Lajos Rónyai joint work with Gábor Ivanyos, Ádám D. Lelkes ... · Splitting full matrix algebras over number elds Lajos Rónyai joint work with Gábor Ivanyos, Ádám D. Lelkes,

Splitting full matrix algebras over number �elds

Lajos Rónyai

joint work with Gábor Ivanyos, Ádám D. Lelkes, and JosefSchicho

AANT, RICAM, Linz, November 28, 2013

Lajos Rónyai joint work with Gábor Ivanyos, Ádám D. Lelkes, and Josef SchichoSplitting full matrix algebras over number �elds

Contents

Background on algorithms for algebras

Orders

Splitting matrix algebras over QExtension to number �elds

Applications

Some improvements

Open problems


The computational model

Exact (symbolic) computations of substructures

Input and results (algebras) will be given by bases

Ground �elds:

Algebraic number �elds: K = Q[x ]/(f ), f (x) ∈ Z[x ] irreducibleFinite �elds: K = Fq.Global function �elds: K = Fq(x), and �nite extensions

Size= number of bits representing the object

We consider polynomial time algorithms (deterministic,randomized, �)


Structure constants representation

Let A be an algebra over K with basis a1, . . . , am. Thenmultiplication can be speci�ed by giving

ai · aj = γij1a1 + γij2a2 + · · ·+ γijmam.

The γijk ∈ K are the structure constants. Multiplication table ofthe basis.

Essentially the regular representation (possibly via Dorroh'sextension): one may assume

A ≤ Mm+1(K).


Computational tools

arithmetic over Klinear algebra (solving systems of linear equations) over Kfactoring univariate polynomials over K (Berlekamp, LLL, vanHoeij)

approximate computations with real algebraic numbers

approximate lattice basis reduction (LLL, Buchmann)


Computing the radical of A ≤ Mn(K)

An a ∈ A is nilpotent, if ad = 0 for some d . Rad(A) is the largestideal of A consisting of nilpotent elements.

Polynomial time algorithms based on linear algebra

Theorem (Dickson'23)

Assume charK = 0. Then

A is nilpotent⇔ ∀a ∈ A, Tr(a) = 0.

Rad(A) = {a ∈ A| ∀b ∈ A ∪ {I}, Tr(ba) = 0}.

Corollary

Assume charK = 0, B = (a basis of A) ∪ {I}. Then

Rad(A) = {a ∈ A| ∀b ∈ B, Tr(ba) = 0}.


Computing the radical of A ≤ Mn(K)

Over K = Fp [Friedl, R '85]: Compute a sequence of ideals

A = A0 ⊇ A1 ⊇ A2 ⊇ . . . ⊇ Adlogp ne = Rad(A)

Extensions

K = Fq [Eberly '89]

K = Fq(x) [Ivanyos, R, Szántó '94]

Arbitrary K [Cohen, Ivanyos, Wales '96]Ai −→ Ai+1 : system of semilinear equations over KEquations: from coe�cients of the char. polynomial

Radical (nil, solvable) algorithms for Lie algebras [R'90]


Wedderburn decomposition

Assume Rad(A) = (0). Then A is the direct sum of its minimalideals:

A = A1 ⊕A2 ⊕ · · · ⊕ Ak .

This is re�ected in the central part:

Z (A) = Z (A1)⊕ Z (A2)⊕ · · · ⊕ Z (Ak).

Let 0 6= ei ∈ Z (Ai ). Then

Ai = eiA.

Finding the Z (Ai ): reduced in polynomial time to factoringunivariate polynomials over K


Algorithms for Wedderburn decomposition

Over �nite �elds and number �elds: [Friedl, R '85] � aniterative algorithm

Over su�ciently large �elds [Eberly `91] � one round,randomizedPick a random a ∈ Z (A), let f be the min. pol. of a. Then

Z (A) ∼= K[x ]/(f (x))

factors of f ↔Wedderburn components of A

Over global function �elds:[Ivanyos, R, Szántó '94] reduced to factoring polynomials overthe prime �eld


Simple algebras over K

Theorem (Wedderburn)

A simple K-algebra A is isomorphic to Mk(D), for a (skew)�eld Dwith Z (D) ≥ K and positive integer k.

Explicit isomorphism problem: given A, �nd k , D and ∼=.

K is �nite:� [R '90] randomized poly time� [Eberly, Giesbrecht '97] randomized ≈ O(n3)� [Ivanyos, Karpinski, R, Saxena '11] deterministic polynomialtime for dimA bounded

K = R or K = C:� [Eberly '91] randomized poly time� derandomized [R; de Graaf, Ivanyos '93, '98]


Explicit isomorphism over number �elds

Di�cult: (� factoring integers) [R '88], [Voight '10]

dim eAe ≺ factorization [Ivanyos, R '93]

Problem: is explicit isomorphism ≺ factorization?

�-algorithm: allowed to call an oracle for factoring integers,and for factoring univariate polynomials over �nite �elds

Yes for K = Q, dimA = 4 [Ivanyos, Szántó '94]

Yes for A = Mn(K), with K and n small [Ivanyos, R, Schicho'12]

Improvement for K = Q and n small [Lelkes, R '13]

Non conventional applications: rational parametrization of varieties,n-descent for elliptic curves [Cremona, Fisher, O'Neil, Simon, Stoll,'08, '09, '11]


Orders in semisimple algebras over K

Integral element: root in A of a monic polynomial with integercoe�cients

Order: Large subring of A consisting of integral elements

Full lattice in A: �nitely generated Z-submodule M withQM = AOrder: a subring Λ ≤ A with 1 of A in Λ such that Λ is a fulllattice in AMaximal order: maximal w.r.t. inclusion (not unique)


Examples of orders

The sublattice generated by the basis elements and 1, if thestructure constants are integers

ZG in the group algebra QG for a �nite group G

Algebraic integer matrices: K number �eld,R = {alg. integers in K}Mn(R) in Mn(K) is a maximal order

Stabilizers of lattices: M a full lattice in AO(M) = {x ∈ A|xM ⊆ M}


Constructing maximal orders

Given a semisimple algebra A over K, �nd a maximal order Λ ≤ A

[Ivanyos, R '93] � a polynomial time �-algorithm

Starts with an initial order Γ and then enlarges it into a maximalorder Λ ≥ Γ

Suppose that A ∼= Mn(D). From Λ one can e�ciently determine n

Use of factoring: we need only the prime factors of d(Γ)


Golden rings


Golden algebra, golden ring, golden code

Ground �eld: K = Q(i) � the Gaussian rationals

GA = K(x , u; x2 = 5, u2 = i , xu = −ux)

GO = Z[i ][x , u] ≤ GA the golden ring

It leads to a highly e�cient space-time radio code (Yao, Wornell;Dayal, Varanasi; Bel�ore, Rekaya, Viterbo)

GO is a lattice with very small discriminant consisting ofnonsingular integral matrices


Golden rush

Hollanti, Lahtonen, Ranto, Vehkalahti: �nd orders similar to GO inother cyclic algebras

IR-algorithm: given an order Γ ≤ A in a �nite dimensional simplealgebra A over Q; it computes a maximal order Λ ≥ Γ

Polynomial time �-algorithm

de Graaf: MAGMA -implementation

With that better codes can be obtained. For example in

GA+ = K(x , u; x2 = 2 + i , u2 = i , xu = −ux)

the order Γ = Z[i ][x , u] is not maximal. The max. order Λ ⊃ Γobtained by the IR-algorithm gives a better code

Engineering application of non commutative algebraic numbertheory


The main result

Theorem

Let K be an algebraic number �eld of degree d and discriminant ∆over Q. Let A be an associative algebra over K given by structure

constants such that A ∼= Mn(K) for some positive integer n.

Suppose that d, n and |∆| are bounded. Then an isomorphism

A → Mn(K) can be constructed by a polynomial time �-algorithm.

The time bound of our algorithm depends polynomially on |∆| andexponentially on n and d.

Proof will be outlined for the case K = Q

The general case follows by extending the argument via thecanonical embedding K→ Rd


Small zero divisors in maximal orders

Theorem

Let Λ be a maximal order in A ≤ Mn(R), A ∼= Mn(Q). Then there

exists an element C ∈ Λ which has rank 1 as a matrix, and whose

Frobenius norm ‖C‖ is less than n.

The Frobenius norm of a matrix X ∈ Mn(R) is ‖X‖ =√Tr(XTX )

We haveΛ = PMn(Z)P−1

for some P ∈ GLn(R)

Dividing by | detP|1/n we may assume P ∈ Mn(R), detP = ±1


Proof

Let ρ be the left ideal of Mn(Z) of all matrices which are 0 exceptin the �rst column

ρ is a lattice of covolume 1 in the space S ∼= Rn of all real matriceshaving all zeros outside the �rst column

L = Pρ is a sublattice of S , with covolume 1

Apply Minkowski's theorem to L in S and to the ball of radius√n

in S centered at the zero matrix

The volume of the ball is more than 2n � contains 2n internallydisjoint copies of the n-dimensional unit cube

There exists nonzero B ∈ ρ such that PB has length <√n


Proof, concluded

B and hence PB is a rank 1 matrix

Next consider the transpose of this argument with P−1 in the placeof P

There exists a nonzero integer matrix B ′, which is 0 everywhereexcept in the �rst row, such that B ′P−1 has length <

√n

C := PBB ′P−1 meets the requirements

It is in Λ because BB ′ ∈ Mn(Z), and

‖C‖ = ‖(PB)(B ′P−1)‖ ≤ ‖PB‖ · ‖B ′P−1‖ < (√n)2 = n

FinallyrankBB ′ = rankC = 1.


Hermite's constant

γn = supL

λ1(L)2

(det L)2/n,

where L is over all full lattices in Rn, and λ1(L) is the length of theshortest nonzero vector of L.√γn: diameter of spheres at the densest lattice based packing.


n 1 2 3 4 5 6 7 8 24γnn 1 4

32 4 8 64

364 28 424

For large n

12πe≤ γn

n≤ 1.745

2πe.

In fact, we proved also ‖C‖ ≤ γn, and even ‖C‖ ≤ γ′n,where γ′n is the Bergé-Martinet constant.


The Bergé-Martinet constant

γn = supL

λ1(L)λ1(L∗)

(det L)2/n,

where L is over all full lattices in Rn, and L∗ is the dual lattice of L

L∗ = {y ∈ Rn, 〈x, y〉 ∈ Z for every x ∈ L}.

Clearly γ′n ≤ γn.


Small is beautiful

Lemma (Fisher '07)

Let X ∈ Mn(C) be a matrix such that detX is an integer, and

‖X‖ <√n. Then X is a singular matrix.

Proof. Let X = QR be the QR decomposition of X , Q unitary, Rupper triangular with diagonal entries r1, r2, . . . , rn. We have

| detX |2/n = (|r1|2|r2|2 · · · |rn|2)1/n ≤

≤ 1n

(|r1|2 + |r2|2 + · · · |rn|2) ≤ 1n‖R‖2 =

1n‖X‖2 < 1.

Note that ‖X‖ =√Tr(X ∗X ) =

√Tr(R∗R) as Q∗Q = I . We

conclude that detX = 0. �


Lemma

Let X ∈ Mn(Q) be a matrix whose characteristic polynomial has

integral coe�cients, and ‖X‖ < 1. Then X is a nilpotent matrix.

Proof. [G. Kós] The eigenvalues of X are algebraic integers, hencethe eigenvalues of X t are algebraic integers as well, for any t ∈ N+.We infer that X t has char. poly. with integral coe�cients. Thenorm condition implies

X t → O as t →∞,

hence X t = O for t large. �

The bound is sharp:

X =

(1n

)n

i ,j=1

.


Reduced bases

Γ is a full lattice in Rm with basis b1, . . . ,bm over Z such that

|b1| · |b2| · · · |bm| ≤ cm · det(Γ)

The LLL algorithm achieves cm = 2m(m−1)/4

The approximate version of LLL by [Buchmann '94] gives

cm := (γm)m2

(32

)m

2m(m−1)

2

γm is Hermite's constant


Small coe�cients

Lemma (H. W. Lenstra '83)

Γ is a full lattice in Rm with basis b1, . . . ,bm over Z such that

|b1| · |b2| · · · |bm| ≤ cm · det(Γ)

holds for a real number cm > 0. Suppose that

v =m∑i=1

αibi ∈ Γ, αi ∈ Z.

Then we have |αi | ≤ cm|v ||bi |

for i = 1, . . . ,m.

The proof is Cramer's rule and Hadamard's bound.


The algorithm for K = Q

Input: A ∼= Mn(Q), given by a basis (of m = n2 vectors) andstructure constants over Q

Output: C ∈ A, rankC = 1

Note that M = AC is an n dimensional A module, hence leftmultiplication on M gives an A → Mn(Q) isomorphism


The algorithm for K = Q

1 Use the IR algorithm to construct a maximal order Λ in A.2 Compute an A ↪→ Mn(R). Use the derandomization from [de

Graaf, Ivanyos '00] of [Eberly '91]. This gives a ‖ · ‖ on A.3 Compute a su�ciently precise rational approximation A of our

basis B of Λ by the method of [Schönhage '82].4 Compute a reduced basis b1, . . . ,bm of the lattice Λ ⊂ Rm by

the LLL algorithm on A. We have the bigger cm.5 If a bi is singular, then there are two cases. If rankbi = 1,

then we STOP with the output C := bi . Else, if1 < rankbi < n, then we compute the the right identityelement e of Abi , set A := eAe and go back to Step 1.

6 Here |bi | ≥√n for every i . Generate C =

∑mi=1 αibi , where

αi ∈ Z,|αi | ≤ cm

n

|bi |≤ cm

√n

until a rank 1 C is found. Output this C .


Correctness

If we have su�cient (polynomial) precision at 3 , then by[Buchmann '94] after 4 we have

|b1||b2| · · · |bm| ≤ (γm)m2

(32

)m

2m(m−1)

2 .

When at 5 , we have rank e = k , then it is easy to see that

eAe ∼= Mk(Q).

A rank 1 element of eAe has rank 1 in A as well.

The theorem and the lemma (latter applied for v := C and|v | ≤ n) show that an element C with rank one exists at 6 .


Timing

1 runs in polynomial time as an �-algorithm.

2 , 4 and 5 can be done in deterministic polynomial time.

At 3 the precision parameter is polynomial in the input size, henceSchönhage's approximation algorithm runs in polynomial time.

The number of jumps back to 1 is bounded, hence each Step iscarried out in a bounded number of times.

Finally, the number of elements C enumerated at 6 is at most(2cm

√n + 1)m.


The general case

K be a number �eld of degree d over Q, the maximal order of K isR , the discriminant of R is ∆.

Let A be a central simple algebra over K such that A ∼= Mn(K),and let Λ be a maximal order in A.

Λ is isomorphic to

Λ′ :=

R · · ·R J−1

.... . .

...R · · ·R J−1

J · · · J R

,

where J is a fractional ideal in K.


Let σ1, . . . , σr be the embeddings of K into R andσr+1, σr+1, . . . , σr+s , σr+s be the non-real embeddings of K into C;we have d = r + 2s.

For 1 ≤ i ≤ r + s consider an embedding φi of A into Mn(C),which extends σi (for i ≤ r we require φi (A) ≤ Mn(R)). Suchembeddings are poly. time computable.

Set

b =

((2π

)2sn

∆n

) 1

nd

=

(2π

) 2sd

∆1

d .


Small zero divisors in Λ

Theorem

There exists a rank one element x ∈ Λ such that the entries of the

matrices φi (x) for i = 1, . . . , s + r all have absolute value ≤ b.

The proof is similar to the case K = Q. Here we employMinkowski's Thm to a lattice in Rnd .

Some special cases:1. K = Q, R = Z, then ∆ = 1, s = 0, hence b = 1. We havex ∈ Λ which has rank 1 as a matrix from Mn(Q), and with respectto A ↪→ Mn(R) has elements of absolute value at most 1.2. If D > 0 is a squarefree integer, K = Q(

√D), then ∆ = D, if D

is congruent to 1 modulo 4, and ∆ = 4D, if D is congruent to 3modulo 4.Then s = 0, d = 2, hence b ≤ 2

√D.


A general variant of Fischer's Lemma.

Lemma

Let y ∈ Λ be an element such that ‖φi (y)‖ ≤√n holds for

i = 1, . . . , r + s. Then y is a zero divisor in A.

The algorithm works with a reduced basis of the lattice Γ of vectors

(φ1(y), . . . , φr (y),<(φr+1(y)),=(φr+1(y)), . . . ,<(φr+s(y)),=(φr+s(y))),

where y ∈ Λ. We have Γ ⊂ Rn2d .


Set m := n2d

At the �nal step we have a reduced basis b1, . . . ,bm of Γ with|bi | ≥

√n for every i .

We generate all integer linear combinations

w =m∑i=1

αibi ,

with

|αi | ≤ cm

(2π

) 2sd

∆1

d

√n(r + s)

until a w is found such that rank x = 1 holds for the correspondingx ∈ Λ.


Applications

Corollary

K is a number �eld, A is given by structure constants, and

A ∼= Mn(K) for an n > 1. Then there exists a zero divisor x ∈ Awhich admits polynomially bounded coordinates with respect to the

input basis of A. Such an x can be computed in pspace.

Proof. Let c1, . . . , cn2 be the basis of Λ given by the IR algorithm.Express the element x of the Theorem in this basis:

x = α1c1 + α2c2 + · · ·+ αn2cn2

with αi ∈ Z. Using that ‖x‖ ≤ bn, and that the vectors c i havepolynomial size, Cramer's rule implies a polynomial bound on thesize of the αi . �


Norm equations

There is an important connection between split cyclic algebras andrelative norm equations.

Corollary

Let K be a number �eld, L be cyclic extension of K, and a ∈ K. If

the a norm equation

NL/K(x) = a

is solvable, then there is a solution whose standard representation

has polynomial size (in terms of the size of the standard

representation of a and a basis of L). Furthermore, for �xed K and

�xed degree |L : K|, a solution can be found by a polynomial time

�-algorithm.


Isomorphism of central simple algebras

Corollary

K is a number �eld of degree d and discriminant ∆ over Q. Let

A,B be isomorphic central simple algebras over K of dimension n2,

given by structure constants. Suppose that d, n and |∆| arebounded. Then an isomorphism A → B can be constructed by a

polynomial time �-algorithm. The running time is polynomial in

|∆| and exponential in n and d.

Proof. We have A ∼= B if and only if

A⊗K Bop ∼= Mn2(K).

Let V be a simple A⊗K Bop module. Then dimKV = n2. So V , asleft A-module, is isomorphic to the regular left A module.Similarly, V is a regular right Bop-module.Let v ∈ V be a generating element of V as an A-module, and alsoas a Bop module.


Then φ : a 7→ av is a left A-module isomorphism from A to V .Also, ψ : b 7→ vb is a right Bop-module isomorphism from Bop to V .

Then σ = ψ−1φ is a K algebra isomorphism from A and B.

It is clear that σ is K linear.

For a ∈ A, σa is the unique element b ∈ B with

av = vb.

Therefore σ(a1a2) is the unique b ∈ B with a1a2v = vb. Buta1a2v = a1v(σa2) = v(σa1)(σa2), whence σ(a1a2) = (σa1)(σa2).�


Tensor products of lattices

Let L ⊂ Rm and M ⊂ Rn be lattices. We have

L⊗Z M ↪→ Rm ⊗R Rn.

This allows one to de�ne the lattice L⊗M.

L⊗M can be viewed as the set of m by n matrices over R which areintegral linear combinations of dyads xyT , where x ∈ L and y ∈ M.

Rm ⊗ Rn is an Euclidean space with

〈x1 ⊗ y1, x2 ⊗ y2〉 = 〈x1, x2〉〈y1, y2〉.

The norm on Rm ⊗ Rn is the Frobenius norm on Mm,n(R).We have

Λ ∼= PMn(Z)P−1 ∼= QZn ⊗ (QZn)∗.


Improvement when K = Q and n ≤ 43

Results of Y. Kitaoka (1977) on tensor products of lattices implythat the shortest nonzero element C in the ring PMn(Z)P−1 musthave rank one, when P is an invertible real matrix, and n ≤ 43.

Above fails badly for n ≥ 292.

Kitaoka's result leads to a simpler and faster method over Q, whenn ≤ 43.


An improved algorithm for K = Q, n ≤ 43

1 Use the IR algorithm to construct a maximal order Λ in A.2 Compute an A ↪→ Mn(R). Use the derandomization from [de

Graaf, Ivanyos '00] of [Eberly '91]. This gives a ‖ · ‖ on A.3 Compute a su�ciently precise rational approximation A of our

basis B of Λ by the method of [Schönhage '82].4 Compute a reduced basis b1, . . . ,bm of the lattice Λ ⊂ Rm by

the LLL algorithm on A. We have the bigger cm.5 Generate C =

∑mi=1 αibi , where αi ∈ Z,

|αi | ≤ cmmin{γ′n, |b1|}

|bi |≤ cm

until a rank 1 C is found. Output this C .


A Kitaoka type bound

Assume that 2 ≤ n ≤ 8, and set Λ = PMn(Z)P−1, where P is aninvertible real matrix. Let m be the minimal norm of the nonzeroelements of Λ.

Theorem

Let D ∈ Λ with rank at least 2. Then ‖D‖ ≥√

32m.

Allows to reduce further the search space at step 5 .

The proof is a re�nement of Kitaoka's argument.


Theorem

Let L and M be lattices. For every tensor v ∈ L⊗M of rank r we

have

‖v‖ ≥√

r

γ2rλ1(L⊗M).

Lemma

Let A,B ∈ Mn(R) be positive de�nite real symmetric matrices.

Then Tr(AB) ≥ nn√detA n

√detB.

Proof of the Theorem. Write v as

v =r∑

i=1

xi ⊗ yi .

Let L1 be the lattice generated by {x1, . . . , xr}, and M1 be thelattice spanned by {y1, . . . , yr}. The rank of these sublattices is r .


‖v‖2 = ‖∑r

i=1 xi ⊗ yi‖2 =∑r

i ,j=1〈xi , xj〉〈yi , yj〉 =Tr([〈xi , xj〉]ri ,j=1 · [〈yi , yj〉]ri ,j=1).

By the Lemma

‖v‖2 ≥ r (det[〈xi , xj〉] · det[〈yi , yj〉])1/r .

Now assume for contradiction that ‖v‖2 < (r/γ2r )λ1(L⊗M)2.It follows that

‖v‖2 < r

γ2r(λ1(L)λ1(M))2 ≤ r

γ2r(λ1(L1)λ1(M1))2.


Combining the two inequalities

r <r

γ2r· λ1(L1)2

(det[〈xi , xj〉])1/r· λ1(M1)2

(det[〈yi , yj〉])1/r≤ r

γ2rγ2r = r ,

as [〈xi , xj〉]ri ,j=1 and [〈yi , yj〉]ri ,j=1 are Gram matrices for L1 and M1,respectively. Contradiction.

The minimum of r/γ22 for 2 ≤ r ≤ 8 is 32, attained at r = 2.


Improvement for K = Q(√−d), n = 2, d = 1 or 3

Assume that A ∼= M2(K), Λ is a maximal order in A. Let φ be anembedding of A into M2(C).

Kitaoka-type results of R. Coulangeon over Q(√−d) give:

Proposition

For d = 1, at least one of the smallest nonzero elements of φ(Λ)wrt the Frobenius norm has rank one.

Proposition

For d = 3, the smallest nonzero elements of φ(Λ) have rank one.

Application: parametrization of Del Pezzo surfaces of degree 8 (W.A. de Graaf, J. Pílniková, J. Schicho, 2009).


Problems for future work

In the case A ∼= Mn(K) allow n or/and K vary (even with d

�xed)

Study the case Mn(D), where D is a skew�eld over KDevelop a practical algorithm

Extend the improvement to small number �elds, say imaginaryquadratic �elds with small discriminant


Papers:

G. Ivanyos, L. R., J. Schicho:Splitting full matrix algebras over algebraic number �elds, Journalof Algebra, 354(2012), 211-223.

http://arxiv.org/abs/1106.6191

G. Ivanyos, Á. D. Lelkes, L. R.:Improved algorithms for splitting full matrix algebras, JP Journal of

Algebra, Number Theory and Applications, 28(2013), 141�156.

http://arxiv.org/abs/1211.1356


Thank you for your attention!


Documents

Lajos Rónyai joint work with Gábor Ivanyos, Ádám D. Lelkes ... · Splitting full matrix algebras over number elds Lajos Rónyai joint work with Gábor Ivanyos, Ádám D. Lelkes,