Characterizing Triviality of the Exponent Lattice of A Polynomial through Galois and Galois-Like Groups
CCharacterizing Triviality of the ExponentLattice of A Polynomial through Galois andGalois-Like Groups (cid:63)
Tao Zheng
School of Mathematical Sciences, Peking UniversityBeijing, China [email protected]
Abstract.
The problem of computing the exponent lattice which con-sists of all the multiplicative relations between the roots of a univariatepolynomial has drawn much attention in the field of computer algebra.As is known, almost all irreducible polynomials with integer coefficientshave only trivial exponent lattices. However, the algorithms in the liter-ature have difficulty in proving such triviality for a generic polynomial.In this paper, the relations between the Galois group (respectively, theGalois-like groups ) and the triviality of the exponent lattice of a poly-nomial are investigated. The Q -trivial pairs, which are at the heart ofthe relations between the Galois group and the triviality of the exponentlattice of a polynomial, are characterized. An effective algorithm is devel-oped to recognize these pairs. Based on this, a new algorithm is designedto prove the triviality of the exponent lattice of a generic irreduciblepolynomial, which considerably improves a state-of-the-art algorithm ofthe same type when the polynomial degree becomes larger. In addition,the concept of the Galois-like groups of a polynomial is introduced. Someproperties of the Galois-like groups are proved and, more importantly, asufficient and necessary condition is given for a polynomial (which is notnecessarily irreducible) to have trivial exponent lattice.
Keywords: polynomial root, multiplicative relation, exponent lattice,trivial, Galois group, Galois-like.
Set Q ∗ to be the set of nonzero algebraic numbers. Suppose that n ∈ ZZ > . Forany v ∈ (Q ∗ ) n , define the exponent lattice of v to be R v = { u ∈ ZZ n | v u = 1 } , where v u = (cid:81) ni =1 v ( i ) u ( i ) with v ( i ) the i -th coordinate of v and u ( i ) the one of u . For a univariate polynomial f ∈ Q[ x ] (with f (0) (cid:54) = 0) of degree n , denoteby (cid:126)Ω ∈ (Q ∗ ) n the vector formed by listing all the complex roots of f withmultiplicity in some order. For convenience, we call R (cid:126)Ω the exponent lattice ofthe polynomial f and use the notation R f instead of R (cid:126)Ω , if no confusion iscaused. Moreover, we define R Q f = { u ∈ ZZ n | (cid:126)Ω u ∈ Q } . (cid:63) This work is supported partly by NSFC under grants 61732001 and 61532019. a r X i v : . [ c s . S C ] M a y T. Zheng
The exponent lattice has been studied extensively from the perspective ofnumber theory and algorithmic mathematics since the year 1997 (see [12,15,16,19]and [20,22,23,25,26]). There are applications of the exponent lattice to manyother areas or problems concerning, for example, linear recurrence sequences,loop invariants, algebraic groups, compatible rational functions and differenceequations (see [1,4,5,15,17,18]). Many of the applications involve computing theexponent lattice of a polynomial in Q[ x ]. A lattice R ⊂ ZZ n or a linear subspace R ⊂ Q n is called trivial if any v ∈ R satisfies v (1) = · · · = v ( n ). It is proved in[10] that almost all irreducible monic polynomials in ZZ [ x ] have trivial exponentlattices. However, the state-of-the-art algorithms ( FindRelations in [12,15] and
GetBasis in [26]), which compute the lattice R v for a general v ∈ (Q ∗ ) n , havedifficulty in proving the triviality of R f for an irreducible monic polynomials f in ZZ [ x ]. Recently, an algorithm called FastBasis is introduced in [25] to efficientlyprove the triviality of the exponent lattice of a given generic polynomial.In Section 2, the relations between the Galois group and the triviality of theexponent lattice of an irreducible polynomial is studied. By characterizing the socalled Q- trivial pairs from varies points of view (Proposition 1, 2 and 3), we de-sign an algorithm (Algorithm 1) recognizing all those Q-trivial pairs derived fromtransitive Galois groups. Base on this, an algorithm called
FastBasis + is obtainby adjusting the algorithm FastBasis in [25]. It turns out that
FastBasis + ismuch more efficient than FastBasis in proving the triviality of the exponentlattice of a generic irreducible polynomial when the degree of the polynomial islarge.In Section 3, we define the Galois-like groups of a polynomial since the Galoisgroup of a polynomial does not contain enough information to decide whetherthe exponent lattice is triviality or not (Example 1). We prove that a Galois-likegroup of a polynomial is a subgroup of the automorphism group of the multi-plicative group generated by the polynomial roots (Proposition 7). Furthermore,almost all conditions on the Galois group assuring the triviality of the exponentlattice can be generalized to correspondent ones on the Galois-like groups (see § Set G to be a finite group, H a subgroup of G . Set g = gH for any g ∈ G ,then G can be regarded as a permutation group on the set of the left co-sets G/H = { g | g ∈ G } via acting sg = sg . The pair ( G, H ) is called faithful,primitive, imprimitive, doubly transitive, doubly homogeneous, etc. , when thepermutation representation of G on G/H has the respective property. The groupalgebra Q[ G ] = { (cid:80) s ∈ G a s s | s ∈ G, a s ∈ Q } is defined as usual and the Q-vectorspace Q[ G/H ] = (cid:8) (cid:80) s ∈ G/H a s s | s ∈ G, a s ∈ Q (cid:9) becomes a left Q[ G ]-module attice Triviality through Groups 3 via acting λt = (cid:80) s ∈ G a s st , with λ = (cid:80) s ∈ G a s s ∈ Q[ G ] and t ∈ G/H . We set ZZ [ G/H ] = (cid:8) (cid:80) s ∈ G/H a s s | s ∈ G, a s ∈ ZZ (cid:9) for convenience.A subset M of Q[ G/H ] is called Q- admissible if there is an element µ ∈ Q[ G ]with stabilizer G µ = H so that mµ = 0 for any m ∈ M (Definition 3 of [13]).Set V = (cid:8) a (cid:80) s ∈ G/H s | a ∈ Q (cid:9) . Then the pair ( G, H ) is called Q- trivial if0 and V are the only two Q[ G ]-submodules that are Q-admissible (Definition7 of [13]). A polynomial f ∈ Q[ x ] ( f (0) (cid:54) = 0) without multiple roots is called non-degenerate if the quotient of any two roots of f is not a root of unity, and degenerate otherwise. The relations between the Q-triviality of a pair and thetriviality of an exponent lattice is given below: Proposition 1.
Suppose that L is a finite Galois extension of the field Q withGalois group G , and that H < G is a subgroup of G so that G operates on the set G/H faithfully. Then the pair ( G, H ) is Q -trivial iff for any f ∈ Q[ x ] ( f (0) (cid:54) = 0) satisfying all the following conditions, R f is trivial:(i) f is irreducible over Q and non-degenerate;(ii) the splitting field of f equals L and its Galois group G f = G ;(iii) H is the stabilizer of a root of f .Proof. “If”: Suppose on the contrary that the pair ( G, H ) is not Q-trivial. Thenthere is a Q-admissible Q[ G ]-submodule M of Q[ G/H ] containing an element v = (cid:80) s ∈ G/H v s s ∈ ZZ [ G/H ] so that there are s (cid:54) = s ∈ G/H satisfying v s (cid:54) = v s . By definition, there is an element µ ∈ Q[ G ] with G µ = H such that vµ = 0.Since Q is an algebraic number field, [13] Proposition 4 indicates that M isadmissible in the multiplicative sense. Now by [13] Proposition 2, there is analgebraic number α ∈ L ∗ with stabilizer G α = H and the element v is a non-trivial multiplicative relation between the conjugations of α . What’s more, anyquotient of two conjugations of α cannot be a root of unity. These mean thatthe minimal polynomial f of α over the field Q is non-degenerate and the lattice R f is nontrivial. Denote by F the splitting field of f over Q. Then F is asubfield of L and the Galois group G f of f is isomorphic to G/Gal ( L/F ). Since G α = H , G g ( α ) = gHg − for any g ∈ G . Hence the fixed field of the group gHg − is Q[ g ( α )]. Note that Q[ g ( α )] ⊂ F , gHg − ⊃ Gal(
L/F ) by Galois theory.Thus ∩ g ∈ G gHg − ⊃ Gal(
L/F ). Since the subgroup ∩ g ∈ G gHg − of G operatestrivially on the set G/H and the group G operates faithfully on this set, ∩ g ∈ G gHg − = 1. Hence Gal( L/F ) = 1 and G f (cid:39) G . In fact, L = F and G f = G . Sothe existence of f leads to a contradiction.“Only If”: Assume that there is an irreducible non-degenerate polynomial f ∈ Q[ x ] ( f (0) (cid:54) = 0) satisfying the condition (iii) with splitting field equal to L and exponent lattice R f nontrivial. Suppose that the set of the roots of f is Ω and α ∈ Ω ⊂ L ∗ is with stabilizer G α = H . Thus there is a bijection τ : G/H → Ω, g (cid:55)→ g ( α ) through which the permutation representations of G on these two sets are isomorphic and we have the ZZ -module isomorphism ZZ Ω (cid:39) ZZ [ G/H ]. By [13] Proposition 2, the lattice R f ⊂ ZZ Ω (cid:39) ZZ [ G/H ]provides an admissible subset M of ZZ [ G/H ] in the multiplicative sense. Thenby Proposition 3 and Definition 3 in [13], one sees that M is a Q-admissible T. Zheng subset. Since R f is nontrivial, the Q[ G ]-module generated by M in Q[ G/H ] isneither 0 nor V . Hence the pair ( G, H ) is not Q-trivial, which is a contradiction. (cid:117)(cid:116)
For any irreducible non-degenerate polynomial f ∈ Q[ x ] with Galois group G and a root stabilizer H < G , Proposition 1 gives the weakest sufficient conditionon the pair (
G, H ) for R f to be trivial ( i.e. , ( G, H ) being Q-trivial). However,the pair (
G, H ) does not contain all the information needed to decide whetherthe lattice R f is trivial. This is shown in the following example. Example 1.
Set g ( x ) = x − x + 4 x + 6, then g is irreducible in Q[ x ]. By the Unitary-Test algorithm in [24], one proves that g is non-degenerate. Set L tobe the splitting field of g over the rational field and G = Gal( L/ Q) its Galoisgroup. Denote by α = (2 . · · · ) + √− · (0 . · · · )one of the roots of g , and set H = G α to be its stabilizer. Computing withAlgorithm 7.16 in [15], we obtain R g = ZZ · ( − , , , − T , which is nontrivial(thus ( G, H ) is not Q-trivial by Proposition 1).Set f ( x ) = g ( x − f is irreducible over Q with splitting field L andGalois group G . Moreover, the number α + 1 is a root of f with stabilizer H .We note that the polynomials g and f share the same pair ( G, H ). However,computing with [15] Algorithm 7.16, we obtain that the lattice R f = { } istrivial. Suppose that G is a finite group and H < G is a subgroup sothat G operates faithfully on G/H . Denote by GH the character of the permuta-tion representation of G on the set G/H . Then the pair ( G, H ) is Q -trivial iffthe character GH − is Q -irreducible.Proof. By [13] Proposition 12, the pair (
G, H ) is Q-trivial iff (
G, H ) is primitiveand the character 1 GH − GH − G, H ) is primitive. (cid:117)(cid:116)
Throughout the paper, a root of rational refers to an algebraic number α suchthat there is a positive rational integer k ensuring α k ∈ Q. Remark 1.
In the settings of Proposition 1, when (
G, H ) is Q-trivial, (
G, H ) isprimitive. This is equivalent to the condition that H is a maximal subgroup of G . A polynomial f satisfying the conditions (ii) and (iii) in Proposition 1 hasa root α with stabilizer G α = H and the fixed field Q[ α ] of the group H is aminimal intermediate field of the extension L/ Q by Galois theory. Note that f is irreducible over Q. If f is degenerate with no root being a root of rational,then there is an integer k (cid:54) = 0 so that 1 < deg( α k ) < deg( α ). Thus Q (cid:36) Q[ α k ] (cid:36) attice Triviality through Groups 5 Q[ α ], which contradicts the minimality of the field Q[ α ]. Thus, when ( G, H ) isQ-trivial, a polynomial f satisfying the conditions (ii) and (iii) is either non-degenerate or with all roots being roots of rational. Hence the condition (i) inProposition 1 can be replaced by the condition that “ f is irreducible over Q withno root being a root of rational”. Proposition 3.
Let ( G, H ) be as in Proposition 2. Then, regarded as a permu-tation group operating on the set G/H , the group G satisfies exactly one of thefollowing conditions iff the pair ( G, H ) is Q -trivial:(i) G is doubly transitive;(ii) G is of affine type (but not doubly transitive) of degree p d for someprime p and G = M (cid:111) H , where M (cid:39) IF dp is the socle of G and the subgroup H is isomorphic to a subgroup of GL ( d, p ) ; moreover, let Z be the center of thegroup GL ( d, p ) and regard H as a subgroup of GL ( d, p ) , the group HZ/Z is atransitive subgroup of
P GL ( d, p ) operating on the projective points;(iii) G is almost simple (but not doubly transitive) of degree q ( q − , where q = 2 f ≥ and q − is prime, and either G = P SL ( q ) or G = P Γ L ( q ) withthe size of the nontrivial subdegrees q + 1 or ( q + 1) f , respectively.Proof. This is a combination of Theorem 3 and Theorem 12 in [7] together withCorollary 1.6 in [2]. (cid:117)(cid:116)
Denote by P the set of prime numbers and by P ω the set of prime powers { p d | p ∈ P , d ∈ ZZ ≥ } . A useful corollary is as follows: Corollary 1.
Suppose that a polynomial f ∈ Q[ x ] ( f (0) (cid:54) = 0) is irreducible withGalois group G and a root stabilizer H . If the number deg( f ) is NOT in the set S = P ω ∪ (cid:8) f − (2 f − (cid:12)(cid:12) f ∈ ZZ ≥ , f − ∈ P (cid:9) , (1) then the pair ( G, H ) is Q -trivial iff it is doubly transitive. Besides the doubly transitive pairs (
G, H ), the author provided some other par-ticular Q-trivial pairs in [13] Proposition 13–15. A permutation group G on aset S is called doubly homogeneous if for any two subsets { s , s } , { t , t } of S ,there is some g ∈ G so that { g ( s ) , g ( s ) } = { t , t } . In this subsection, we provethat any doubly homogeneous pair ( G, H ) is also Q-trivial.
Proposition 4.
Suppose that G is a finite group and H < G is a subgroup sothat G operates faithfully on G/H . If the pair ( G, H ) is doubly homogeneous,then it is Q -trivial.Proof. When the pair (
G, H ) is doubly transitive, the character 1 GH − G, H ) is doublyhomogeneous but not doubly transitive, then G is of odd order (Exe. 2.1.11 of[8]). Then by [11] and TEOREMA 7 of [21], G is the Galois group of a finiteGalois extension of the rational field. T. Zheng
Let f ∈ Q[ x ] ( f (0) (cid:54) = 0) be any polynomial satisfying the conditions (i)–(iii)in Proposition 1. Then f is irreducible and non-degenerate with Galois group G . Since the pair ( G, H ) is doubly homogeneous and the condition (iii) holds, G operates in a doubly homogeneous way on the set Ω of the roots of f . Doublyhomogeneousness naturally requires that deg( f ) = | Ω | = | G/H | ≥
2. Hence f has no root being a root of rational since it is non-degenerate. By [25] Theorem3.2, the lattice R Q f is trivial and so is the lattice R f . Finally, according toProposition 1, the pair ( G, H ) is Q-trivial. (cid:117)(cid:116)
The following example shows that a Q-trivial pair need not be doubly homoge-neous.
Example 2.
Set L to be the splitting field of the irreducible polynomial f = x − x − x + 3 x + 3 x − G the Galois group. Infact, G (cid:39) C is the cyclic group of order 5, and the stabilizer of any root of f istrivial. The faithful pair ( C ,
1) is Q-trivial by Proposition 5. Nevertheless, thepair ( C ,
1) is not doubly homogeneous.
Proposition 5.
Let ( G, H ) be as in Proposition 1. If the cardinality of the set G/H is a prime, then ( G, H ) is Q -trivial.Proof. When | G/H | = 2, the pair ( G, H ) is doubly homogeneous and we aredone. When | G/H | is an odd prime, the proposition is a straightforward resultof [9] Theorem 1 and Proposition 1. (cid:117)(cid:116) The figure below shows the relations between different classes of Q-trivialpairs. This is based on Theorem 3 of [7], Corollary 1.6 of [2] and Proposition 3.1of [14].
Fig. 1.
Classification of Q-trivial pairs.
Assume that f ∈ Q[ x ] ( f (0) (cid:54) = 0) is irreducible with Galois Group G and aroot stabilizer H . In this subsection, we develop an algorithm deciding whethera pair ( G, H ) is Q-trivial for such a polynomial f . Moreover, numerical resultsshow that the algorithm is quite efficient compared with some other relativealgorithms (see Table 3 and 4). All numerical results are obtained on a desktopof WINDOWS 7 SYSTEM with 8GB RAM and a 3.30GHz Intel Core i5-4590processor with 4 cores. attice Triviality through Groups 7 The “
IsQtrivial ” Algorithm
Algorithm 1 shown below is designed accordingto § § G ]-submodule B generated by u in Step 6 iscontained in the Q[ G ]-submodule V = (cid:110) (cid:88) t ∈ G/H a t t (cid:12)(cid:12)(cid:12) a t ∈ Q , (cid:88) t ∈ G/H a t = 0 (cid:111) with character 1 GH − f ) −
1. So the correctness of Step7–10 follows from Proposition 2.
Algorithm 1:
IsQtrivial
Input:
An irreducible polynomial f ∈ Q[ x ] with f (0) (cid:54) = 0; Output: “ True ” if the pair (
G, H ) is Q-trivial and “
False ” otherwise. if ( f is reducible or f (0) == 0) then { Return “Error!” } end if (deg( f ) is a prime) then { Return True ; } end Compute the Galois group G of f ; if ( G is doubly transitive) then { Return True ; } end if (deg( f ) / ∈ S as defined in (1)) then { Return False ; } end Compute B = Q[ G ] (cid:104) u (cid:105) with u = s − ∈ Q[ G/H ] for an s / ∈ H ; if (dim( B ) == deg( f ) − and B is Q -irreducible ) then Return True ; end Return False ; Algorithm 1 is implemented with Magma and random examples are generatedto test it. A random polynomial f of degree n with f (0) (cid:54) = 0 is generated in thefollowing way: First, generate its leading coefficient and its constant term bypicking integer numbers randomly from the set {± , . . . , ± } , then pick therest of the coefficients of f in the set {− , − , . . . , } randomly. Second, checkwhether f is irreducible: if it is, we are done; otherwise, go back to the first step.The numerical results are shown in Table 1.In Table 1 (and throughout the section), the notation “ PolynomialWithGaloisGroup , which provides polynomialswith all types of transitive Galois groups of degree between 2 and 15. The resultsare shown in Table 2.In both tables, the “GaloisFail” columns show, for each degree, the num-bers of the polynomials with Galois groups computed unsuccessfully in Algo-rithm 1 Step 3, which is implemented by the Magma functions
GaloisGroup
T. Zheng and
GaloisProof . There are more “GaloisFail” cases in Table 2. The problemis: in those “GaloisFail” cases, though the Galois groups can be computed bythe first function (which does not provide proven results), the second functionreturns error and fails to support the result. The “Average Time” in Table 2excludes the “GaloisFail” examples, i.e. , it only counts in the “Qtrivial” and the“NotQtrivial” cases. We see that the algorithm is still efficient when the Galoisgroup is successfully computed.
Table 1.
Random Test for
IsQtrivial
Deg
Table 2.
Testing
IsQtrivial by Different Galois GroupsDeg
Ensuring Lattice Triviality
By [10] Theorem 2, almost all irreducible poly-nomials f with f (0) (cid:54) = 0 has trivial lattice R f . However, the general algorithms, FindRealtions in [15,12] and
GetBasis in [26], dealing with the general inputwhich are arbitrarily given nonzero algebraic numbers instead of all the roots of attice Triviality through Groups 9
Table 3. “ IsQtrivial ” Ensuring Triviality EfficientlyDeg Polynomial Runtime (s)
FindRelations GetBasis IsQtrivial + IsROR f (1) f (2) f (3) g (1) OT OT 0.000 g (2) OT OT 0.000 g (3) OT OT 0.0009 h (1) OT OT 0.047 h (2) OT OT 0.078 h (3) OT OT 0.047 a certain polynomial, are not every efficient in proving exponent lattice trivialityin the latter case.For a randomly generated irreducible polynomial f with f (0) (cid:54) = 0, if thefunction IsQtrivial ( f ) returns True and f is proved to have no root being aroot of rational by Algorithm 5 in [27] (named “ RootOfRationalTest ” therein,we call it “
IsROR ” here instead), then R f is trivial by Proposition 1 and Remark1. We call this the “ IsQtrivial + IsROR ” procedure. Table 3 shows the efficiencyof this procedure to prove the triviality of the exponent lattice of a randomlygenerated polynomial. The polynomials used here are with integer coefficientspicked randomly from the set {− , − , . . . , } .The notation “OT” in Table 3 (and throughout the section) means the com-putation is not finished within two hours. As is shown in Table 3, it is time con-suming for the general algorithms FingRelations and
GetBasis to prove the ex-ponent lattice triviality of a generic polynomial. Thus the “
IsQtrivial + IsROR ”procedure can be used before running either of the two general algorithms, whenthe inputs are all the roots of a certain polynomial. If the procedure fails to provethe triviality, then one turns to the general algorithms.
The “
FastBasis + ” Algorithm Similar to the “
IsQtrivial + IsROR ” proce-dure, Theorem 3.2 in [25] allows one to prove lattice triviality of a polynomial byproving doubly homogeneousness of its Galois group and by checking the condi-tion that none of its roots is a root of a rational. Based on this, the algorithm
FastBasis is designed to compute the lattice R f fast for any f in a generic set E ⊂ Q[ x ] (Definition 5.1 of [25]).Similarly, we can define another set E + ⊂ Q[ x ] to be the set of polynomials f for which both the following two conditions hold:( i ) ∃ c ∈ Q ∗ , g ∈ Q[ x ] , k ∈ ZZ ≥ so that f = cg k , g is irreducible and x doesnot divide g ( x ) ; ( ii ) all the roots of g are roots of rational or IsQtrivial ( g ) = True .Then, by Proposition 4 and Example 2, one claims that E + (cid:37) E . Thus E + isalso generic in the sense of [25]. Moreover, an algorithm similar to FastBasis ,which will be called “
FastBasis + ”, can be obtained by replacing Steps 6–7 inAlgorithm 6.1 of [25] (namely, FastBasis ) by the following step:... if ( IsQtrivial ( g ) == False ) then { return F } end if ;...Like FastBasis , the algorithm
FastBasis + computes the lattice R f for any f ∈ E + while returning a special symbol “F” when f / ∈ E + .The algorithm FastBasis + is implemented with Magma while the algorithm FastBasis is implemented with Mathematica by the author of [25]. In Table4 we compare these two algorithms by applying them to a great deal of ran-dom polynomials of varies degree. For an f ∈ Q[ x ] of degree at most n , wedefine h ( f ) = max ≤ i ≤ n | c f,i | with c f,i the coefficient of the term x i of f . Thepolynomials in Table 4 are picked randomly from the classes ZZ ,n [ x ] = { f ∈ ZZ [ x ] | h ( f ) ≤ , deg( f ) ≤ n } . In Table 4, the notation “ E + within two hours. The average timeonly counts in all the “Success” examples. We can see from Table 4 and Fig. 2that for the small inputs with n <
15, the algorithm
FastBasis is slightly moreefficient while for those lager inputs with n >
15, the algorithm
FastBasis + ismuch more efficient. This allows one to handle inputs with higher degree thatwere intractable before. As is shown in Example 1, provided only the pair (
G, H ), one may not be able todecide whether the lattice R f is trivial or not. Here f is an irreducible polynomialwith Galois group G and a root stabilizer H . In this section, the concept of aGalois-like group is introduced. An equivalent condition for the lattice R f to betrivial is given through the concept of a Galois-like group. Set f ∈ Q[ x ] ( f (0) (cid:54) = 0) to be a polynomial with no multiple roots. Denote by Σ the symmetry group operating on the set Ω = { r , . . . , r n } of the roots of f . In the sequel, we denote by (cid:126)Ω = ( r , . . . , r n ) T a vector of the roots and by σ ( (cid:126)Ω ) = ( σ ( r ) , . . . , σ ( r n )) T a permutation of (cid:126)Ω with σ ∈ Σ . Definition 1.
A Galois-like group of the polynomial f refers to any one of thefollowing groups: attice Triviality through Groups 11 Fig. 2.
Comparing average runtime of two algorithms.
Log r i t h m i c A v e r age T i m e FastBasisFastBasis + A v e r age T i m e ( s ) FastBasisFastBasis + Table 4.
FastBasis v.s.
FastBasis + Class + n = 6 10000 9011 989 0 0.007304 0.025499 0 989 9011 n = 8 10000 9064 936 0 0.018372 0.055328 0 936 9064 n = 9 10000 9113 887 0 0.029044 0.069996 0 887 9113 n = 15 10000 9227 773 0 0.305941 0.301110 0 773 9227 n = 20 10000 9243 757 0 1.502110 0.700131 0 757 9243 n = 28 10000 9279 721 0 9.29961 1.527806 0 721 9279 n = 40 100 93 7 0 76.3928 6.788000 0 7 93 n = 50 100 96 4 0 315.523 15.70400 0 4 96 n = 60 35 33 1 1 1291.38 31.27800 0 1 34 n = 81 40 15 1 24 5539.67 104.515 0 1 39 n = 90 40 0 2 38 – 224.413 0 2 38 n = 120 40 – – – – 2058.228 0 2 38 (i) G f = { σ ∈ Σ | ∀ v ∈ Z n , (cid:126)Ω v = 1 ⇒ σ ( (cid:126)Ω ) v = 1 } ;(ii) G Bf = { σ ∈ Σ | ∀ v ∈ Z n , (cid:126)Ω v ∈ Q ⇒ σ ( (cid:126)Ω ) v = (cid:126)Ω v } ;(iii) G Q f = { σ ∈ Σ | ∀ v ∈ Z n , (cid:126)Ω v ∈ Q ⇒ σ ( (cid:126)Ω ) v ∈ Q } ; To verify the terms used above in the definition, we need to prove that anysubset of Σ defined in Definition 1 is indeed a group: Proposition 6.
Suppose that f ∈ Q[ x ] ( f (0) (cid:54) = 0) is a polynomial with nomultiple roots. Set G = G f , G Bf or G Q f , then G is a subgroup of Σ .Proof. Any σ ∈ Σ results in a coordinate permutation ˆ σ operating on the space C n with n = | Ω | in a manner so that for any vector v = ( c , . . . , c n ) T ∈ C n ,ˆ σ ( v ) = ( b , . . . , b n ) T with b i = c j whenever σ ( r i ) = r j . Then one observes thatthe equalities (cid:100) σ − = (ˆ σ ) − , σ ( (cid:126)Ω ) ˆ σ ( v ) = (cid:126)Ω v and (cid:126)Ω ˆ σ ( v ) = σ − ( (cid:126)Ω ) v hold for any σ ∈ Σ and any v ∈ Z n .Set G = G f ( G Bf or G Q f respectively) and R = R f ( R Q f respectively). Then,by definition, ˆ σ − ( v ) ∈ R for any σ ∈ G and any v ∈ R . Hence the set ˆ σ − ( R ) = { ˆ σ − ( v ) | v ∈ R} is a subset of the lattice R . Noting that ˆ σ − operates linearly,one concludes that ˆ σ − ( R ) is also a lattice. Thus ˆ σ − ( R ) is a sub-lattice of R .Since ˆ σ − is linear and non-singular, any basis of R is transformed into a basis ofˆ σ − ( R ) by ˆ σ − . Hence rank( R ) = rank(ˆ σ − ( R )). Since ˆ σ − is orthogonal on thespace R n and orthogonal operations preserve the lattice volume, R = ˆ σ − ( R ).Thus ˆ σ ( R ) = R .So ˆ σ ( v ) ∈ R for any σ ∈ G and any v ∈ R . If G = G f (or G Q f ) and R = R f (or R Q f respectively), then (cid:126)Ω ˆ σ ( v ) = 1 (or (cid:126)Ω ˆ σ ( v ) ∈ Q respectively). Equivalently, attice Triviality through Groups 13 σ − ( (cid:126)Ω ) v = 1 (or σ − ( (cid:126)Ω ) v ∈ Q). Hence σ − ∈ G for any σ ∈ G . Now supposethat G = G Bf and R = R Q f . Since ˆ σ ( v ) ∈ R Q f and (cid:126)Ω ˆ σ ( v ) ∈ Q, σ ( (cid:126)Ω ) ˆ σ ( v ) = (cid:126)Ω ˆ σ ( v ) follows from the definition of G Bf . The left side of this equality equals (cid:126)Ω v whileits right side equals σ − ( (cid:126)Ω ) v . Hence σ − ( (cid:126)Ω ) v = (cid:126)Ω v for any σ ∈ G Bf and v ∈ R Q f .Thus σ − ∈ G Bf .The closure of the multiplication in the subset G of Σ and the fact that 1 ∈ G are straightforward. Thus G is a group. (cid:117)(cid:116) Define groups (cid:104) Ω (cid:105) = { (cid:126)Ω v | v ∈ Z n } and (cid:104) Ω (cid:105) Q = { c (cid:126)Ω v | c ∈ Q ∗ , v ∈ Z n } . The following proposition asserts that the Galois-like groups G f and G Bf of a polynomial f are subgroups of the automorphism groups of (cid:104) Ω (cid:105) and (cid:104) Ω (cid:105) Q respectively. Proposition 7.
The following relations hold:(i) G f (cid:39) { η ∈ Aut ( (cid:104) Ω (cid:105) ) | ∀ r i ∈ Ω, η ( r i ) ∈ Ω } ;(ii) G Bf (cid:39) (cid:8) η ∈ Aut ( (cid:104) Ω (cid:105) Q ) (cid:12)(cid:12) ∀ r i ∈ Ω, η ( r i ) ∈ Ω ; η | Q ∗ = id Q ∗ (cid:9) .Proof. Denote by Aut Ω ( (cid:104) Ω (cid:105) ) the group in the right side of the formula in (i)and by Aut Q Ω ( (cid:104) Ω (cid:105) Q ) the one in the right side of the formula in (ii).Set G = G f (or G Bf ) and A = Aut Ω ( (cid:104) Ω (cid:105) ) (or Aut Q Ω ( (cid:104) Ω (cid:105) Q ) respectively). Forany σ ∈ G , we define an element E σ in A in the following way: for any (cid:126)Ω v ∈ (cid:104) Ω (cid:105) , E σ ( (cid:126)Ω v ) = σ ( (cid:126)Ω ) v (or, for any c (cid:126)Ω v ∈ (cid:104) Ω (cid:105) Q , E σ ( c (cid:126)Ω v ) = cσ ( (cid:126)Ω ) v ). From thedefinition of G , we see that σ ( (cid:126)Ω ) v = σ ( (cid:126)Ω ) v (cid:48) whenever (cid:126)Ω v = (cid:126)Ω v (cid:48) ∈ (cid:104) Ω (cid:105) (orthat c σ ( (cid:126)Ω ) v = c σ ( (cid:126)Ω ) v (cid:48) whenever c (cid:126)Ω v = c (cid:126)Ω v (cid:48) ∈ (cid:104) Ω (cid:105) Q ). Thus the map E σ : (cid:104) Ω (cid:105) → (cid:104) Ω (cid:105) (or E σ : (cid:104) Ω (cid:105) Q → (cid:104) Ω (cid:105) Q ) is well defined. It is trivial to verify the factthat E σ is an automorphism of (cid:104) Ω (cid:105) (or of (cid:104) Ω (cid:105) Q ) and the property that for all r i ∈ Ω , E σ ( r i ) = σ ( r i ) ∈ Ω (or, moreover, E σ ( c ) = c for any c ∈ Q ∗ ). Hence E σ is indeed in the set A . Thus E • is a map from G to A .For any η ∈ A , η is injective and η ( Ω ) ⊂ Ω . Since Ω is finite, η ( Ω ) = Ω .Hence η | Ω ∈ Σ . Because η is an automorphism (or an automorphism fixing everyrational number), η | Ω ( (cid:126)Ω ) v = 1 whenever (cid:126)Ω v = 1 (or η | Ω ( (cid:126)Ω ) v = (cid:126)Ω v whenever (cid:126)Ω v ∈ Q). Thus η | Ω ∈ G . Define R η = η | Ω , then R • is a map from A to G .It is clear that both the maps E • and R • are group homomorphisms. Thatis, E σ σ = E σ E σ and R η η = R η R η for any σ , σ ∈ G and any η , η ∈ A .One also verifies easily that E R • = id A and R E • = id G . Hence G (cid:39) A . (cid:117)(cid:116) By definition, a Galois-like group of a polynomial f is the group of the per-mutations between its roots that preserve all the multiplicative relations betweenthem. Since any element in the Galois group of f preserves all polynomial rela-tions between the roots, the following relations between the Galois group and aGalois-like group of f is straightforward: Proposition 8.
Suppose that f ∈ Q[ x ] ( f (0) (cid:54) = 0) has no multiple roots. Then,regarded as a permutation group operating on the roots of f , the Galois group of f is a subgroup of any Galois-like group of f . Besides, the following relations between the Galois-like groups is straightforwardbut noteworthy:
Proposition 9.
Let f be as in Proposition 8. Then G Bf ≤ G f and G Bf ≤ G Q f . With the help of the concept of Galois-like groups, we can generalize manysufficient conditions that implying triviality of exponent lattices.
Lemma 1.
Set f ∈ Q[ x ] ( f (0) (cid:54) = 0) to be a polynomial without multiple roots.Denote by (cid:126)Ω = ( α , . . . , α s , γ , . . . , γ t ) T the vector of all the roots of f with α i the roots that are not roots of rational. Suppose that the Galois-like group G f is doubly transitive, then any multiplicative relation v ∈ R (cid:126)Ω = R f satisfies thefollowing condition: v (1) = · · · = v ( s ) = v (1) + · · · + v ( s + t ) s + t . (2) Lemma 2.
Let f and (cid:126)Ω be as in Lemma 1. Suppose that the Galois-like group G Bf or G Q f is doubly transitive, then any multiplicative relation v ∈ R Q (cid:126)Ω = R Q f satisfies the condition (2) . The proofs of those two propositions above are both almost the same to the oneof Theorem 3 in [3], because of which we do not give any of them here. A directcorollary of these propositions are as follows:
Proposition 10.
Set f ∈ Q[ x ] ( f (0) (cid:54) = 0) to be a polynomial without mul-tiple roots and none of its roots is a root of rational. If the group G Bf or G Q f ( respectively, G f ) is doubly transitive, then the lattice R Q f ( respectively, R f ) istrivial. This is a generalization of Theorem 3 in [3]. The essential idea is that the proofof Theorem 3 in [3] relies only on the properties of Galois-like groups ( i.e. ,preserving all the multiplicative relations) but not on those properties that arepossessed uniquely by the Galois groups.Noting that G Bf ≤ G f , one concludes form Proposition 10 that both thelattices R Q f and R f are trivial whenever the group G Bf is doubly transitive andnone of the roots of f is a root of rational. More generally, we have the followingproposition and Corollary 2: Proposition 11.
Set f ∈ Q[ x ] ( f (0) (cid:54) = 0) to be a polynomial without multipleroots. Define W f = Q ⊗ R f and W Q f = Q ⊗ R Q f with “ ⊗ ” the tensor product of Z -modules. Set V = { v ∈ Q n | (cid:80) ni =1 v ( i ) = 0 } and V = { c (1 , . . . , n ) T | c ∈ Q } with n = deg( f ) . Suppose that G Bf is transitive, then the following conclusionshold:(i) W Q f ∩ V = W f ∩ V ;(ii) W Q f = W f + V and thus W Q f = W f iff f (0) ∈ { , − } . attice Triviality through Groups 15 Proof.
The proof is almost the same to the one of Lemma 1 in [6], except that werequire the transitivity of the Galois-like group G Bf instead of the the transitivityof the Galois group of f . (cid:117)(cid:116) Corollary 2.
Let f be as in Proposition 11 such that the group G Bf is transitive,then the lattice R f is trivial iff the lattice R Q f is.Proof. Since the group G Bf is transitive, W Q f = W f + V by Proposition 11. So W Q f is trivial iff W f is trivial. Hence R f is trivial ⇐⇒ W f is trivial ⇐⇒ W Q f is trivial ⇐⇒ R Q f is trivial. (cid:117)(cid:116) Similar to Proposition 10, we have the following result:
Proposition 12.
Let f ∈ Q[ x ] be as in Proposition 10. If the group G Bf or G Q f is doubly homogeneous, then the lattice R Q f is trivial. This is a generalization of [25] Theorem 3.2. The proof of this proposition isalmost the same with the one given in [25], hence we omit it. Another general-ization trough Galois-like groups of the “Only If” part of Proposition 1 is givenbelow:
Proposition 13.
Set f ∈ Q[ x ] ( f (0) (cid:54) = 0) to be a polynomial without multipleroots and one of its roots is not a root of rational. Set G = G f ( respectively, G = G Bf or G Q f ) and R = R f ( resp., R = R Q f ) . If G is transitive and the pair ( G , H ) , with H a root stabilizer, is Q -trivial, then R is trivial.Proof. Let V , V and W = Q ⊗ R be as in Proposition 11. For any σ ∈ G , wedefine a coordinate permutation ˆ σ as in the proof of Proposition 6. Then W is aQ[ G ]-submodule of Q n by the definition of a Galois-like group (for any v ∈ Q n or v ∈ W , a group element σ operates in the way so that it maps v to the vectorˆ σ − ( v )).Since the pair ( G , H ) is Q-trivial, Q n can be decomposed into two irreducibleQ[ G ]-submodules: Q n = V ⊕ V ([13] Proposition 12). Since G is transitive, V and V are the only two irreducible Q[ G ]-submodules of Q n :Suppose that V (cid:54) = V is an irreducible Q[ G ]-submodules and assume that V ∩ V (cid:37) { } . Then V (cid:37) V ∩ V or V (cid:37) V ∩ V . This contradicts the fact thatboth V and V are irreducible, since V ∩ V (cid:37) { } is a proper Q[ G ]-submodulesof at least one of them. So we have V ∩ V = { } . Noting that the Q-dimensionof V is n −
1, one concludes that dim Q ( V ) = 1. Set v ∈ V\{ } , then v / ∈ V and (cid:80) ni =1 v ( i ) (cid:54) = 0. Thus V (cid:51) (cid:80) σ ∈G ˆ σ − ( v ) = |G| n ( (cid:80) ni =1 v ( i ))(1 , . . . , n ) T (cid:54) = follows from the transitivity of G . Hence V = V .Thus all the Q[ G ]-submodules of Q n are { } , V , V and Q n itself. If V ⊂ W ,then W ⊂ W Q f and V ⊂ W Q f imply that W Q f = Q n , which contradicts theassumption that f has a root that is not a root of rational. Hence V (cid:54)⊂ W ,which means W = { } or W = V . Thus W is trivial and so is the lattice R . (cid:117)(cid:116) In this subsection, we characterize those polynomials f with a trivial exponentlattice by giving a necessary and sufficient condition through the concept of aGalois-like group. Theorem 1.
Set f ∈ Q[ x ] ( f (0) (cid:54) = 0) to be a polynomial without multiple roots.Denote by β , . . . , β t the rational roots of f (if there are any) and by β therational number which is the product of all non-root-of-rational roots of f (ifthere are any). Then the lattice R f is trivial iff all the following conditions hold:(i) the Galois-like group G f = Σ ;(ii) any root of f is rational or non-root-of-rational;(iii) the lattice R v f is trivial with the vector v f given by: v f = ( β , β , . . . , β t ) T , if f has both rational andnon-root-of-rational roots , ( β , . . . , β t ) T , if any root of f is rational , ( β ) , if any root of f is non-root-of-rational . Proof. “If”: When deg( f ) = 1, R f is trivial and we are done. Suppose in thefollowing that deg( f ) ≥
2. Then the pair ( G f , H ) = ( Σ, H ) is doubly homoge-neous thus also Q-trivial for any root stabilizer H . If f has a root that is not aroot of rational, then R f is trivial by Proposition 13. When all the roots of f are rational, the lattice R f = R v f is trivial.“Only If”: Now that R f is trivial, the condition (i) is straightforward. Sup-pose that f has a root r which is a root of rational but not a rational number.Then the conjugations of r , say, { r = r (1) , r (2) , . . . , r ( s ) } , with s ≥
2, are all theroots of f . Then there is a positive integer m so that ( r/r (2) ) m = 1. Thus R f isnon-trivial, which contradicts the assumption. So the condition (ii) holds. Sinceany nontrivial multiplicative relation of the vector v f results in a nontrivial mul-tiplicative relation between the roots of f , the condition (iii) holds. (cid:117)(cid:116) From the “If” part of the proof we observe that, when restricted to polynomi-als f with degree higher than one, the condition (i) in Theorem 1 can be replacedby the statement “ G f is transitive and the pair ( G f , H ) is Q-trivial for any rootstabilizer H ”. An interesting result follows directly from this observation: Corollary 3.
Let f be as in Theorem 1. If deg( f ) ≥ and the conditions (ii) –(iii) in Theorem 1 hold, then the following conditions are equivalent to eachother:(i) G f = Σ ;(ii) G f is doubly transitive;(iii) G f is doubly homogeneous;(iv) G f is transitive and the pair ( G f , H ) is Q -trivial for any root stabilizer H . attice Triviality through Groups 17 Proof.
The implications ( i ) ⇒ ( ii ) and ( ii ) ⇒ ( iii ) are trivial. The implication( iv ) ⇒ ( i ) follows from the “If” part of the proof of 1. Now we prove theimplication ( iii ) ⇒ ( iv ): If deg( f ) = 2 and f has only rational roots, the lattice R f = R v f is trivial. So G f = Σ . If deg( f ) = 2 but f has a root that is not aroot of rational, f is irreducible over Q and Σ = G f ≤ G f with G f the Galoisgroup of f . In either case G f is transitive. When deg( f ) ≥
3, the transitivity of G f follows from [8] Theorem 9.4A. Now the Q-triviality of the pair ( G , H ) followsfrom Proposition 4. (cid:117)(cid:116) Thus the condition (i) of Theorem 1 can be replaced by any one of the conditions(ii)–(iv) in Corollary 3.For the lattice R Q f , we have a similar result: Theorem 2.
Set f ∈ Q[ x ] ( f (0) (cid:54) = 0) to be a polynomial without multiple roots.Then the lattice R Q f is trivial iff all the following conditions hold:(i) the Galois-like group G Bf = Σ ;(ii) either deg( f ) = 1 or any root of f is not a root of rational;(iii) f is irreducible over Q .Proof. “If”: When deg( f ) = 1, this is trivial. Suppose in the following thatdeg( f ) ≥ f is not a root of rational. Then G Bf = Σ is tran-sitive and doubly homogeneous. Thus the pair ( G Bf , H ) is Q-trivial for any rootstabilizer H by Proposition 4. So R Q f is trivial by Proposition 13.“Only If”: Now that R Q f is trivial, it is clear that G Bf = Σ . Suppose on thecontrary that f is reducible and g , g are two of its factors. Let α , . . . , α s denotethe roots of g and γ , . . . , γ t the ones of g . Then for any two distinct integers k and l , ( α . . . α s ) k ( γ . . . γ t ) l ∈ Q. This contradicts the assumption that R Q f is trivial. So f is irreducible. Assume that deg( f ) ≥ r of f is a root of rational. The conjugations of r , say, { r = r (1) , r (2) , . . . , r ( n ) } ( n ≥
2) are exactly all the roots of f . Then there is a positive integer m so that( r/r (2) ) m = 1. Thus R f is non-trivial and so is the lattice R Q f . This contradictsthe assumption. (cid:117)(cid:116) Remark 2.
Theorem 2 still holds when the equality G Bf = Σ is replaced by G Q f = Σ in the condition (i). The proof is almost the same. Moreover, fromthe “If” part of the proof we observe that, when restricted to polynomials f with degree higher than one, the condition (i) in Theorem 2 can be replaced bythe statement “ G Bf is transitive and the pair ( G Bf , H ) is Q-trivial for any rootstabilizer H ” or the statement “ G Q f is transitive and the pair ( G Q f , H ) is Q-trivialfor any root stabilizer H ”.The counterpart of Corollary 3 in this case is given below: Corollary 4.
Let f be as in Theorem 2 and G ∈ {G Bf , G Q f } . If deg( f ) ≥ andthe conditions (ii) – (iii) in Theorem 2 hold, then the following conditions areequivalent to each other: (i) G = Σ ;(ii) G is doubly transitive;(iii) G is doubly homogeneous;(iv) G is transitive and the pair ( G , H ) is Q -trivial for any root stabilizer H .Proof. The proof is similar to the one of Corollary 3. (cid:117)(cid:116)
Theorem 1 and 2 characterize, for the first time, the the polynomial f with atrivial exponent lattice R f or R Q f with the help of the concept of a Galois-likegroup. The conditions (ii)–(iii) in both theorems can be decided every efficiently(by § § f , equals the symmetry group Σ or not is not available at present. We characterize the polynomials with trivial exponent lattices through the Galoisand the Galois-like groups. Based on the algorithm
IsQtrivial , we extensivelyimprove the main algorithm in [25] proving triviality of the exponent latticeof a generic polynomial (when the polynomial degree is large). In addition, asufficient and necessary condition is given with the help of the concept of aGalois-like group, which turns out to be essential in the study on multiplicativerelations between the roots of a polynomial. Further study on Galois-like groupsseems to be interesting and promising.
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