on isomorphism of modules over non commutative pidgnavarro/files/issac2014_slides.pdf · quantum...

On isomorphism of modules over non–commutative PIDJ. Gómez-Torrecillas?, F. J. Lobillo? and G. Navarro‡

?Department of Algebra, University of Granada‡Dep. of Computer Sciences and AI, University of Granada

ISSAC 2014, July 25th, 2014

F. J. Lobillo (UGR) On isomorphism of modules over non–commutative PID ISSAC 2014, July 25th, 2014 (1)

Motivation

Skew/Ore polynomials appears in (non exhaustive list)Algebraic treatment of differential equations.Solvable Lie algebras.Quantum groups.Convolutional codes with additional structure.. . . need of efficient algorithms over skew/Ore polynomial. We take care of the univariate case.


Ore polynomialsBackground (Ore, 1933)R = D [x ; σ, δ ] Ore polynomial ring

D is a division ring, σ : D → D automorphism, δ : D → D σ-derivation, i.e. δ(ab) = σ (a)δ(b) + δ(a)bR is a left D-vector space with basis {x i with i ≥ 0}, so elements of R are polynomials, a0 + a1x + · · ·+ anx

nProduct defined by xnxm = xn+m and xa = σ (a)x + δ(a)Basic propertiesNon–commutative ringLeft and right Euclidean division (Euclidean domain)Left and right greatest common divisor and least common multiple. . . so left and right non-commutative PID


Non unique factorization in the non–commutative caseLet F4 = {0, 1, a, a + 1} and σ (a) = a2 = a + 1, and R = F4[x ; σ ]. Hence

x2 + 1 = (x + 1)(x + 1) = (x + a)(x + a + 1),two different factorizations of x2 + 1

Let R = C[x , ]̄. Hencex2 − 1 = (x + γ)(x − γ̄) for all γ ∈ C such that |γ| = 1.There exists infinitely many different factorizations.

Let R = Q(t)[X ; d/dt ] the ring of differential operators over Q(t)x2 = (x + 1

t + z

)(x − 1

t + z

), (z ∈ Q)

There exist infinitely many different factorizations of x2


Similar elementsSimilarityTwo elements a, b ∈ R = D [x ; σ, δ ] are similar (a ∼ b) if

R/Ra ∼= R/Rb as left R–modules or, equivalently,R/aR ∼= R/bR as right R–modules.

As a consequence of Jordan-Hölder’s Theorem, we deduceTheorem 1 (Ore, Jacobson)Any element in R may be written as a product of irreducibles, and this decomposition in unique up to reorderingand similarity

Non trivial problemDecide whether two elements a and b are similar


Index

1 Reduction to a commutative problem2 Similarity test3 Isomorphisms of modules


BackgroundPrincipal Ideal Domains (Jacobson, 1943)

R is a non commutative (left and right) PIDTwo-sided ideals are αR = Rα , (α is called two-sided, normal, . . . )R(a, b)r = Ra + Rb right greatest common divisorR [a, b]l = Ra ∩ Rb left least common multiplea ∼ b ⇔ R/Ra ∼= R/Rb

Theorem 2 (Structure theorem of finitely generated modules)Any finitely generated left R-module M is of the form

M ∼= R s︸︷︷︸free part

⊕ R

Ra1⊕ · · · ⊕ R

Ran︸︷︷︸torsion cyclic modules


BackgroundAnnihilatorLet M be a left R-module AnnR (M) = {a ∈ R : am = 0 ∀m ∈ M}.AnnR (M) is a two-sided idealDefinition 3A left R-module is bounded if AnnR (M) 6= 0

Corollary 4Any finitely generated and bounded left R-module is a direct sum of bounded cyclic modules


Reduction to the centerLet C = C (R) be the center of RDefinition 5A finitely generated bounded left R-module M is centrally bounded if any maximal two-sided ideal containing itsannihilator is generated by an element in the center, i.e. if AnnR (M) ⊆ P , P = Rα with α ∈ C .Theorem 6 (Main Theorem)Assume that R is free of finite rank over its center C , and M and N two centrally bounded left R-modules. Thefollowing are equivalent

M ∼= N as left R-modulesM ∼= N as left C-modules

Proof uses Krull–Schmidt Theorem.


Index



A big enough center in skew polynomialsRestrictionsIn this work,[D : C (D)] <∞

R = D [x ; σ ], i.e. we focus on the case δ = 0.σ µ is inner, for some µ, i.e. σ µ(a) = uau−1 for some u ∈ D

Jacobson, CohnLet R = D [x ; σ ], then the center C = C (R) = K [u−1xµ ], whereK = Dσ ∩ C (D) (invariants by σ in the center)µ first power doing σ innerσ µ(a) = uau−1

Under these conditions R has finite rank over its center.F. J. Lobillo (UGR) On isomorphism of modules over non–commutative PID ISSAC 2014, July 25th, 2014 (11)

Similarity of polynomialsWhat we would like to obtainLet f , g ∈ D [x ; σ ], are the following equivalent?

R/Rf ∼= R/Rg as R-modules⇑ Main Theorem

R/Rf ∼= R/Rg as C-modulesNo, becauseProposition 1R/Rf centrally bounded ⇐⇒ f x-torsionfree ⇐⇒ (f , x)r = 1

Theorem 7Let f , g ∈ D [x ; σ ] x-torsionfree TFAE:

f ∼ g

R/Rf ∼= R/Rg as C-modules


QuestionWhat about non x-torsionfree polynomials?Solution,Proposition 2Let f , g ∈ D [x ; σ ] with f = f ′x r and g = g ′x s where f ′, g ′ x-torsionfree. Then

R

Rf∼= R

Rf ′⊕ R

Rx r,

R

Rg∼= R

Rg ′⊕ R

Rx s,

f ∼ g if and only if r = s and f ′ ∼ g ′.


Similarity of x-torsionfree polynomialsLet f , g ∈ D [x ; σ ] x-torsionfree.Recall that

C = K [z ] commutative polynomial ring with z = u−1xµ

R/Rf , R/Rg are finitely generated torsion C-modulesThere is matrix associated to any finitely generated torsion module (multiplication by z)Two modules are isomorphic if and only if the rational canonical forms are the sameProcedureStep 1. Construct the matrices of R/Rf and R/RgStep 2. Compute their rational canonical formsStep 3. Check equality


Matrix constructionB = {u0, . . . , uµ−1} basis of D over KC = K [u−1xµ ] for some u ∈ D

Function 1 Matrix of R/Rf as C-module1: function MatrixConstruction(f )2: n = deg(f )3: B1 = {uix j : 0 ≤ i < µ, 0 ≤ j < n}4: Let M be an "empty matrix"5: for 0 ≤ j < n do6: for 0 ≤ i < ρ do7: r =remainder of the left division of u−1xµuixj by f8: Add to M a new row with the coefficients of r with respect to B19: return M


Similarity testAlgorithm 2 Similarity in generic settingInput: f , g ∈ R = D [x ; σ ] and a basis {u0, . . . , uµ−1} of D over K .Output: True if they are similar or False otherwise.1: if deg(f ) 6= deg(g ) then2: return False3: Decompose f = cf ′x r and g = dg ′x s (f ′, g ′ monic, x-torsionfree)4: if r 6= s then5: return False6: Let Rf ′ = rational canonical form of MatrixConstruction(f ′)7: Let Rg ′ = rational canonical form of MatrixConstruction(g ′)8: if Rf ′ 6= Rg ′ then9: return False10: return True


ImprovementCorollary 8Let f , g ∈ R = D [x ; σ ] be x–torsionfree elements and S a (commutative) algebra with C ⊆ S ⊆ R , TFAE:

f ∼ g

R/Rf ∼= R/Rg as (left) S–modules.

Corollary 9Let f , g ∈ R = F[x ; σ ] be x–torsionfree and µ = [F : Fσ ], i.e. the order of σ , TFAE:

f ∼ g

R/Rf ∼= R/Rg as modules over C = Fσ [z ], with z = xµ

R/Rf ∼= R/Rg as modules over T = F[z ], with z = xµ

R/Rf ∼= R/Rg as modules over S = Fσ [x ]


Module structure over TT = F[z ], where z = xµ and µ = [F : Fσ ]T–module structure multiplying by z = xµ

Function 3 Matrix of R/Rf as T–module1: function MatrixConstructionT(f )2: n = deg(f )3: B2 = {1, x , . . . , xn−1}4: Let M be an "empty matrix"5: for 0 ≤ i < n do6: r =remainder of the left division of xµ+i by f7: Add to M a new row with the coefficients of r with respect to B28: return M


Similarity test using T

Algorithm 4 Similarity over a finite field using T

Input: f , g ∈ R = F[x ; σ ]Output: True if they are similar or False otherwise.1: if deg(f ) 6= deg(g ) then2: return False3: Decompose f = cf ′x r and g = dg ′x s (f ′, g ′ monic, x-torsionfree)4: if r 6= s then5: return False6: Let Rf ′ = rational form of MatrixConstructionT(f ′)7: Let Rg ′ = rational form of MatrixConstructionT(g ′)8: if Rf ′ 6= Rg ′ then9: return False10: return True


Are similar?F28 [x ; τ2], hence µ = 4

a primitive element of F28

f = x8 + a125x7 + a36x6 + a122x5 + a218x4 + a50x3 + a238x2 + a202x + a21

g = x8 + a51x7 + a238x6 + a51x5 + x4 + a153x3 + a34x2 + a51x + 1

Matrix construction for fx i+4 Corresponding rowx4 −→ 0 0 0 0 1 0 0 0x5 −→ 0 0 0 0 0 1 0 0x6 −→ 0 0 0 0 0 0 1 0x7 −→ 0 0 0 0 0 0 0 1x8 −→ a21 a202 a238 a50 a218 a122 a36 a125

x9 −→ a11 a245 a37 a93 a145 a245 a231 a41

x10 −→ a185 a5 a164 a178 a138 a137 a233 a228

x11 −→ a168 a128 a146 a129 a129 a207 a54 a160


Mf =

0 0 0 0 1 0 0 00 0 0 0 0 1 0 00 0 0 0 0 0 1 00 0 0 0 0 0 0 1a21 a202 a238 a50 a218 a122 a36 a125

a11 a245 a37 a93 a145 a245 a231 a41

a185 a5 a164 a178 a138 a137 a233 a228

a168 a128 a146 a129 a129 a207 a54 a160

Mg =

0 0 0 0 1 0 0 00 0 0 0 0 1 0 00 0 0 0 0 0 1 00 0 0 0 0 0 0 11 a51 a34 a153 1 a51 a238 a51

a204 0 a85 a238 a68 0 1 a204

a51 a34 a85 a238 a85 a187 a34 a221

a119 a51 a204 a34 a136 1 a136 a17


Rf =

0 0 0 0 0 0 0 11 0 0 0 0 0 0 00 1 0 0 0 0 0 10 0 1 0 0 0 0 a85

0 0 0 1 0 0 0 10 0 0 0 1 0 0 10 0 0 0 0 1 0 a85

0 0 0 0 0 0 1 a170

Rg =

0 0 0 0 0 0 0 11 0 0 0 0 0 0 00 1 0 0 0 0 0 10 0 1 0 0 0 0 a85

0 0 0 1 0 0 0 10 0 0 0 1 0 0 10 0 0 0 0 1 0 a85

0 0 0 0 0 0 1 a170


Module structure over SS = Fσ [x ]a primitive element of F{1, a, . . . , aµ−1} is a basis of F over FσS–module structure is given by left multiplication by x

Function 5 Matrix of R/Rf as S–module1: function MatrixConstructionS(f )2: n = deg(f )3: B3 = {aix j : 0 ≤ i < µ, 0 ≤ j < n}4: Let M be a matrix with no rows and nµ columns.5: for 0 ≤ j < n do6: for 0 ≤ i < µ do7: r =remainder of the left division of xaix j = σ (ai )x1+j by f8: Add to M a new row with the coefficients of r with respect to B39: return M


Similarity test using S

Algorithm 6 Similarity over a finite field using S

Input: f , g ∈ R = F[x ; σ ]Output: True if they are similar or False otherwise.1: if deg(f ) 6= deg(g ) then2: return False3: Decompose f = cf ′x r and g = dg ′x s (f ′, g ′ monic, x-torsionfree)4: if r 6= s then5: return False6: Let Rf ′ = rational form of MatrixConstructionS(f ′)7: Let Rg ′ = rational form of MatrixConstructionS(g ′)8: if Rf ′ 6= Rg ′ then9: return False10: return True


Are similar?F4[x ; τ ], hence µ = 2

f = x6 + (a + 1) x5 + x3 + ax2 + ax + 1

g = x6 + (a + 1) x5 + (a + 1) x3 + ax2 + (a + 1) x + a

Matrix construction for f (σ (a) = a + 1)xaix j Corresponding rowx −→ 0 0 1 0 0 0 0 0 0 0 0 0σ (a)x −→ 0 0 1 1 0 0 0 0 0 0 0 0x2 −→ 0 0 0 0 1 0 0 0 0 0 0 0σ (a)x2 −→ 0 0 0 0 1 1 0 0 0 0 0 0x3 −→ 0 0 0 0 0 0 1 0 0 0 0 0σ (a)x3 −→ 0 0 0 0 0 0 1 1 0 0 0 0x4 −→ 0 0 0 0 0 0 0 0 1 0 0 0σ (a)x4 −→ 0 0 0 0 0 0 0 0 1 1 0 0x5 −→ 0 0 0 0 0 0 0 0 0 0 1 0σ (a)x5 −→ 0 0 0 0 0 0 0 0 0 0 1 1x6 −→ 1 0 0 1 0 1 1 0 0 0 1 1σ (a)x6 −→ 1 1 1 0 1 0 1 1 0 0 0 1


Mf =

11 1

11 1

11 1

11 1

11 1

1 1 1 1 1 11 1 1 1 1 1 1

Mg =

11 1

11 1

11 1

11 1

11 1

1 1 1 1 1 1 1 11 1 1 1 1


Rf =

11

11

11

11

11

1 11

Rg =

11

1 11

1 11

11

1 11

1 11

.


ComparisonParameters

µ = [F : Fσ ]Matrix multiplication ∈ O(nω)Theoretical efficiency

Comm. Algebra Timea Spaceb Space (sparse)C = Fσ [xµ ] O

(µωnω log n log log n

)O(µ2n2) −−

T = F[xµ ] O(µnω log n log log n

)O(µn2) O(µ2n)

S = Fσ [x ] O(µωnω log n log log n

)O(µ2n2) O(µn)

aOperations over FσbElements in Fσ


Index



Isormorphic modulesProblemLet R = D [x ; σ ], A and B finitely generated left R-modules.How to decide if A ∼= B? (efficiently, if possible)ConsiderationsGiven a finitely generated left R-module A

Rm −→ A surjective, for some mhence, A ∼= Rm/L for some m

ProblemDecide if Rm/L1∼= Rn/L2


Reduction to cyclic modulesJacobson (1943)Let A = Rm/L a left R-module,

1 Find h1, h2, . . . , ht generators of L2 Construct M ∈Mt×m(R), rows are coordinates of hi3 Diagonalize matrix M , i.e. find D diagonal such that PMQ = D with Q,P regular.

I Check if M00 divides (by the left) all elements in its row and column(M00 →↓ M ′

)I If not, divide, put remainder at position (0, 0) and repeat(

b1 00 M ′

)I Repeat with matrix M ′I Return diagonal elements of D = diag{b1, b2, . . . , br , 0, . . . , 0}

4 Drop those bi with zero degree and factor out bi = b′ixei where b′i is x–torsionfree.

5 A = Rm−r ⊕ R

Rxe1⊕ · · · ⊕ R

Rxep′⊕ R

Rb′1⊕ · · · ⊕ R

Rb′p(Rough decomposition)


Reduction to the centerTheorem 10Let A,B finitely generated R-modules with rough decomposition

A = R s ⊕ R

Rxe1⊕ · · · ⊕ R

Rxep′⊕ R

Rb1⊕ · · · ⊕ R

Rbpand B = R t ⊕ R

Rxd1⊕ · · · ⊕ R

Rxdq′⊕ R

Rc1⊕ · · · ⊕ R

Rcq

where bi , cj are x–torsionfree. Hence, A is isomorphic to B if and only if1 s = t ,2 p′ = q′,3 [e1, . . . , ep′ ] is equal to [d1, . . . , dq′ ] up to reordering,

4 and R

Rb1⊕ · · · ⊕ R

Rbe

∼= R

Rf1⊕ · · · ⊕ R

Rfdas C-modules


Algorithm 7 Isomorphism of finitely generated modulesInput: A = Rn/L1 and B = Rm/L2 finitely generated left R–modulesOutput: True if they are isomorphic; False otherwise.

1: Find rough decomposition A = R s ⊕ R

Rxe1⊕ · · · ⊕ R

Rxep′⊕ R

Rb1⊕ · · · ⊕ R

Rbp

2: Find rough decomposition B = R t ⊕ R

Rxd1⊕ · · · ⊕ R

Rxdq′⊕ R

Rc1⊕ · · · ⊕ R

Rcq3: if not (1, 2 and 3 of Theorem 10) then4: return False5: Compute the matrices Mbi =MatrixConstruction(bi )6: Compute the matrices Mci =MatrixConstruction(ci )7: Create block matricesMA =

Mb1 0. . .0 Mbp

and MB = Mc1 0. . .

0 Mcq

8: if the rational canonical form of MA and MB are different then9: return False10: return True


Example

F28 [x ; τ2], a primitive element, µ = 4Matrix for A 1 0 0 x + a3

0 x + a11 x2 + a116x + a12 00 0 x5 + a7x4 + a170x + a177 0

x + a78 0 0 x2 + a42x + a7 + a81

Matrix for B (

x + a2 x2 + a247x + a5 + a69 x2 + a196x + a225

x2 + a206x + a36 x5 + a25x4 + x3 + a253x2 + a149x + a122 x3 + a52x2 + a56x + a4

)


F28 [x ; τ2], a primitive element, µ = 4Diagonal matrix for A 1 0 0 00 x + a11 0 00 0 x5 + a7x4 + a170x + a177 00 0 0 0

Diagonal matrix for B (

x + a2 0 00 x5 + a25x4 + a170x + a195 0

)


F28 [x ; τ2], a primitive element, µ = 4Rough decomposition for AA ∼= R ⊕ R

R(x + a11) ⊕ R

R(x5 + a7x4 + a170x + a177)Rough decomposition for B

B ∼= R ⊕ R

R(x + a2) ⊕ R

R(x5 + a25x4 + a170x + a195)


F28 [x ; τ2], a primitive element, µ = 4Matrix of the diagonal polynomials for A

MA =

a170 0 0 0 0 00 0 0 0 0 10 a177 a170 0 0 a7

0 a205 0 a170 0 a35

0 a62 0 0 a170 a147

0 1 0 0 0 1

Matrix of the diagonal polynomials for B

MB =

a170 0 0 0 0 00 0 0 0 0 10 a195 a170 0 0 a25

0 a40 0 a170 0 a125

0 a185 0 0 a170 a15

0 1 0 0 0 1


F28 [x ; τ2], a primitive element, µ = 4Rational form of MA

RA =

a170 0 0 0 0 0

0 a170 0 0 0 0

0 0 a170 0 0 0

0 0 0 a170 0 0

0 0 0 0 0 1

0 0 0 0 1 1

Rational form of MB

RB =

a170 0 0 0 0 0

0 a170 0 0 0 0

0 0 a170 0 0 0

0 0 0 a170 0 0

0 0 0 0 0 1

0 0 0 0 1 1


on isomorphism of modules over non commutative pidgnavarro/files/issac2014_slides.pdf · quantum...

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