZ to Zp and then back to Z/pZ enables morecomputation, e.g. the algorithms of Bostan et al. and Lercier et al.using p-adic differential equations ;

Kedlaya’s and Lauder’s counting-point algorithms via p-adiccohomology ;


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Studying polynomial computations over p-adics

A building blockAt ISSAC 2015, we have studied the p-adic stability of somecomputations in linear algebra.

This year, we study basic operationsrelated to polynomial computations.

More motivationsUnderstanding basic operations related to field extensions, inparticular division and quotients.Understanding the behaviour of precision during factorisation: overQp or kJT K, or as an intermediate to factorisation over Q.

Today’s highlightsOptimal tracking of precision for modular multiplication anddiffused digits.Precision for slope factorization algorithms.

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A building blockAt ISSAC 2015, we have studied the p-adic stability of somecomputations in linear algebra. This year, we study basic operationsrelated to polynomial computations.



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More motivationsUnderstanding basic operations related to field extensions, inparticular division and quotients.

Understanding the behaviour of precision during factorisation: overQp or kJT K, or as an intermediate to factorisation over Q.


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Today’s highlightsOptimal tracking of precision for modular multiplication anddiffused digits.

Precision for slope factorization algorithms.

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What are p-adic numbers?

p refers to a prime number

p-adic numbers are numbers written in p-basis of the shape:a = . . . ai . . . a2 a1 a0 , a−1 a−2 . . . a−n

with 0 ≤ ai < p for all i .

Addition and multiplication on these numbers are defined by applyingSchoolBook algorithms.

The valuation vp(a) of a is the smallest v such that av ̸= 0.

The p-adic numbers form the field Qp.

A p-adic number with no digit after the comma is ap-adic integer.

The p-adic integers form a subring Zp of Qp.

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Summary on p-adics

Proposition

Zp/pZp = Z/pZ.

∀k ∈ N,Zp/pkZp = Z/pkZ.

A first ideaQp is an extension of Q where one can perform calculus, as simplyas over R.We are closer to arithmetic : we can reduce modulo p.

Remark

Qp R Zp Zp

Q Z/pZ Z/pZ

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Summary on p-adics

Proposition

Zp/pZp = Z/pZ.



Remark

Qp R Zp Zp

Q Z/pZ Z/pZ

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Summary on p-adics

Proposition

Zp/pZp = Z/pZ.


A first ideaQp is an extension of Q where one can perform calculus, as simplyas over R.

We are closer to arithmetic : we can reduce modulo p.

Remark

Qp R Zp Zp

Q Z/pZ Z/pZ

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Summary on p-adics

Proposition

Zp/pZp = Z/pZ.



Remark

Qp R Zp Zp

Q Z/pZ Z/pZ

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Summary on p-adics

Proposition

Zp/pZp = Z/pZ.



Remark

Qp R Zp Zp

Q Z/pZ Z/pZ

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Definition of the precision

Finite-precision p-adics

Elements of Qp can be written∑+∞

i=k aipi , with ai ∈ J0, p − 1K, k ∈ Zand p a prime number.While working with a computer, we usually only can consider thebeginning of this power serie expansion: we only consider elements of thefollowing form

∑d−1i=l aipi + O(pd) , with l ∈ Z.

Definition

The order, or the absolute precision of∑d−1

i=k aipi + O(pd) is d .

ExampleThe order of 3 ∗ 7−1 + 4 ∗ 70 + 5 ∗ 71 + 6 ∗ 72 + O(73) is 3.

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Definition




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Definition




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Definition




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p-adic precion vs real precisionThe quintessential idea of the step-by-step analysis is the following :

Proposition (p-adic errors don’t add)Indeed,

(a + O(p k )) + (b + O(p k )) = a + b + O(p k ).

That is to say, if a and b are known up to precision O(pk), then so isa + b.

RemarkIt is quite the opposite to when dealing with real numbers, because ofRound-off error :

(1 + 5 ∗ 10−2) + (2 + 6 ∗ 10−2) = 3 + 1 ∗ 10−1 + 1 ∗ 10−2.

That is to say, if a and b are known up to precision 10−n, then a + b is

known up to 10(−n +1 ).

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(a + O(p k )) + (b + O(p k )) = a + b + O(p k ).



(1 + 5 ∗ 10−2) + (2 + 6 ∗ 10−2) = 3 + 1 ∗ 10−1 + 1 ∗ 10−2.



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(a + O(p k )) + (b + O(p k )) = a + b + O(p k ).



(1 + 5 ∗ 10−2) + (2 + 6 ∗ 10−2) = 3 + 1 ∗ 10−1 + 1 ∗ 10−2.



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(a + O(p k )) + (b + O(p k )) = a + b + O(p k ).



(1 + 5 ∗ 10−2) + (2 + 6 ∗ 10−2) = 3 + 1 ∗ 10−1 + 1 ∗ 10−2.



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Precision formulae

Proposition (addition)

(x0 + O(pk0)) + (x1 + O(pk1)) = x0 + x1 + O(pmin(k0,k1))

Proposition (multiplication)

(x0 + O(pk0)) ∗ (x1 + O(pk1)) = x0 ∗ x1 + O(pmin(k0+vp(x1),k1+vp(x0)))

Proposition (division)

xpa + O(pb)ypc + O(pd) = x ∗ y−1pa−c + O(pmin(d+a−2c,b−c))

In particular, 1pcy + O(pd) = y−1p−c + O(pd−2c)

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Precision formulae








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Precision formulae








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Interesting behaviour

Large modular multiplicationSetting: 10 significant digits in Q2.

Modulus P = X 2 + 2.

A = 5 − 2X + 4X 2.

Compute iteratively (naively) An = An mod P.

Behaviour of significant digits

n 100 101 102 103 104 105 106

relative precision of lc(An) 10 9 8 7 6 5 4

Question:How can we obtain satisfying / optimal behaviour regarding toprecision?

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A = 5 − 2X + 4X 2.



n 100 101 102 103 104 105 106



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A = 5 − 2X + 4X 2.



n 100 101 102 103 104 105 106



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A = 5 − 2X + 4X 2.



n 100 101 102 103 104 105 106



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A = 5 − 2X + 4X 2.



n 100 101 102 103 104 105 106



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A = 5 − 2X + 4X 2.



n 100 101 102 103 104 105 106



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Division and Slope Factorization of p-Adic PolynomialsDivision and Differential Precision

p-Adic Precision

Table of contents




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p-Adic Precision

The Main lemma of p-adic differential precision

Lemma (CRV14)Let f : Qn

p → Qmp be a (strictly) differentiable mapping.

Let x ∈ Qnp. We assume that f ′(x) is surjective.

Then for any ball B = B(0, r) small enough,

f (x + B) = f (x) + f ′(x) · B.

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p-Adic Precision






f (x + B) = f (x) + f ′(x) · B.

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p-Adic Precision






f (x + B) = f (x) + f ′(x) · B.

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p-Adic Precision






f (x + B) = f (x) + f ′(x) · B.

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p-Adic Precision

Geometrical meaning

Interpretation

x

B

f (x)

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p-Adic Precision

Geometrical meaning

Interpretation

x

B

f (x)

f ′(x)

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p-Adic Precision

Geometrical meaning

Interpretation

x

B

f (x)

f ′(x)

f ′(x) · B

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p-Adic Precision

Geometrical meaning

Interpretation

xx + B

B

f (x)

f ′(x)

f ′(x) · B

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p-Adic Precision

Geometrical meaning

Interpretation

xx + B

f

B

f (x)

f ′(x)

f ′(x) · B

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p-Adic Precision

Geometrical meaning

Interpretation

xx + B

f

B

f (x)

f ′(x)

f (x) + f ′(x) · B

f ′(x) · B

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p-Adic Precision

Lattices

LemmaLet f : Qn




f (x + ) = f (x) + f ′(x) · .

RemarkThis allows more models of precision, like

(x , y) = (1 + O(p10), 1 + O(p)).

RemarkOur framework can be extended to (complete) ultrametric K -vectorspaces (e.g. being Fp((X ))n, Q((X ))m, R((ε))s).

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p-Adic Precision

Lattices

LemmaLet f : Qn




f (x + B) = f (x) + f ′(x) · B.


(x , y) = (1 + O(p10), 1 + O(p)).


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p-Adic Precision

Lattices

LemmaLet f : Qn



Then for any ball B = B(0, r) small enough, for any open Zp-latticeH ⊂ B

f (x + H) = f (x) + f ′(x) · H.


(x , y) = (1 + O(p10), 1 + O(p)).


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p-Adic Precision

Lattices

LemmaLet f : Qn




f (x + H) = f (x) + f ′(x) · H.


(x , y) = (1 + O(p10), 1 + O(p)).


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p-Adic Precision

Lattices

LemmaLet f : Qn




f (x + H) = f (x) + f ′(x) · H.


(x , y) = (1 + O(p10), 1 + O(p)).


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p-Adic Precision

Higher differentials

What is small enough ?How can we determine when the lemma applies ?When f is locally analytic, it essentially corresponds to

+∞∑k=2

1k! f (k)(x) · Hk ⊂ f ′(x) · H.

This can be determined with Newton-polygon techniques.

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p-Adic Precision


What is small enough ?How can we determine when the lemma applies ?

When f is locally analytic, it essentially corresponds to

+∞∑k=2

1k! f (k)(x) · Hk ⊂ f ′(x) · H.


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p-Adic Precision



+∞∑k=2

1k! f (k)(x) · Hk ⊂ f ′(x) · H.


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p-Adic Precision



+∞∑k=2

1k! f (k)(x) · Hk ⊂ f ′(x) · H.


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Study of the division

Table of contents




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Handling the precision

Lattice precisionTo each polynomial, we attach a Zp-lattice (given by a basis of thislattice).

Lattice and divisionA and B are known with precision lattice HA and HB . Then (HQ , HR) aregiven by the Euclidean division of HA − QHB by B.

Proof.

A = BQ + R,

A + δA = (B + δB)(Q + δQ) + (R + δR),δA = QδB + BδQ + δR (at first order),

δA − QδB = BδQ + δR.

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Proof.

A = BQ + R,



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Proof.

A = BQ + R,

A + δA = (B + δB)(Q + δQ) + (R + δR),

δA = QδB + BδQ + δR (at first order),δA − QδB = BδQ + δR.

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Proof.

A = BQ + R,



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Proof.

A = BQ + R,



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Modular Multiplication

Table of contents




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Lattice precision and modular multiplication

Composed derivativesIt is easy to handle in an optimal way the modular multiplication byapplying:

For A × B,δ(A × B) = AδB + BδA.

For remainder R in A = BQ + R,

δ(R) = δA − QδB mod B.

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For A × B,

δ(A × B) = AδB + BδA.



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Toward numerical understanding

Displaying precision

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Displaying precisionLet H ⊂ Qn

p be a lattice.

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p be a lattice.

H0 = π1(H) ⊕ · · · ⊕ πn(H)

provides the best precision oncoordinates.It is the smallest diagonal latticecontaining H.

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p be a lattice.

H0 = π1(H) ⊕ · · · ⊕ πn(H)

provides the best precision oncoordinates.It is the smallest diagonal latticecontaining H.

Diffused digitsThe number of diffused digits of precisionof H is the length of H0/H.

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More on diffused digits

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Another definitionHere, the number of diffused digits is:

− logp (|H0/H|) .

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Diffused digits: example

Diffused digits: the latticeLet

H =⟨ 1

pp2

,

pp2

2p4

,

2pp2

2p3

⟩.

Then

H0 =

1p

p2

.

Number of diffused digits

The SNF of H0/H is:

1p

p

. Hence 2 diffused digits.

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H =⟨ 1

pp2

,

pp2

2p4

,

2pp2

2p3

⟩.

Then

H0 =

1p

p2

.


The SNF of H0/H is:

1p

p


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H =⟨ 1

pp2

,

pp2

2p4

,

2pp2

2p3

⟩.

Then

H0 =

1p

p2

.


The SNF of H0/H is:

1p

p

.

Hence 2 diffused digits.

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H =⟨ 1

pp2

,

pp2

2p4

,

2pp2

2p3

⟩.

Then

H0 =

1p

p2

.


The SNF of H0/H is:

1p

p


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Comparison: ∏ni=1 Ai mod M

Modulus M nGain of precision

Jagged Lattice(not dif. + dif.)

X 5 + X 2 + 1(Irred. mod 2)

10 0.2 0.2 + 0.050 4.2 4.2 + 0.0100 11.2 11.2 + 0.0

X 5 + 1(Sep. mod 2)

10 0.4 0.9 + 6.050 5.6 11.1 + 42.0100 13.6 27.0 + 87.0

X 5 + 2(Eisenstein)

10 6.2 6.2 + 0.050 44.0 44.0 + 0.0100 92.5 92.5 + 0.0

(X + 1)5 + 2(Shift Eisenstein)

10 0.6 4.7 + 1.450 7.1 42.6 + 1.4100 15.1 91.8 + 1.4

X 5 + X + 2(Two slopes)

10 1.7 7.9 + 9.850 8.1 70.7 + 59.8100 16.1 152.6 + 125.9

Figure: Precision for modular multiplication

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Qualitative understanding: two possibilities

Input: P, A1, . . . , An ∈ Zp[X ]d known at precision O(pN)Output: the product A1A2 · · · An mod P

1. A = 1 + O(pN)2. for i in 1, 2, . . . n:3. A = A · Ai

4. return A

×Ai

Zp[X ] Zp[X ]

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Qualitative understanding: two possibilitiesInput: P, A1, . . . , An ∈ Zp[X ]d known at precision O(pN)Output: the product A1A2 · · · An mod P


4. return A

×Ai

Zp[X ] Zp[X ]

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4. return A

×Ai

Zp[X ] Zp[X ]

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4. return A

×Ai

Zp[X ] Zp[X ]

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4. return A

×Ai

Zp[X ] Zp[X ]

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4. return A

×Ai

Zp[X ] Zp[X ]

no diffused digits

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4. return A

×Ai

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4. return A

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4. return A

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4. return A

×Ai

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diffused digits

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Qualitative understanding: long-termInput: P, A1, . . . , An ∈ Zp[X ]d known at precision O(pN)Output: the product A1A2 · · · An mod P


4. return A

Zp[X ] Zp[X ]

without lattice with lattice

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4. return A

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without lattice with latticei = 0

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4. return A

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4. return A

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4. return A

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4. return A

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4. return A

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4. return A

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4. return A

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4. return A

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4. return A

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Division and Slope Factorization of p-Adic PolynomialsNewton Polygons

Basics

Table of contents




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Basics

Newton polygon of a polynomial

DefinitionLet f (X ) = a0 + · · · + anX n ∈ Qp[x ] (remark: QpJxK would also be fine).

Let U = {(0, v(a0)), . . . , (n, v(an))}.We define the Newton polygon of f , NP(f ), as the lower convex hullof U.By lower convex hull, we mean the points of the convex hull of U belowthe straight line from (0, (a0)) to (n, v(an)).

PropositionLet Newtf be the (point-wise) biggest convex mapping below U. Thenthe graph of Newtf is (the lower frontier of) NP(f ).

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Basics


DefinitionLet f (X ) = a0 + · · · + anX n ∈ Qp[x ] (remark: QpJxK would also be fine).Let U = {(0, v(a0)), . . . , (n, v(an))}.

We define the Newton polygon of f , NP(f ), as the lower convex hullof U.By lower convex hull, we mean the points of the convex hull of U belowthe straight line from (0, (a0)) to (n, v(an)).


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Basics


DefinitionLet f (X ) = a0 + · · · + anX n ∈ Qp[x ] (remark: QpJxK would also be fine).Let U = {(0, v(a0)), . . . , (n, v(an))}.We define the Newton polygon of f , NP(f ), as the lower convex hullof U.

By lower convex hull, we mean the points of the convex hull of U belowthe straight line from (0, (a0)) to (n, v(an)).


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Basics


DefinitionLet f (X ) = a0 + · · · + anX n ∈ Qp[x ] (remark: QpJxK would also be fine).Let U = {(0, v(a0)), . . . , (n, v(an))}.We define the Newton polygon of f , NP(f ), as the lower convex hullof U.By lower convex hull, we mean the points of the convex hull of U belowthe straight line from (0, (a0)) to (n, v(an)).


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Basics


DefinitionLet f (X ) = a0 + · · · + anX n ∈ Qp[x ] (remark: QpJxK would also be fine).Let U = {(0, v(a0)), . . . , (n, v(an))}.We define the Newton polygon of f , NP(f ), as the lower convex hullof U.By lower convex hull, we mean the points of the convex hull of U belowthe straight line from (0, (a0)) to (n, v(an)).


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Basics

例

A Newton polygon

1

2

0 1 2 3 4

P = 2 + 3X + 7X 2 + 9X 3 + 3X 4, over Q3

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Basics

例

A Newton polygon

1

2

0 1 2 3 4

P = 2 + 3X + 7X 2 + 9X 3 + 3X 4, over Q3

NP(P)

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Basics

Vocabulary

DefinitionA slope of the Newton polygon of f is an element of Newt ′

f ([0, n]).

DefinitionIf λ is a slope of Newtf , we call segment of slope λ of Newtf the set{(x , Newtf (x))⧸Newt ′

f (x) = λ}.

DefinitionThe length of this slope is the length of its projection on the x -axis.

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Basics

Fundamental theorem of Newton polygons

Theoremf has a root of valuation λ iff −λ is a slope of Newtf .

Moreover, the number of roots of f (with multiplicity) of valuation λ, isthe length of the segment of slope −λ of Newtf .

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Basics

Fundamental theorem of Newton polygons

Theoremf has a root of valuation λ iff −λ is a slope of Newtf .Moreover, the number of roots of f (with multiplicity) of valuation λ, isthe length of the segment of slope −λ of Newtf .

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Basics

例

A Newton polygon

1

2

0 1 2 3 4

P = 2 + 3X + 7X 2 + 9X 3 + 3X 4, over Q3

NP(P)

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Basics

例

A Newton polygon

1

2

0 1 2 3 4

P = 2 + 3X + 7X 2 + 9X 3 + 3X 4, over Q3

NP(P)

One slope for 0 of length 2, one slope 1/2 of length 2.

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Basics

例

A Newton polygon

1

2

0 1 2 3 4

P = 2 + 3X + 7X 2 + 9X 3 + 3X 4, over Q3

NP(P)

One slope for 0 of length 2, one slope 1/2 of length 2.

Two roots of valuation 0, two of valuation −1/2.

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Basics

Basic operations

Proposition (Addition)If f and g are two polynomials, then the Newton polygon of f + g can belower-bounded by taking the lower convex hull for the vertices of Newtfand Newtg .

Proposition (Multiplicativity)If f and g are two polynomials, then the Newton polygon of fg has forslopes that of f and g, with length the sum of that of f and g.

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Basics

Basic operations

Proposition (Addition)If f and g are two polynomials, then the Newton polygon of f + g can belower-bounded by taking the lower convex hull for the vertices of Newtfand Newtg .

Proposition (Multiplicativity)If f and g are two polynomials, then the Newton polygon of fg has forslopes that of f and g, with length the sum of that of f and g.

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Basics

Some remarks

PropositionIf P ∈ Qp[X ] is irreducible, then all its roots have the same valuation.Hence, NewtP has only one slope.

RemarkThe converse is false. For instance, (X − 1)(X − 2) over Q5.

CorollaryIf NewtP has more than one slope, P is not irreducible.

RemarkThere are good irreducibility criterion based on testing whether one slopecan be obtained by multiplication (namely, Dumas and Eisenstein).

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Basics

Some remarks





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Basics

Some remarks





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Euclidean division

Table of contents




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Euclidean division

Euclidean division and Newton polygon

Lemma (Division lemma)

What is the Newton polygon of the remainder in the division of Aby B (in Qp[X ])?What is the Newton polygon of the quotient in the division of A byB (in Qp[X ])?

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Euclidean division



Division of puX n by B : puX n = BQ + R

valuation

orderdd − 1

NP(B)

n

u puX n

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Euclidean division




valuation

orderdd − 1

NP(B)

n

u puX n

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Euclidean division




valuation

orderdd − 1

NP(B)

n

u puX n

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Euclidean division




valuation

orderdd − 1

NP(B)

n

u puX n

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Euclidean division




valuation

orderdd − 1

NP(B)

n

u puX n

NP(R)

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Euclidean division




valuation

orderdd − 1

NP(B)

n

u puX n

NP(R)

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Euclidean division




valuation

orderdd − 1

NP(B)

n

u puX n

NP(R)

n − d

NP(Q)

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Euclidean division



Division of A by B : A = BQ + R

valuation

orderdd − 1

NP(B)NP(A)

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Euclidean division




valuation

orderdd − 1

NP(B)NP(A)

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Euclidean division




valuation

orderdd − 1

NP(B)NP(A)NP(R) NP(Q)

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Precision: Return on Modular Multiplication

Table of contents




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Lattice precisionA and B are known with precision lattice HA and HB . Then (HQ , HR) aregiven by the Euclidean division of HA − QHB by B.

Newton precisionFor A = BQ + R with A known with precision-polygon φ, we can applythe previous construction to φ divided by B to obtain the precision on Qand R.

RemarkThis proved to be useful to handle precision for the computation of thecharacteristic polynomial.

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Comparison: ∏ni=1 Ai mod M

Modulus M nGain of precision

Jagged Newton Lattice(not dif. + dif.)

X 5 + X 2 + 1(Irred. mod 2)

10 0.2 0.2 0.2 + 0.050 4.2 4.2 4.2 + 0.0100 11.2 11.2 11.2 + 0.0

X 5 + 1(Sep. mod 2)

10 0.4 0.4 0.9 + 6.050 5.6 5.6 11.1 + 42.0100 13.6 13.6 27.0 + 87.0

X 5 + 2(Eisenstein)

10 6.2 6.2 6.2 + 0.050 44.0 44.0 44.0 + 0.0100 92.5 92.5 92.5 + 0.0

(X + 1)5 + 2(Shift Eisenstein)

10 0.6 0.6 4.7 + 1.450 7.1 7.1 42.6 + 1.4100 15.1 15.1 91.8 + 1.4

X 5 + X + 2(Two slopes)

10 1.7 1.7 7.9 + 9.850 8.1 8.1 70.7 + 59.8100 16.1 16.1 152.6 + 125.9

Figure: Precision for modular multiplication

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Division and Slope Factorization of p-Adic PolynomialsSlope factorization

A Newton scheme

Table of contents




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A Newton scheme

Factoring respecting slopes

TheoremLet f ∈ Qp[X ]. Then,

We can write f =∏

i fi .The fi ’s are all of one slope.They have respectively different slopes.

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A Newton scheme



We can write f =∏

i fi .

The fi ’s are all of one slope.They have respectively different slopes.

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A Newton scheme



We can write f =∏

i fi .The fi ’s are all of one slope.

They have respectively different slopes.

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A Newton scheme



We can write f =∏

i fi .The fi ’s are all of one slope.They have respectively different slopes.

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A Newton scheme

例

A Newton polygon

1

2

0 1 2 3 4

P = 2 + 3X + 7X 2 + 9X 3 + 3X 4, over Q3.

NP(P)

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A Newton scheme

例

A Newton polygon

1

2

0 1 2 3 4

P = 2 + 3X + 7X 2 + 9X 3 + 3X 4, over Q3.

NP(P)

NP(f1)

NP(f2)

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A Newton scheme

例

A Newton polygon

1

2

0 1 2 3 4

P = 2 + 3X + 7X 2 + 9X 3 + 3X 4, over Q3.

NP(P)

NP(f1)

NP(f2)

P = (2 + 3X + X 2) × (1 + 3X 2)

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A Newton scheme

例

A Newton polygon

1

2

0 1 2 3 4

P = 2 + 3X + 7X 2 + 9X 3 + 3X 4, over Q3.

NP(P)

NP(f1)

NP(f2)

P = (2 + 3X + X 2) × (1 + 3X 2)

Remark: 2 + 3X + X 2 = (1 + X )(1 + 2X ).

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A Newton scheme

A Newton iteration

The iterationAlready found in Polynomial root finding over local rings and applicationto error correcting codes by Berthomieu, Lecerf, Quintin:

Ai+1 := Ai + (ViP mod Ai)Bi+1 := P \quo Ai+1

Vi+1 := (2Vi − V 2i Bi+1) mod Ai+1

The resultAi , Bi , Vi converge quadratically to A, B, V such that AB = P, V is theinverse of B modulo A and A, B have the desired Newton polygons.

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A Newton scheme

Ideas on the proof

About the proofWe monitor Ri = Ai+1 − Ai and Si = P mod Ai .

We prove that both NP(Ri) and NP(Si) goes to infinity.

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A Newton scheme

Illustration

First step.

κ

0 dd−1 d+1

A0

NF (P)

slope = λ1

slope = λ0φ

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A Newton scheme

Illustration

First step. Euclidean division.

κ

0 dd−1 d+1

A0

NF (P)

slope = λ1

slope = λ0φ

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A Newton scheme

Illustration


κ

0 dd−1 d+1

A0

NF (P)

slope = λ1

slope = λ0φ

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A Newton scheme

Illustration


κ

0 dd−1 d+1

A0

NF (P)

slope = λ1

slope = λ0φ

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A Newton scheme

Illustration


κ

0 dd−1 d+1

A0

NF (P)

slope = λ1

slope = λ0φ

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A Newton scheme

Illustration


κ

0 dd−1 d+1

A0

NF (P)

slope = λ1

slope = λ0φ

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A Newton scheme

Illustration

First step.

κ

0 dd−1 d+1

A0

NF (P)

NP(R0)NP(S0)

slope = λ1

slope = λ0φ

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A Newton scheme

Illustration

First step.

κ

0 dd−1 d+1

A0

NF (P)

NP(R0)NP(S0)

slope = λ1

slope = λ0φ

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A Newton scheme

Illustration

First step.

κ

0 dd−1 d+1

A0

NF (P)

NP(R0)NP(S0)

slope = λ1

slope = λ0φ

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A Newton scheme

Illustration

Second Step.

κ

0 dd−1 d+1

A0

NF (P)

slope = λ1

slope = λ0φ

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A Newton scheme

Illustration

Second Step.

2κκ

0 dd−1 d+1

A0

NF (P)

NP(R1)NP(S1)

slope = λ1

slope = λ0φ

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A Newton scheme

Illustration

Second Step.

2κκ

0 dd−1 d+1

A0

NF (P)

NP(R1)NP(S1)

slope = λ1

slope = λ0φ

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A Newton scheme

Illustration

Third Step.

2κ

4κ

κ

0 dd−1 d+1

A0

NF (P)

NP(R2)NP(S2)

slope = λ1

slope = λ0φ

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A Newton scheme

Illustration

Third Step.

2κ

4κ

κ

0 dd−1 d+1

A0

NF (P)

NP(R2)NP(S2)

slope = λ1

slope = λ0φ

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Applying differential precision

Table of contents




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What about precision?

SettingLet FA : P 7→ A(1) be the application such that P = A(1)B(1), withA(1), B(1) corresponding to the slopes before/after the breakpoint d .

DifferentialThe application FA : P 7→ A(1) is of class C1. Its differential at somepoint P is the linear mapping

dP 7→ dA(1) = (V (1) dP) mod A(1)

where A(1)B(1) = P and V (1) is the inverse of B(1) modulo A(1).

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What about precision?

SettingLet FA : P 7→ A(1) be the application such that P = A(1)B(1), withA(1), B(1) corresponding to the slopes before/after the breakpoint d .

DifferentialThe application FA : P 7→ A(1) is of class C1. Its differential at somepoint P is the linear mapping

dP 7→ dA(1) = (V (1) dP) mod A(1)

where A(1)B(1) = P and V (1) is the inverse of B(1) modulo A(1).

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Some numerical results: A 7→ AB 7→ A.

Polynomials PrecisionMean gain of precision

Jagged Newton Lattice

dA B absolute −14.5 −14.7 0.0

relative −1.5 −3.6 0.0

dA B absolute 0.0 0.0 0.0

relative −0.3 −1.2 0.0

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To sum up

On p-adic precision

Step-by-step analysis: as a first step. Can show differentiability andnaïve loss in precision during the computation.Differential calculus: intrinsic and can handle both gain and loss.Lattice precision: achieving and understandig the best precision.

On polynomial computationsDiffused digits for modular multiplication.Newton iteration for slope factorisation.

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To sum up

On p-adic precisionStep-by-step analysis: as a first step. Can show differentiability andnaïve loss in precision during the computation.

Differential calculus: intrinsic and can handle both gain and loss.Lattice precision: achieving and understandig the best precision.


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To sum up

On p-adic precisionStep-by-step analysis: as a first step. Can show differentiability andnaïve loss in precision during the computation.Differential calculus: intrinsic and can handle both gain and loss.

Lattice precision: achieving and understandig the best precision.


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To sum up

On p-adic precisionStep-by-step analysis: as a first step. Can show differentiability andnaïve loss in precision during the computation.Differential calculus: intrinsic and can handle both gain and loss.Lattice precision: achieving and understandig the best precision.


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To sum up


On polynomial computations

Diffused digits for modular multiplication.Newton iteration for slope factorisation.

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To sum up


On polynomial computationsDiffused digits for modular multiplication.

Newton iteration for slope factorisation.

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To sum up



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References

Initial article

Xavier Caruso, David Roe and Tristan Vaccon Tracking p-adicprecision, ANTS XI, 2014.

Linear Algebra

Xavier Caruso, David Roe and Tristan Vaccon p-adic stability inlinear algebra, ISSAC 2015.

Polynomial Computations

Xavier Caruso, David Roe and Tristan Vaccon Division andSlope Factorization of p-Adic Polynomials, ISSAC 2016.

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Thank you for your attentionThanks

x + B

f

B

f ′(x)

f (x) + f ′(x) · B

f ′(x) · B

f̃

x

x + O(pN′) y + O(pM′) ⊂ f (x) + O(pN)

Documents

Division and Slope Factorization of p-Adic Polynomials · Division and Slope Factorization of p-Adic Polynomials Introduction Why should one work with p-adic numbers ? p-adic methods