Separating Variables in Bivariate Polynomial Ideals
aa r X i v : . [ c s . S C ] J un Separating Variables in Bivariate Polynomial Ideals
Manfred Buchacher ∗ , Manuel Kauers † , Gleb Pogudin ‡ Abstract
We present an algorithm which for any given ideal I ⊆ K [ x, y ] finds all elements of I that havethe form f ( x ) − g ( y ), i.e., all elements in which no monomial is a multiple of xy . One of the fundamental problems in computer algebra and applied algebraic geometry is the problemof elimination. Here, we are given a polynomial ideal I ⊆ K [ x , . . . , x n , y , . . . , y m ] and the task is tocompute generators of the ideal I ∩ K [ x , . . . , x n ]. The resulting ideal of K [ x , . . . , x n ] consists of allelements of I that do not contain any terms that are a multiple of any of the variables y i . It is well-known that this problem can be solved by computing a Gr¨obner basis with respect to an elimination orderthat assigns higher weight to terms involving y , . . . , y m than to terms not involving these variables.It is less clear how to use Gr¨obner bases (or any other standard elimination techniques) for findingideal elements that do not contain any terms which are a multiple of certain prescribed terms ratherthan certain prescribed variables. The problem considered in this paper is an elimination problem of thiskind. Here, given an ideal I ⊆ K [ x , . . . , x n , y , . . . , y m ], we are interested in all elements of I that do notinvolve any terms which are multiples of any of the terms x i y j ( i = 1 , . . . , n , j = 1 , . . . , m ). Note that,these are precisely the elements of I which can be written as the sum of a polynomial in x , . . . , x n onlyand a polynomial in y , . . . , y m only, so the problem under consideration is as follows. Problem 1.1 (Separation) .Input
An ideal I ⊆ K [ x , . . . , x n , y , . . . , y m ] ; Output
Description of all f − g ∈ I such that f ∈ K [ x , . . . , x n ] and g ∈ K [ y , . . . , y m ] . At first glance, it may seem that there should be a simple way to solve this problem with Gr¨obnerbases, similarly as for the classical elimination problem. However, we were not able to come up with suchan algorithm. The obstruction seems to be that there is no term order that ranks the term xy higherthan both x and y .We ran into the need for such an algorithm when we tried to automatize an interesting non-standardelimination step which appears in Bousquet-M´elou’s “elementary” solution of Gessel’s walks [9]. Dealingwith certain power series, say u ∈ K [ x ][[ t ]] and v ∈ K [ x − ][[ t ]], she finds polynomials f, g such that f ( u ) − g ( v ) = 0, and then concludes that f ( u ) and g ( v ) must in fact belong to K [[ t ]]. Deriving a pair( f, g ) automatically from known relations among u, v amounts to the problem under consideration.The problem also arises when one wants to compute the intersection of two K -algebras. For example,suppose that for given u, v ∈ K [ t , . . . , t n ] one wants to compute K [ u ] ∩ K [ v ]. This can be done by findingall pairs ( f, g ) such that f ( u ) = g ( v ), i.e., all pairs ( f, g ) with f ( x ) − g ( y ) ∈ h x − u, y − v i ∩ K [ x, y ].See [3, 13] for a discussion of this and similar problems. Definition 1.2.
Let p ∈ K [ x , . . . , x n , y , . . . , y m ] .1. p is called separated if there exist f ∈ K [ x , . . . , x n ] and g ∈ K [ y , . . . , y m ] such that p = f − g .2. p is called separable if there is a q ∈ K [ x , . . . , x n , y , . . . , y m ] such that qp is separated. ∗ [email protected], Institute for Algebra, Johannes Kepler University, Linz, Austria † [email protected], Institute for Algebra, Johannes Kepler University, Linz, Austria ‡ [email protected], Department of Computer Science, Higher School of Economics, Moscow, Russia andLIX, CNRS, ´Ecole Polytechnique, Institut Polytechnique de Paris, France roposition 1.3. Let I be an ideal in K [ x , . . . , x n , y , . . . , y m ] . Then A ( I ) := { ( f, g ) ∈ K [ x , . . . , x n ] × K [ y , . . . , y m ] : f − g ∈ I } is a unital K -algebra with respect to component-wise addition and multiplication and component-wisemultiplication by elements of K . We refer to A ( I ) as the algebra of separated polynomials of I .Proof. We just note that A ( I ) is clearly a K -vector space, and that it is closed under component-wisemultiplication, as for any ( f, g ) , ( f ′ , g ′ ) ∈ A ( I ) we have f − g ∈ I and f ′ − g ′ ∈ I , so ( f − g ) f ′ + g ( f ′ − g ′ ) = f f ′ − gg ′ ∈ I . It is unital, because we always have (1 , ∈ A ( I ).Given ideal generators of I , we want to determine K -algebra generators of A ( I ). This is in generaltoo much to be asked for, because, as shown in Example 5.1, A ( I ) may not be finitely generated. On thepositive side, it is known that A ( I ) is finitely generated if I is a principal ideal in the ring of bivariatepolynomials (see [15]).The main result of the paper is Algorithm 4.3 for computing generators of the algebra A ( I ) for a givenbivariate ideal I ⊆ K [ x, y ]. In particular, it implies that such an algebra is always finitely generated andyields an algorithm to compute a minimal separated multiple of a bivariate polynomial [15, Definition 4.1].An implementation of the algorithm in Mathematica can be found on the website of the second author.The general structure of the algorithm is the following. Every bivariate ideal is the intersection of azero-dimensional ideal and a principal ideal. We solve the separation problem for the zero-dimensionalcase (Section 2) and for the principal case (Section 3) separately. Then we show how to compute theintersection of the resulting algebras in Section 4. We conclude with discussing the case of more thantwo variables in Section 5.In the context of separated polynomials, many deep results have been obtained for some kind of“inverse problem” to the problem considered here, i.e., the study of the shape of factors of polynomialsof the form f ( x ) − g ( y ), see [6, 7, 10, 11, 12, 14, 15] and references therein. We use techniques developedin [10] in our proofs (see Section 3).We assume throughout that the ground field K has characteristic zero and that for a given elementof an algebraic extension of K we can decide whether it is a root of unity. This is true, for example, forevery number field (see Section 3.3).It is an open question whether the assumption on the characteristic of K can be eliminated. Inpositive characteristic, additional phenomena have to be taken into account. For example, separablepolynomials need not be squarefree, as the example ( x + y ) ∈ Z [ x, y ] shows, which is separable because( x + y )( x + y ) = ( x + y ) = x + y . When I ⊆ K [ x, y ] has dimension zero, it is easy to separate variables. In this case, there are nonzeropolynomials p, q with I ∩ K [ x ] = h p i and I ∩ K [ y ] = h q i . Clearly, these univariate polynomials p and q are separated. Also all K [ x ]-multiples of p and all K [ y ]-multiples of q are separated elements of I .An arbitrary pair ( f, g ) ∈ K [ x ] × K [ y ] belongs to A ( I ) if and only if ( f + up, g + vq ) belongs to A ( I )for all u ∈ K [ x ] and v ∈ K [ y ]. In particular, we have ( f, g ) ∈ A ( I ) ⇐⇒ (rem x ( f, p ) , rem y ( g, q )) ∈ A ( I ).It is therefore sufficient to find all pairs ( f, g ) ∈ A ( I ) with deg x f < deg x p and deg y g < deg y q . Thesepairs can be found with linear algebra. Algorithm 2.1.
Input: I ⊆ K [ x, y ] of dimension zero.Output: generators of the K -algebra A ( I ) ⊆ K [ x ] × K [ y ] if I = h i , return { (1 , , ( x, , (0 , , (0 , y ) } . compute p ∈ K [ x ] and q ∈ K [ y ] such that I ∩ K [ x ] = h p i and I ∩ K [ y ] = h q i . make an ansatz h = P deg x p − i =0 a i x i − P deg y q − j =0 b j y j with undetermined coefficients a i , b j . compute the normal form of h with respect to a Gr¨obner basis of I and equate its coefficients to zero. solve the resulting linear system over K for the unknowns a i , b j and let ( f , g ) , . . . , ( f d , g d ) be thepairs of polynomials corresponding to a basis of the solution space. return ( f , g ) , . . . , ( f d , g d ) , ( p, , . . . , ( x deg x p − p, , (0 , q ) , . . . , (0 , y deg y q − q ) . Proposition 2.2.
Algorithm 2.1 is correct. roof. It is clear by construction that all returned elements belong to A ( I ). It remains to show that theygenerate A ( I ) as K -algebra. This is clear if I = h i , because then A ( I ) = K [ x ] × K [ y ]. Now supposethat I = h i and let ( f, g ) ∈ A ( I ). Because of I = h i , we have deg x p, deg y q >
0. Then h p i ⊆ K [ x ] isgenerated as a K -algebra by p, xp, . . . , x deg x p − p . To see this, we just note that, by performing repeatedlydivision by p on a polynomial and the resulting quotients, any u ∈ h p i can be written u = k X i =1 r i p i where r i are polynomials with deg r i < deg p . Hence, h p i is a subset of the algebra generated by p, xp, . . . , x deg x p − p , and clearly, the reverse inclusion holds as well. For the same reason, h q i is gen-erated as K -algebra by q, xq, . . . , x deg x q − q .Hence ( f, g ) can be expressed in terms of the given generators if and only if (rem x ( f, p ) , rem y ( g, q )) canbe expressed in terms of the given generators. Because of deg x (rem x ( f, p )) < deg x ( p ) and deg y (rem y ( g, q )) < deg y ( q ), the pair (rem x ( f, p ) , rem y ( g, q )) is a K -linear combination of ( f , g ) , . . . , ( f d , g d ), as required. Example 2.3.
Consider the -dimensional ideal I = h x y − , y + y + xy + x i . We have I ∩ K [ x ] = h x + x − x − i and I ∩ K [ y ] = h y + y − y − i . Every separated polynomial of I therefore has the form f ( x ) + u ( x )( x + x − x − − g ( y ) − v ( y )( y + y − y − for certain f ( x ) , g ( y ) of degree less than and some u ( x ) , v ( y ) . To find the pairs ( f, g ) , compute thenormal form of h = P i =0 a i x i − P i =0 b j y j with respect to a Gr¨obner basis of I . Taking a degrevlexGr¨obner basis, this gives ( a + a − b ) + ( a − b ) y + ( a + b ) xy + · · · . Equate the coefficients with respect to x, y to zero and solve the resulting linear system for the unknowns a , . . . , a , b , . . . , b . The following pairs of polynomials ( f, g ) correspond to a basis of the solution space: (1 , , ( x − x , y − y ) , ( x , y + y − , ( x + x , − y − y )( x , − y + y + 1) , ( x − x , y − y ) , ( x , y + y − x + x , − y − y ) , ( x , − y ) . These pairs together with the pairs ( x i ( x + x − x − , and (0 , y i ( y + y − y − for i = 0 , . . . , form a set of generators of A ( I ) . For an ideal I ⊆ K [ x, y ] to be zero-dimensional means that its codimension as K -subspace of K [ x, y ]is finite. Note that, in this case, also A ( I ) has finite codimension as K -subspace of K [ x ] × K [ y ]. Since wewill need this feature later, let us record it as a lemma. Lemma 2.4. If I ⊆ K [ x, y ] has dimension zero, then there is a finite-dimensional K -subspace V of K [ x ] × K [ y ] such that the direct sum V ⊕ A ( I ) is equal to K [ x ] × K [ y ] . Moreover, we can compute a basisof such a V , and for every ( f, g ) ∈ K [ x ] × K [ y ] we can compute a ( ˜ f , ˜ g ) ∈ V such that ( f, g ) − ( ˜ f , ˜ g ) ∈ A ( I ) .Proof. Let p, q, ( f , g ) , . . . , ( f d , g d ) be as in Algorithm 2.1. Note that, as a K -vector space, A ( I ) has thebasis { ( f , g ) , . . . , ( f d , g d ) } ∪ { ( x k p,
0) : k ∈ N } ∪ { (0 , y k q ) : k ∈ N } . Using row-reduction, it can be arranged that the f i have pairwise distinct degrees. Note that, all f i arenonzero by the choice of q . Let V be the K -subspace of K [ x ] × K [ y ] generated by the pairs ( x k ,
0) forall k < deg x ( p ) which are not the degree of some f i and the pairs (0 , y k ) for all k < deg y ( q ). We have V ⊕ A ( I ) = K [ x ] × K [ y ].Given ( f, g ) ∈ K [ x ] × K [ y ], we compute (rem x ( f, p ) , rem y ( g, q )), and then eliminate all terms from thefirst component whose exponent is the degree of an f i . The resulting pair ( ˜ f , ˜ g ) is an element of V with( f, g ) − ( ˜ f , ˜ g ) ∈ A ( I ). 3 Principal Ideals
We now consider the case where I = h p i is a principal ideal of K [ x, y ]. If p ∈ K [ x ] ∪ K [ y ], the algebra A ( I ) of separated polynomials is finitely generated, as we have seen in the proof of Proposition 2.2. Itwas shown in [15, Theorem 4.2] that, if p is separable, there is a separated multiple f ( x ) − g ( y ) of p thatdivides any other separated multiple of it. We refer to f ( x ) − g ( y ) as the minimal separated multiple of p .Moreover, [15, Theorem 2.3] implies that if p K [ x ] ∪ K [ y ], then ( f, g ) is an algebra generator for A ( I ).We note that, [15, Theorem 2.3] was reproven in [8], and generalized further in [1, 19]. The proof of [15,Theorem 4.2] was not constructive. In the following we provide a criterion that allows to decide if p isseparable, and if it is, to compute its minimal separated multiple.Our criterion is based on considering the highest graded component of the polynomial with respect toa certain grading. The separability of the highest component is a necessary but not a sufficient conditionfor the separability of a polynomial itself. Surprisingly, there is a weaker converse, that is, the minimalseparated multiple of the highest component is equal to the highest component of the minimal separatedmultiple of p if the latter exists (see Theorem 3.5). This allows us to reduce the problem for a generalnot necessarily homogeneous polynomial to the same problem for a homogeneous polynomial (which issolved in Section 3.1) and solving a linear system. The resulting algorithm is presented in Section 3.3.Since the case p ∈ K [ x ] ∪ K [ y ] is trivial, for the rest of the section, we assume that p ∈ K [ x, y ] \ ( K [ x ] ∪ K [ y ]). Definition 3.1.
1. A function ω from the set of monomials in x and y to R is called a weight function if there exist ω x , ω y ∈ Z > such that ω ( x i y j ) = ω x i + ω y j for every i, j ∈ Z ≥ .2. Two weight functions are considered to be equivalent if they differ by a constant non-zero factor.3. For a weight function ω and a nonzero polynomial p ∈ K [ x, y ] , ω ( p ) is defined to be the maximumof the weights of the monomials of p .4. For a weight function ω and a polynomial p ∈ K [ x, y ] , we define the ω -leading part of p (denotedby lp ω ( p ) ) as the sum of the terms of p of weight ω ( p ) . In this subsection, we consider the case of p being homogeneous with respect to some weight function ω , that is, lp ω ( p ) = p . Proposition 3.2.
Let ω be a weight function, and let p ∈ K [ x, y ] \ ( K [ x ] ∪ K [ y ]) satisfy lp ω ( p ) = p . Then p is separable if and only if1. p involves a monomial only in x , and2. all the roots of p ( x, in the algebraic closure K of K are distinct and the ratio of every two of themis a root of unity.Moreover, if p is separable and N is the minimal number such that the ratio of every pair of roots of p ( x, is an N -th root of unity, then the weight of the minimal separated multiple of p is N ω x .Proof. Assume that p is separable, and let P be a separated multiple. Replacing P with lp ω ( P ) ifnecessary, we will further assume that P = lp ω ( P ). Since P / ∈ K [ x ] ∪ K [ y ] and is separated, P involves amonomial in x only, and hence, so does p .Since P is ω -homogeneous and separated, it is of the form ax m − by n for some a, b ∈ K \ { } , so p ( x, | ax m − b . All roots of the latter are distinct and the ratio of each of them is an m -th root ofunity. Hence, the same is true for p ( x, N be as in the statement of the proposition, and γ ∈ K be a root of p ( x, ω -homogeneous Puiseux polynomial P := x N − γ N y Nω x /ω y . We perform Euclidean division of P by p over the field F of Puiseux series in y over K . This will yield arepresentation P = qp + r , where q and r are also ω -homogeneous. Since P ( x,
1) is divisible by p ( x, r ( x,
1) = 0. However, the ω -homogeneity of r implies that each of its coefficients with respect4o x is a Puiseux monomial in y . Thus, r = 0. Next, assume that N ω x /ω y is not an integer. Then thereis an automorphism σ of the Galois group of F over K ( y ) that moves y Nω x /ω y . Then p | P − σ ( P ) ∈ F, which is impossible. Therefore, P is a separated polynomial divisible by p of weight N ω x .Of course, because of symmetry, the statements of Proposition 3.2 also hold for y instead of x . We will start with a necessary condition for p being separable. Lemma 3.3.
Let p ∈ K [ x, y ] \ ( K [ x ] ∪ K [ y ]) be separable.1. There exists a unique (up to a constant factor) weight function ω such that lp ω ( p ) involves at leasttwo monomials.2. The polynomial lp ω ( p ) is separable.Proof. Let q ∈ K [ x, y ] \ { } be such that qp is separated. Let deg x qp = m and deg y qp = n . Define ω ( x i y j ) = ni + mj . If lp ω ( p ) contains only one monomial, then every monomial in lp ω ( qp ) is divisible byit. This is impossible since lp ω ( qp ) involves both x m and y n .To prove the uniqueness, assume that there are two nonequivalent weight functions ω and ω withthis property. Since lp ω i ( qp ) = lp ω i ( q ) lp ω i ( p ) for i = 1 ,
2, we have that both lp ω ( qp ) and lp ω ( qp )contain at least two monomials. However, the only monomials of qp that can appear in the leading partare x m and y n , and there is a unique weight function so that they have the same weight.The second claim of the lemma follows from lp ω ( q ) lp ω ( p ) = lp ω ( qp ).There is an analogous version of Lemma 3.3 with the lowest homogeneous part in place of the leadinghomogeneous part. However, even when both the lowest and the leading homogeneous part are separable,the whole polynomial need not be separable, as the following example shows. Example 3.4.
For p = ( x + x y + xy + y ) + y ∈ Q [ x, y ] , the relevant weightfunction for the leading homogeneous part as in Lemma 3.3 is given by ω x = ω y = 1 .It leads to the leading homogeneous part x + x y + xy + y . Analogously, the relevantweight function for the lowest homogeneous part is given by ω x = 2 , ω y = 3 . It leadsto the lowest homogeneous part x + y . Both the leading and the lowest homogeneous part are separable.We claim that p is not separable. xy Let ω be the weight function defined by ω ( x i y j ) = 2 i + 3 j , so that the lowest homogeneous part of p is x + y (weight 6), and the next-to-lowest part is x y (weight 7). With respect to ω , any separatedpolynomial involving both variables only consists of homogeneous parts ax n + by m whose weight n = 3 m is a multiple of 6.Assume that p is separable and let q ∈ Q [ x, y ] \{ } be such that qp is separated. Write q = q + q + · · · ,where q , q , . . . are the lowest, the next-to-lowest, etc. homogeneous parts of q with respect to ω . Thelowest homogeneous part of pq is then q ( x + y ) , and since it must be separated and involve both variables,we have ω ( q ) = 0 mod 6 .Because of ω ( q x y ) = ω ( q ( x + y )) + 1 = 1 mod 6 , none of the terms of q x y can appear in qp , sothey must all be canceled by something. We must therefore have ω ( q ) = ω ( q )+1 and q x y + q ( x + y ) =0 . This implies that x + y divides q , which in turn implies that the lowest homogeneous part q ( x + y ) of pq has a multiple factor. On the other hand, q ( x + y ) = ax n + by m for some a, b = 0 , and everysuch polynomial is squarefree. This is a contradiction. The main result of the section is the following “partial converse” of Lemma 3.3.
Theorem 3.5.
Let p ∈ K [ x, y ] \ ( K [ x ] ∪ K [ y ]) be a separable polynomial. Let ω be the weight functiongiven by Lemma 3.3, and let P be the minimal separated multiple of p . Then lp ω ( P ) is the minimalseparated multiple of lp ω ( p ) . Before proving the theorem, we will establish some combinatorial tools for dealing with divisors ofseparated polynomials extending the results of Cassels [10].5 otation 3.6.
Consider a separated polynomial f ( x ) − g ( y ) with deg x f = m and deg y g = n , where m, n > , and a weight function ω ( x i y j ) = in + jm . We introduce a new variable t and consider twoauxiliary equations f ( x ) = t and g ( y ) = t. We solve these equations with respect to x and y in K ( t ) , the algebraic closure of K ( t ) . Let the solutions be α , . . . , α m − and β , . . . , β n − , respectively. Then every element π of Gal( K ( t ) / K ( t )) , the Galois groupof K ( t ) over K ( t ) , acts on Z m × Z n by π ( i, j ) := ( i ′ , j ′ ) ⇐⇒ ( π ( α i ) , π ( β j )) = ( α i ′ , β j ′ ) . Let G ⊆ S m × S n be the group of permutations induced on Z m × Z n by this action. Notation 3.7.
For a subset T ⊆ Z m × Z n , and ( i, j ) ∈ Z m × Z n , we introduce T i, ∗ := { k | ( i, k ) ∈ T } and T ∗ ,j := { k | ( k, j ) ∈ T } . Lemma 3.8.
Let T ⊆ Z m × Z n be a G -invariant subset. Then | T , ∗ | = | T , ∗ | = . . . = | T m − , ∗ | and | T ∗ , | = | T ∗ , | = . . . = | T ∗ ,n − | .Proof. We show that | T , ∗ | = | T , ∗ | , the rest is analogous. First, we observe that f ( x ) − t is irreducibleover K ( t ). If it was not, it would be reducible over K [ t ] due to Gauss’s lemma. The latter is impossiblebecause f ( x ) − t is linear in t and does not have factors in K [ x ]. The irreducibility of f ( x ) − t impliesthat its Galois group acts transitively on the roots. In particular, there exists π ∈ Gal( K ( t ) / K ( t )) suchthat π ( α ) = α . Hence, π maps T , ∗ to T , ∗ , and we have | T , ∗ | | T , ∗ | . The reverse inequality isanalogous. Lemma 3.9 (cf. [10, p. 9-10]) . Let T ⊆ Z m × Z n be a G -invariant subset. There exists a divisor p of f ( x ) − g ( y ) , unique up to a multiplicative constant, such that T = { ( i, j ) ∈ Z m × Z n | p ( α i , β j ) = 0 } . (1) Proof. Existence.
Let T , ∗ = { j , . . . , j s } . Since f ( α ) = t , we have K ( α ) ⊇ K ( t ), so every elementof Gal( K ( t ) / K ( α )) leaves T invariant. If α is fixed, then β j , . . . , β j s are permuted. Therefore, thepolynomial ( y − β j )( y − β j ) . . . ( y − β j s ) is invariant under the action of Gal( K ( t ) / K ( α )). Hence, by thefundamental theorem of Galois theory, it is a polynomial in K ( α )[ y ]. Since, by construction, it divides f ( α ) − g ( y ) over K ( α ), and α and y are algebraically independent, it in fact belongs to K [ α , y ].Replacing α by x , we find a polynomial p ∈ K [ x, y ], which divides f ( x ) − g ( y ) in K [ x, y ].Let ( i, j ) ∈ Z m × Z n . Since Gal( K ( t ) / K ( t )) acts transitively on the roots of f ( x ) − t (see the proof ofLemma 3.8), there is an automorphism π with π ( α i ) = α . Let β j ′ = π ( β j ). We then have p ( α i , β j ) = 0 ⇐⇒ p ( α , β j ′ ) = 0 ⇐⇒ j ′ ∈ T , ∗ ⇐⇒ ( i, j ) ∈ T. Uniqueness.
It remains to prove that p is unique up to a multiplicative constant. Assume that ˜ p isanother divisor of f ( x ) − g ( y ) such that ˜ p ( α i , β j ) = 0 for all ( i, j ) ∈ T . The same argument which provedthat p is a divisor of f ( x ) − g ( y ) applies to show that p is a divisor of ˜ p in K [ x, y ], and vice versa. Hence,they only differ by a multiplicative constant. Lemma 3.10.
Let T ⊆ Z m × Z n be a G -invariant subset. The unique factor p corresponding to T ⊆ Z m × Z n (see Lemma 3.9) is separated if and only if ∀ i, j ∈ Z m : ( T i, ∗ ∩ T j, ∗ = ∅ ) or ( T i, ∗ = T j, ∗ ) (2) Proof.
Assume that T satisfies (2), and let T , ∗ = { j , . . . , j s } . Consider the corresponding polynomial p constructed in the proof of Lemma 3.9, which is of the form p ( x, y ) = y s + a s − ( x ) y s − + · · · + a ( x ) , where, for every 0 i < s and 0 j < m , a i ( α j ) is (up to sign) the s − i -th elementary symmetricpolynomial in { β k | k ∈ T j, ∗ } .Since p | f ( x ) − g ( y ), we have lp ω ( p ) | lp ω ( f ( x ) − g ( y )) = ax m − by n , with a, b ∈ K \ { } . Hence, y s belongs to lp ω ( p ), and so ω ( a i ( x ) y i ) ω ( y s ) = ms for all i ∈ { , . . . , s − } , This impliesdeg x a i ( x ) ms − min = ( s − i ) mn .
01 12 23 3 • • • •• •• • x + y + 1separated • • •• •• • • x − y − • • • •• •• •• • •• •• • • x − ( y + 1) separatedFigure 1: The factors of x − ( y + 1) in Q [ x, y ] and the sets T ⊆ Z corresponding to them.Since T is the disjoint union of the T i, ∗ ’s and of the T ∗ ,j ’s, respectively, whose cardinality, byLemma 3.8, does not depend on i and j , and T , ∗ , by definition, consists of s elements, we find that ms = | T | = n | T ∗ ,j | , in particular ℓ := | T ∗ ,j | = msn . Hence there exist 0 = i < i < . . . < i ℓ < m suchthat j ∈ T i , ∗ ∩ . . . ∩ T i ℓ , ∗ and so, by (2), T i , ∗ = . . . = T i ℓ , ∗ . This shows that the polynomial a j ( x ) − a j ( α )has at least ℓ pairwise distinct roots, α i , . . . , α i ℓ , while it has degree less than ℓ for 0 < j < s . Hence, itis the zero polynomial, and a j ( x ) is a constant (which we denote by a j ). Therefore, p is separated andof the form p ( x, y ) = f ( x ) − g ( y ) with f ( x ) = a ( x ) and g ( y ) = − ( y s + a s − y s − + · · · + a y ).To prove the other implication, let p ( x, y ) = f ( x ) − g ( y ) be a separated factor of f ( x ) − g ( y ). It issufficient to show that ( i, j ) , ( i ′ , j ) , ( i, j ′ ) ∈ T = ⇒ ( i ′ , j ′ ) ∈ T. Indeed, ( i, j ) , ( i ′ , j ) ∈ T implies that f ( α i ) = f ( α i ′ ), so that f ( α i ) − g ( β j ′ ) = 0 implies that f ( α i ′ ) − g ( β j ′ ) = 0, i.e. ( i ′ , j ′ ) ∈ T .Lemma 3.10 motivates the following definition. Definition 3.11.
1. A subset T ⊆ Z m × Z n is called separated if it satisfies (2) , that is ∀ i, j ∈ Z m : ( T i, ∗ ∩ T j, ∗ = ∅ ) or ( T i, ∗ = T j, ∗ ) .
2. The intersection of all separated subsets containing T ⊆ Z m × Z n is called the separated closure of T and denoted by T sep . Notice that the separated closure is separated. Example 3.12.
1. Let f ( x ) = x and g ( y ) = y + 2 y + 1 . The group of permutations on pairsof roots of f ( x ) − t and g ( y ) − t is generated by ((0123) , (0123)) , ((0321) , (03)(12)) and ( id, (02)) .According to f ( x ) − g ( y ) having two separated irreducible factors, x − y − and x + y + 1 , wefind that there are two orbits, each of them forming a separated set (Figure 1).2. Let f ( x ) − g ( y ) = x − y . Let t / ∈ C ( t ) be any 6th root of t , and let ǫ be a primitive 6th root ofunity. Then the polynomials f ( x ) − t and g ( y ) − t have the same roots, namely: α i = β i = ǫ i t / , i ∈ { , . . . , } . The Galois group of C ( t ) permutes these elements cyclically, so the induced action on Z is generatedby ((012345) , (012345)) . Figure 2 shows the sets T for the various factors of x − y . Observe that T is separated if and only if the corresponding factor is separated. Observe also that multiplyingtwo factors corresponds to taking the union of the corresponding sets T . Lemma 3.13.
Let T ⊆ Z m × Z n be invariant with respect to G ⊆ S m × S n . Then T sep is also G -invariant.Proof. Let π = ( σ, τ ) ∈ S m × S n , and let S ⊆ Z m × Z n be a separated set. Since π ( S ) i, ∗ = τ ( S σ ( i ) , ∗ ), wefind that π ( S ) is separated as well.Assume that T sep is not G -invariant, that is, there exists a π ∈ G such that π ( T sep ) = T sep . As wehave shown, π ( T sep ) is separated, hence so is S := T sep ∩ π ( T sep ). Observe that, since π ( T sep ) = T sep , S ( T sep . Since T is G -invariant, T ⊆ π ( T sep ), so T ⊆ S . This contradicts the minimality of T sep . Proof of Theorem 3.5.
We use Notation 3.6 with K ( t ) being identified with a subfield of the field F ofPuiseux series in t − over K . Let α , . . . , α m − and β , . . . , β n − denote the roots of f ( x ) − t and g ( y ) − t and α , . . . , α m − and β , . . . , β n − their highest degree terms. Observe that the highest degree terms are7
01 12 23 34 45 5 • • • • • • x − y separated • • • • • • x + y separated • • • • • •• • • • • • x − y separated • • • •• • • •• • • • x + xy + y not separated • • • • • •• • • •• • • •• • • • x − y separated • • • • • •• • • •• • • •• • • • x +2 x y +2 xy + y not separated • • • • • •• • • • • •• • • •• • • •• • • • x + x y − xy − y not separated • • • • •• •• • • • • x − xy + y not separated • • • • • •• • • • •• •• • • • • x − x y +2 xy − y not separated • • • • • •• • • • •• •• • • • • x + y separated • • • • • •• • • • • •• • • • •• •• • • • • x − x y + xy − y not separated • • • •• • • •• • • •• • • • •• •• • • • • x + x y + y not separated • • • • • •• • • •• • • •• • • •• • • • •• •• • • • • x − x y + · · · − y not separated • • • • • •• • • •• • • •• • • •• • • • •• •• • • • • x + x y + · · · + y not separated • • • • • •• • • • • •• • • •• • • •• • • •• • • • •• •• • • • • x − y separated Figure 2: The factors of x − y in Q [ x, y ] and the sets T ⊆ Z corresponding to them. For the unseparatedcases, we highlight one choice of two incompatible rows.proportional to t /n and t /m , and hence they are the roots of lp ω ( f ( x )) − t and lp ω ( g ( y )) − t , respectively.We define T = { ( i, j ) ∈ Z m × Z n | p ( α i , β j ) = 0 } ,T = { ( i, j ) ∈ Z m × Z n | lp ω ( p )( α i , β j ) = 0 } . If lp ω ( P ) were not the minimal separated multiple of lp ω ( p ), by Lemma 3.10, we would have T sep (Z m × Z n . Therefore, it is sufficient to show that T sep = Z m × Z n .Since p ( α i , β j ) = 0 = ⇒ lp ω ( p )( α i , β j ) = 0 , we have T ⊆ T . By assumption, P is the minimal separated multiple of p , so, by Lemma 3.13, T sep = Z m × Z n . Since T sep ⊆ T sep , this implies that T sep = Z m × Z n , and finishes the proof. The algorithm for finding a generator of the algebra of separated polynomials of a principal ideal h p i is based on the results above. First, it uses Theorem 3.5 to reduce the situation to a homogeneouspolynomial for a suitable grading, then, it uses Proposition 3.2 to find a degree bound for the minimalseparated multiple, and finally, it uses linear algebra to determine if such a multiple exists. Algorithm 3.14.
Input: p ∈ K [ x, y ] \ ( K [ x ] ∪ K [ y ]) .Output: a ∈ K [ x ] × K [ y ] such that K [ a ] = A ( h p i ) . The algorithm returns a = (1 , iff A ( h p i ) ∼ = K . let ω x , ω y ∈ N be maximal such that p contains monomials x ω y y and x y ω x . Such parameters existbecause p is not univariate. set h = lp ω ( p ) with ω ( x i y j ) := ω x i + ω y j . if h does not contain x ω y , return (1 , . let { ζ , . . . , ζ m } ⊆ K be the roots of h ( x, ∈ K [ x ] . If any of them is not a simple root, return (1 , . let N ∈ N be minimal such that ( ζ i /ζ j ) N = 1 for all i, j . If no such N exists, return (1 , . make an ansatz f = N X i =0 a i x i , g = Nω x /ω y X j =0 b j y j , compute rem x ( f − g, p ) in K ( a , . . . , a N , b , . . . , b Nω x /ω y , y )[ x ] . The result of the reduction belongs to K [ a , . . . , a N , b , . . . , b Nω x /ω y , y, x ] because the leading coefficient of p is in K . equate the coefficients of rem x ( f − g, p ) with respect to x, y to zero and solve the resulting linear systemfor the unknowns a i , b j . if there is a nonzero solution, return the corresponding pair ( f, g ) , otherwise return (1 , . When K is a number field, Step 5 can be carried out as follows: for each ratio ζ i /ζ j , one should checkwhether the minimal polynomial of this ratio over Q is a cyclotomic polynomial Φ n and, if yes, returnsuch n . This check can be performed using a bound from [18, Theorem 15] that yields the upper boundon n based on the degree of the polynomial. Proposition 3.15.
Algorithm 3.14 is correct.Proof.
The algorithm consists of an application of the results of the previous section and a handling ofdegenerate cases not covered by these results. In Steps 3–5, it is correct to return (1 ,
1) in the indicatedsituations because Proposition 3.2 implies that h is not separable in these cases, which in combinationwith Lemma 3.3 implies that p is not separable either.By Proposition 3.2, when h has a separated multiple at all, it has one of weight N ω x , and by The-orem 3.5, when p has a separated multiple at all, it also has one of weight N ω x . Therefore, if p has aseparated multiple, it will have one of the shape set up Step 6. For f − g to be a separated multipleof p is equivalent to rem x ( f − g, p ) = 0, which we can safely view as univariate division with respect tothe variable x because the leading coefficient of p with respect to x does not contain y (nor any of theundetermined coefficients). It is checked in Step 7 whether there is a way to instantiate the undeter-mined coefficients in such a way that this remainder becomes zero. If so, any such way translates into aseparated multiple, and by [15, Theorem 2.3], it is a generator of A ( I ). If there is no non-zero solution,it is correct to return (1 , The case of an arbitrary ideal I ⊆ K [ x, y ] is reduced to the two cases discussed in Sections 2 and 3. Everyideal I ⊆ K [ x, y ] can be written as I = T ki =1 P i , where the P i ’s are primary ideals. Unless I = { } or I = h i , these primary ideals have dimensions zero or one. Primary ideals in K [ x, y ] of dimension 1 mustbe principal ideals, because dim( P i ) = 1 together with Bezout’s theorem implies that P i cannot containany elements p, q with gcd( p, q ) = 1, and then P i being primary implies that P i is generated by somepower of an irreducible polynomial.The intersection of zero-dimensional ideals is zero-dimensional and the intersection of principal idealsis principal, so there exists a zero-dimensional ideal I and a principal ideal I such that I = I ∩ I .These ideals are obtained as the intersections of the respective primary components of I . When I = h i or I = h i , we have I = I or I = I , respectively, and are in one of the cases already considered. Assumenow that I , I are both different from h i .In order to use the results of Section 3, we have to make sure that the generator of I contains bothvariables. If this is not the case, say if I = h h i for some h ∈ K [ x ] \ K , then the separated polynomialsin I are precisely the elements of I ∩ K [ x ]. If p is such that h p i = I ∩ K [ x ], then the pairs ( x i p,
0) for i = 0 , . . . , deg x p − A ( I ) (see the proof of Proposition 2.2), so this case is settled.Therefore, from now on we assume that the generator of I contains both the variables.We can compute generators of the algebra A ( I ) ⊆ K [ x ] × K [ y ] of separated polynomials in I asdescribed in Section 2 and a generator of the algebra A ( I ) ⊆ K [ x ] × K [ y ] of separated polynomials in I as described in Section 3. Clearly, the algebra A ( I ) ⊆ K [ x ] × K [ y ] of separated polynomials in I is A ( I ) = A ( I ) ∩ A ( I ). It thus remains to compute generators for this intersection. In order to do so,we will exploit that the codimension of A ( I ) as K -subspace of K [ x ] × K [ y ] is finite (Lemma 2.4), andthat A ( I ) = K [ a ] for some a ∈ K [ x ] × K [ y ]. We have to find all polynomials p such that p ( a ) ∈ A ( I ).9olynomials p with a prescribed finite set of monomials can be found with the help of Lemma 2.4 asfollows. Algorithm 4.1.
Input: a ∈ K [ x ] × K [ y ] , A ( I ) and V as in Lemma 2.4, and a finite set S = { s , . . . , s m } ⊆ N .Output: a K -vector space basis of the space of all polynomials p with p ( a ) ∈ A ( I ) such that p involvesonly monomials with exponents in S . for i = 1 , . . . , m , compute r i ∈ V such that a s i − r i ∈ A ( I ) compute a basis B of the space of all ( c , . . . , c m ) ∈ K m with c r + · · · + c m r m = 0 for every element ( c , . . . , c m ) ∈ B , return c t s + · · · + c m t s m . Proposition 4.2.
Algorithm 4.1 is correct.Proof.
If ( c , . . . , c m ) ∈ K m is such that P mi =1 c i a s i ∈ A ( I ), then P mi =1 c i r i ∈ A ( I ), and since r i ∈ V forall i and A ( I ) ∩ V = { } , we have P mi =1 c i r i = 0. Therefore ( c , . . . , c m ) is among the vectors computedin step 2, so the algorithm does not miss any solutions. Conversely, if ( c , . . . , c m ) ∈ K m is such that P mi =1 c i r i = 0, then P mi =0 c i a s i = P mi =0 c i ( a s i − r i ) ∈ A ( I ), so the algorithm does not return any wrongsolutions.To find a set of generators of A ( I ) ∩ A ( I ), we apply Algorithm 4.1 repeatedly. First call it with S = { , . . . , dim V +1 } . Since | S | > dim V , the output must contain at least one nonzero polynomial p . If d is its degree, we can restrict the search for further generators to subsets S of N \ d N , because when q issuch that q ( a ) ∈ A ( I ), then we can subtract a suitable linear combination of powers of p to remove from q all monomials whose exponents are multiples of d . When d = 1, we have A ( I ) ∩ A ( I ) = K [ a ] and aredone. Otherwise, N \ d N is still an infinite set, so we can choose S ⊆ N \ d N with | S | > dim V and callAlgorithm 4.1 to find another nonzero polynomial p , say of degree d . The search for further generatorscan be restricted to polynomials consisting of monomials whose exponents belong to N \ ( d N + d N ). Wecan continue to find further generators of degrees d , d , . . . with d i ∈ N \ ( d N + · · · + d i − N ) for all i .Since the monoid ( N , +) has the ascending chain condition, this process must come to an end.The end is clearly not reached as long as g := gcd( d , . . . , d m ) = 1, because then N \ g N is aninfinite subset of N \ ( d N + · · · + d m N ). Once we have reached g = 1, it is well known [2, 17] that N \ ( d N + · · · + d m N ) is a finite set, and there are algorithms [5] for computing its largest element (knownas the Frobenius number of d , . . . , d m ). We can therefore constructively decide when all generators havebeen found.Putting all steps together, our algorithm for computing the separated polynomials in an arbitraryideal of K [ x, y ] works as follows. We use the notation h d , . . . , d m i for the submonoid d N + · · · + d m N generated by d , . . . , d m in N . Algorithm 4.3.
Input: an ideal I ⊆ K [ x, y ] , given as a finite set of ideal generatorsOutput: a finite set of generators for the algebra A ( I ) of separated polynomials of I if dim I = 0 , call Algorithm 2.1, return the result. compute a zero-dimensional ideal I and a principal ideal I = h h i with I = I ∩ I (for example,using Gr¨obner bases [4] and the remarks at the beginning of this section). if h ∈ K [ x ] , compute p such that h p i = I ∩ K [ x ] , return the pairs ( x i p, for i = 0 , . . . , deg x p − .Likewise if h ∈ K [ y ] . call Algorithm 2.1 to get generators of A ( I ) , and let V be as in Lemma 2.4. call Algorithm 3.14 to get an a ∈ K [ x ] × K [ y ] with A ( I ) = K [ a ] . If A ( I ) ∼ = K , return (1 , . G = ∅ , ∆ = ∅ . while gcd(∆) = 1 , do: select a set S ⊆ N \ h ∆ i with | S | > dim V and call Algorithm 4.1 to find a nonzero polynomial p with p ( a ) ∈ A ( I ) consisting only of monomials with exponents in S . G = G ∪ { p } , ∆ = ∆ ∪ { deg x p } call Algorithm 4.1 with S = N \ h ∆ i (which is now a computable finite set) and add the resultingpolynomials to G . return G An implementation of the algorithm in Mathematica can be found on the website of the second author.Incidentally, the algorithm also shows that A ( I ) is always a finitely generated K -algebra.10 xample 4.4. For the ideal I = h ( x − xy + y )( x − xy − , ( x − xy + y )( y − x y − i we have I = h x − xy − , y − x y − i and I = h x − xy + y i . Algorithm 2.1 yields a somewhatlengthy list of generators for A ( I ) from which it can be read off that a suitable choice for V is the K -vector space generated by (0 , y i ) for i = 0 , . . . , . In particular, dim V = 9 . Algorithm 3.14 yields A ( I ) = K [( x , − y )] .Making an ansatz for a polynomial p of degree at most such that p ( a ) ∈ A ( I ) , we find a solutionspace of dimension 7. Its lowest degree element is t − t , giving rise to the element ( x − x , y − y ) of A ( I ) ∩ A ( I ) . If we discard the other solutions and continue with the next iteration, we search forpolynomials p whose support is contained { x s : s ∈ S } for S = { , , , , , , , , , } . Again, thesolution space turns out to have dimension 7. The lowest degree element is now t − t + 17 . Since gcd(4 ,
5) = 1 , we can exit the while loop. In step 10 of the algorithm, we get S = { , , , , , } , andthis exponent set leads to a solution space of dimension three, generated by the polynomials t − t , t − t + 458 , and t − t + 184564 . The resulting generators of A ( I ) = A ( I ) ∩ A ( I ) are therefore the pairs p (( x , − y )) where p runs through the five polynomials found by the algorithm. It is a natural question whether anything more can be said about the case of several variables. Incidentally,a multivariate version would be needed in order to solve the combinatorial problem that motivated thisresearch in the first place.Algorithm 2.1 for bivariate zero-dimensional ideals works in the same way for zero-dimensional idealsof K [ x , . . . , x n , y , . . . , y m ] for arbitrary n, m . Also Lemma 2.4 generalizes without problems. We believethat with some further work, our results for principal ideals can also be generalized to the case of severalvariables. However, in general, not every polynomial ideal with more than two variables is the intersectionof a principal ideal and a zero-dimensional ideal, so the route taken in Section 4 is blocked. Also, as thenext example shows we cannot expect an algorithm that finds the algebra of separated polynomials foran arbitrary ideal I ⊆ K [ x , . . . , x n , y , . . . , y m ], since it does not need to be finitely generated. Example 5.1 ( A ( I ) is not necessarily finitely generated) . It is shown in [16, Example 1.3] that thealgebra R := C [ t , t , t ] ∩ C [ t , t − t ] ⊂ C [ t , t ] is not finitely generated. Consider the ideal I = h x − t , x − t , x − t ,y − t , y − ( t − t ) i ∩ C [ x , x , x , y , y ]= h x − y , − x + x y − y y , x − y − x y + y i . We claim that A ( I ) ∼ = R as C -algebras, implying that A ( I ) is not finitely generated. We show that φ : A ( I ) → R defined by φ ( f, g ) = f ( t , t , t ) is an isomorphism: • φ is well-defined (the image is contained in R ⊆ C [ t , t , t ] ). To see this, note that, ( f, g ) ∈ A ( I ) means f − g ∈ I , which by definition of I means f ( t , t , t ) = g ( t , t − t ) . Therefore, f ( t , t , t ) ∈ C [ t , t , t ] ∩ C [ t , t − t ] = R . • φ is surjective. For every p ∈ R there exist polynomials f, g with p = f ( t , t , t ) = g ( t , t − t ) .By definition of I we have f ( x , x , x ) − g ( y , y ) ∈ I , hence ( f, g ) ∈ A ( I ) . Now φ ( f ) = p , so p isin the image of φ . • φ is injective. This follows from I ∩ C [ y , y ] = { } . It would still make sense to ask for an algorithm that decides whether A ( I ) is nontrivial. We do nothave such an algorithm, but being able to solve the problem in the bivariate case gives rise to a necessarycondition. 11 roposition 5.2. Let ξ : K [ x , . . . , x n ] → K [ x ] and η : K [ y , . . . , y m ] → K [ y ] be two homomorphisms, and let I ⊆ K [ x , . . . , x n , y , . . . , y m ] be an ideal such that I ∩ K [ y , . . . , y m ] = { } and (id ⊗ η )( I ) ∩ K [ x , . . . , x n ] = { } . If the algebra of separated polynomials of I is non-trivial, then so is the algebra of separated polynomialsof J := ( ξ ⊗ η )( I ) ⊆ K [ x, y ] .Proof. Let ( f, g ) be an arbitrary, non-constant element of A ( I ). If ( ξ ( f ) , η ( g )) ∈ A ( J ) were a K -multipleof (1 , f − η ( g ) were an element of (id ⊗ η )( I ) ∩ K [ x , . . . , x n ], and hence, by ourassumption, that f itself were a constant. So f − g ∈ I ∩ K [ y , . . . , y m ], and hence, by assumption, g = f is a constant as well, contradicting that ( f, g ) is not a constant.The examples below show different reasonable choices for homomorphisms ξ and η . Example 5.3.
Consider the polynomial p = x + xy y + y + y . Let ξ = id and let η be defined by η ( y ) = y , η ( y ) = 2 . Notice that η is just the evaluation of y at . Then ( ξ ⊗ η )( p ) = x + 2 xy + y + 4 ,a polynomial that is not separable. Hence p is not separable. Example 5.4.
Consider the polynomial p = x + xy + y + y . We cannot use the same strategy as inthe previous example because any evaluation of y or y results in a separable polynomial. Nevertheless,the homomorphism defined by ξ ( x ) = x , η ( y ) = y , and η ( y ) = y maps p to ( ξ ⊗ η )( p ) = x + xy + 2 y ,a polynomial which is not separable. So p is not separable either. Acknowledgements.
We thank Erhard Aichinger and Josef Schicho for sharing their thoughts on thetopic and for providing pointers to the literature. We also thank the referees for their careful readingand their valuable suggestions. MB was supported by the Austrian FWF grant F5004. Part of this workwas done during the visit of MB to HSE University. MB would like to thank the Faculty of ComputerScience of HSE for its hospitality. MK was supported by the Austrian FWF grants F5004 and P31571-N32. GP was supported by NSF grants CCF-1564132, CCF-1563942, DMS-1853482, DMS-1853650, andDMS-1760448, by PSC-CUNY grants
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