Dependence of Homogeneous Components of Polynomials with Small Degree of Poisson Bracket
aa r X i v : . [ m a t h . A C ] J a n DEPENDENCE OF HOMOGENEOUS COMPONENTS OF POLYNOMIALSWITH SMALL DEGREE OF POISSON BRACKET
DARIA HOLIK, MAREK KARAŚ
Abstract.
Let
F, G ∈ C [ x , . . . , x n ] be two polynomials in n variables x , . . . , x n over thecomplex numbers field C . In this paper, we prove that if the degree of the Poisson bracket [ F, G ] is small enough then there are strict constraints for homogeneous components of thesepolynomials (see Section 4). We also prove that there is a relationship between the homogeneouscomponents of the polynomial F of degrees deg F − and deg F − (see Section 6) as well someresults about divisibility of the homogeneous component of degree deg F − (see Section 5).Moreover we propose, possibly an appropriate, reformulation of the conjecture of Yu regardingthe estimation of the Poisson bracket degree of two polynomials (see Conjecture 1.1). introduction The research of this paper was inspired by the result of two papers. One of them is the resultof the paper
Polynomial mappings with Jacobian determinant of bounded degree due to F. LeonPritchard [12] and the second one is the result from the paper
Degree estimate for commutators due to Vesselin Drensky and Jie-Tai Yu [1] (see also [11] and [8]).In his paper, Pritchard considered the following problem. For a given polynomial mapping ( F, G ) : C → C with small degree of the Jacobian determinant, i.e. with small degree of thepolynomial ∂F∂x ∂G∂y − ∂F∂y ∂G∂x , where F, G ∈ C [ x, y ] are there some relations between homogeneouscomponents of F and G . The main result of the paper gives such a relation.On the other hand, Drensky and Yu, in their article consider the lower bound for the degreeof the Poisson bracket (in the commutative case) or of the degree of the commutator (in thenon-commutative case) of two polynomials F and G in n variables x , . . . , x n . Their consideredthe conjecture which was formulated by Yu [18]. This conjecture says that, with some additionalassumptions, the following inequality holds: deg[
F, G ] > min { deg F, deg G } . Unfortunately the conjecture was showed to be false and, in the paper [1], there was presentedthe counterexample, to the conjecture of Yu, due to Makar-Limanov (see [1, Example 1.1]). For-tunately this example has some bad property. For instance this property can not be possessedby two polynomials that are components of the same polynomial authomorphism of C n . To beprecise, let F = F + · · · + F d and G = G + · · · + G N , where F i and G j are homogeneous polyno-mials of degree i and j, respectively. In the example of Makar-Limanov the linear components ofboth polynomials F and G are linearly dependend. In the case when F and G are componentsof the same polynomial authomorphism of C n the linear components of F and G must be linearyindependent. Thus, it is natural to state the following, new version of the conjecture of Yu: Conjecture 1.1.
Let F and G be algebraically independent polynomials in C [ x , . . . , x n ] . Let F and G generate their own centralizers in C [ x , . . . , x n ] , respectively. Suppose, also, that deg F ∤ Mathematics Subject Classification.
Primary 13F20, 14R10; Secondary 16W20.
Key words and phrases.
Poisson bracket, degree of Poisson bracket, polynomial ring, Jacobian determinant.
Corresponding author:
Daria Holik, e-mail : [email protected]. deg G, deg G ∤ deg F, F (0) = G (0) = 0 and that linear parts of F and G are linearly independent.Then deg[ F, G ] > min { deg F, deg G } or maybe some other (weaker) inequality holds, for example, deg[ F, G ] > c · min { deg F, deg G } for some constant c which is not dependent of F and G. Let us notice that the assumption that the homogeneous components of maximal degree of F and G are algebraically dependent, which appears in the original conjecture of Yu, is not nec-essary, because if the homogeneous components of maximal degree of F and G are algebraicallyindependent then deg[ F, G ] = deg F + deg G > min { deg F, deg G } . The problem of estimation from below of the degree of the Poisson bracket [ F, G ] is, of course,connected with the results about estimation of the lower bound of the degree of nonconstantelements of subalgebras generated by two polynomials. The first such an estimation was givenby Shestakov and Umirbaev [14]. Their used this estimation and the theorem saying that eachtame automorphism of C admits an elementary reduction or a reduction of some other types,to show that the famous Nagata example(1) σ : C ∋ ( x, y, z ) ( x + 2 y ( y + zx ) − z ( y + zx ) , y − z ( y + zx ) , z ) ∈ C is wild. For information about four types of reductions defined by Shestakov and Umirbaev seefor example [13]. In the paper [9] Shigeru Kuroda showed that there is no tame automorphismthat admits a reduction of type IV.Later in 2009 Leonid Makar-Limanov and Jie-Tai Yu [11] make an improvement of the esti-mation due to Shestakov and Umirbaev. The improved estimation is the following(2) deg P ( f, g ) ≥ D ( f, g ) w deg f, deg g P, where w deg f, deg g P for polynomial P = P ( x, y ) in two variables x and y denotes the weighteddegree of P such that w deg f, deg g x = deg f and w deg f, deg g y = deg g, and(3) D ( f, g ) = 1 − gcd(deg f, deg g ) − (deg( f g ) − deg[ f, g ])deg f deg g . In the paper [11] the authors used, instead of deg[ f, g ] , the equivalent, in this context, notion ofthe degree of the following differential form df ∧ dg. On the other hand in [8] Kuroda generalizedShestakov-Umirbaev inequality to the case of more than two polynomials. In his article the degreeof the Poisson bracket was replaced, which was very natural, by the degree of the differentialform which is obtained as the wedge product of the differentials of these polynomials.The numerous possible results about existence or nonexistence of tame automorphisms P =( P , P , P ) : C → C with a given multidegree of P (by multidegree of P we mean mdeg P =(deg P , deg P , deg P ) ) would be possible to be achieved if the Conjecture 1.1 appears to betrue. For some results about the multidegrees see for example [2, 4, 5, 6, 7, 16].One of a possible way of attacking the above conjecture or maybe some weaker conjecture (withstronger assumptions) is to prove that small degree of deg[ F, G ] implies the relations between F = F + · · · + F d and G = G + · · · + G N (here, by F i and G j we denote the homogeneouscomponent of F of degree i and the homogeneous component of G of degree j, respectively)and then show that such a relation can not be satisfied. On the other hand such a relation canbe viewed as a generalization of the result of Pritchard. Indeed, in the case F, G ∈ C [ x, y ] itis known that deg[ F, G ] = 2 + deg (cid:16) ∂F∂x ∂G∂y − ∂F∂y ∂G∂x (cid:17) . Since in the paper, we study the pair ofpolynomials
F, G with small degree of [ F, G ] we will assume that deg[ F, G ] < deg F + deg G. This implies that [ F d , G N ] = 0 , and so by Lemma 2 in [17] there exist a, b ∈ C , k , k ∈ N anda homogeneous polynomial h such that F d = ah k and G N = bh k . Of course, withuot loss ofgenerality, we can assume that h is not a power of any other polynomial of lower degree, andthat a = 1 , because C is algebraically closed. Thus, in the rest of the paper, we will assume that ependence of Homogeneous Components 3 F d = h k and G N = bh k with b ∈ C ⋆ = C \ { } denoted by a N . Of course, in this setup, we have k = d deg h and k = N deg h . The paper is organized as follows. In Section 2, we recall the notion of Poisson bracket oftwo polynomials, formulate some properties that can be used in the calculations and presentthe main tool of the paper the so-called H -reduction method. In Section 3, we give someauxilliary lemmas that can be used in the next section in computation. Section 4 is devotedfor establishing the general formulas for the homogeneous components of the polynomial G interms of the homogeneous components of the polynomial F. Then, in the following section weuse the above mentioned general formulas to prove that if deg[
F, G ] is small enough then thehomogeneous components of the polynomial F of degree deg F − is divisible by h in the caseof deg h ≤ and by sqrf h in the case deg h > (for definition of sqrf h see Section 5), where h is the polynomial such that the homogeneous components of maximal degree of F and G areproportional to some powers of h. Under the assumption that h is a square-free polynomial weobtain that if deg[ F, G ] is smaller and smaller then F s (the homogeneous component of F ofdegree s, where s = deg F − ) is divisible by higher and higher power of h. The next sectiongive some other relations between homogeneous components of the polynomial F in some specialcases of deg F. More precisely, we show that there is a strong relation between F s − and F s , butwith some assumptions about the degree of F and G, and of course, about deg[ F, G ] . Let us notice that the assumption that h is square-free geometrically means that the compo-nents at infinity of F = 0 (and G = 0 ) all have the same multiplicity. The notion of square-freepolynomials seems to be naturally arrising in the context of Jacobian Conjecture or related topics(see for example [3]).At the end of the introduction let us notice that all given in the present article results arevalid not only for C but for any algebraically closed field of characteristic zero.2. Poisson bracket and H -reduction method In this section we develop the main tool that will be used in the next sections of the paper.We start with the following lemma in which we use the notation C [ x , . . . , x n ] d = { f ∈ C [ x , . . . , x n ] : f is homogeneous of degree d } ∪ { } . Lemma 2.1.
Let H be a nonconstant homogeneous polynomial that is not a power of any otherpolynomial of lower degree (in particular, if H is a square-free polynomial) and let P be anyhomogeneous polynomial such that [ H, P ] = 0 . Then there exist a ∈ C and k ∈ N such that P = aH k . Moreover, if P ∈ C [ x , . . . , x n ] d and deg H ∤ d, then P = 0 . For the convenience of the reader, before we give the proof of the above lemma, let us recallthat for any f, g ∈ C [ x , . . . , x n ] , by [ f, g ] we denote the Poisson bracket of f and g, which is thefollowing sum: X ≤ i Since [ H, P ] = 0 , it follows that H and P are algebraically dependentand so by Lemma 2 in [17] there exist a, b ∈ C , k , k ∈ N and a homogeneous polynomial h suchthat P = ah k and H = bh k . Since H is not a power of any other polynomial of lower degree, we conclude that k = 1 and sowe can take h = H. Thus, in particular, if a = 0 (i.e. P = 0 ), then deg P is divisible by deg H. (cid:3) The mentioned H -reduction method is the statement given in the following corollary. Corollary 2.2. Let H be a nonconstant homogeneous polynomial that is not a power of anyother polynomial of lower degree and let P be any polynomial such that [ H, P ] = 0 . Then P ∈ C [ H ] . Proof. Let d = deg P and let P = P + · · · + P d be the homogeneous decomposition of P. Since [ H, P ] = 0 , it follows that(5) [ H, P i ] = 0 for i = 0 , . . . , d. In particular, [ H, P d ] = 0 . Since P d = 0 (by definition of d ), it follows that d = k deg H for some k ∈ N . By (5) and Lemma 2.1 there exist a , . . . , a k ∈ C , a k = 0 such that P l deg H = a l H l for l = 0 , . . . , k and P i = 0 for i / ∈ { , deg H, H, . . . , k deg H } . (cid:3) We will also use the following fact that is easy to check. Lemma 2.3. Let P ∈ C [ x , . . . , x n ] . For any Q, R ∈ C [ x , . . . , x n ] and α, β ∈ C we have [ P, QR ] = Q [ P, R ] + R [ P, Q ] , [ P, αQ + βR ] = α [ P, Q ] + β [ P, R ] , [ P, Q ] = − [ Q, P ] . In other words, the mappings Q [ P, Q ] and Q [ Q, P ] are C -derivations. ependence of Homogeneous Components 5 Preparatory lemmas We assume that are given the following two polynomials of degree, respectively, d and N with d ≤ N : (6) F = F + · · · + F d and G = G + · · · + G N , where F i and G j are homogeneous polynomials of degree i and j, respectively, with(7) F d = h dt and G N = a N h Nt , where h is a homogeneous polynomial of degree t and a N is a nonzero constant. Of course, wemust have t | d and t | N. Let us notice that: [ F i , F α · · · F α s s h α ] = α F α · · · F α s s h α − [ F i , h ] + s X j =1 α j F α · · · F α s s F − j h α [ F i , F j ] , (8)where s = d − . Lemma 3.1. Assume that polynomials (9) G u = X α =( α ,...,α s ) ∈ Z × N s c ( u ) α F α · · · F α s s h α u = i, . . . , i + s − , with c ( u ) α ∈ C , satisfy the following condition:(C1) ( α k + 1) c ( i + s − l ) α + e k = ( α l + 1) c ( i + s − k ) α + e l , for all ≤ k, l ≤ s and α ∈ Z × N s , where e = (1 , , . . . , , e = (0 , , , . . . , , . . . , e s = (0 , . . . , , . Then, the sum [ F s , G i ] +[ F s − , G i +1 ] + · · · + [ F , G i + s − ] can be written as X α ∈ Z × N + × N s − α α c ( i + s − α − e F α · · · F α s s h α − (10) + X α ∈ Z ×{ }× N + × N s − α α c ( i + s − α − e F α · · · F α s s h α − ... + X α ∈ Z ×{ (0 ,..., }× N + α α s c ( i ) α − e s F α s s h α − , h . Proof. By (8) we see that [ F s , G i ]+ [ F s − , G i +1 ]+ · · · + [ F , G i + s − ] is the sum of the followingsummands: X α ∈ Z × N s F α · · · F α s s h α n ( α k + 1) c ( i + s − l ) α + e k [ F l , F k ] + ( α l + 1) c ( i + s − k ) α + e l [ F k , F l ] o (11)for all ≤ k < l ≤ s and s X k =1 X α ∈ Z × N s α c ( i + s − k ) α F α · · · F α s s h α − [ F k , h ] . (12)Let us notice that (11) equals to zero by (C1). Daria Holik, Marek Karaś We can assume that for all u, c ( u )( α ,...,α s ) = 0 , if α k < for some k = 1 , . . . , s. Then, byrenumbering of the summs in (12), the summand (12) can be rewritten as follows P α ∈ Z × N s α h α − n c ( i + s − α − e F α − F α · · · F α s s [ F , h ] + c ( i + s − α − e F α F α − F α · · · F α s s [ F , h ] (13) + · · · + c ( i ) α − e s F α · · · F α s − s − F α s − s [ F s , h ] o . Assume that α k = 0 for some k ∈ { , . . . , s } and set e = (1 , , . . . , , e = (0 , , , . . . , , . . . ,e s = (0 , . . . , . Then, one can see that h α k c ( i + s − k ) α − e k F α · · · F α s s , h i (14) = α α k c ( i + s − k ) α − e k F α − F α · · · F α s s [ F , h ] + α α k c ( i + s − k ) α − e k F α F α − F α · · · F α s s [ F , h ]+ · · · + α s α k c ( i + s − k ) α − e k F α · · · F α s − s − F α s − s [ F s , h ]= c ( i + s − α − e F α − F α · · · F α s s [ F , h ] + c ( i + s − α − e F α F α − F α · · · F α s s [ F , h ]+ · · · + c ( i ) α − e s F α · · · F α s − s − F α s − s [ F s , h ] . Indeed, by assumption (C1), for l ∈ { , . . . , s } , we have [( α l − 1) + 1] c ( i + s − k )( α − e k − e l )+ e l = [( α k − 1) + 1] c ( i + s − l )( α − e k − e l )+ e k . (15)If we have α l = 0 in the above equality, then ( α − e k − e l ) + e k / ∈ Z × N s and we can assume that c ( i + s − l )( α − e k − e l )+ e k = 0 . Thus, the summand (12) can be written as in (10). Since the summands (11) are equal tozero, the result follows. (cid:3) Lemma 3.2. Assume that h is a homogeneous polynomial that is no power of any other poly-nomial of degree < deg h, polynomials G i , . . . , G i + s − , defined as in Lemma 3.1, satisfy thecondition (C1). If polynomial G i − is such that [ h r , G i − ] + [ F s , G i ] + · · · + [ F , G i + s − ] = 0 , forsome r ∈ N + , then G i − = X α =( α ,...,α s ) ∈ Z × N s c ( i − α F α · · · F α s s h α with c ( i − α ∈ C such that for all α = ( α , . . . , α s ) ∈ Z × N s we have:(P1) if α l > with l ∈ { , . . . , s } , then c ( i − α − re = α rα l c ( i + s − l ) α − e l . Proof. Since [ h r , G i − ] = rh r − [ h, G i − ] = [ h, rh r − G i − ] , it follows by Lemma 3.1 andCorollary 2.2 that G i − = 1 rh r − X α ∈ Z × N + × N s − α α c ( i + s − α − e F α · · · F α s s h α − (16) + X α ∈ Z ×{ }× N + × N s − α α c ( i + s − α − e F α · · · F α s s h α − ... + X α ∈ Z ×{ (0 ,..., }× N + α α s c ( i ) α − e s F α s s h α − + P ( h ) , ependence of Homogeneous Components 7 for some polynomial P in one variable, over k. Thus, if l = 1 (and α > ), then the equation in(P1) is a direct consequence of (16).Assume that l > and α l > . If α = . . . = α l − = 0 , then the equality in (P1) is, as above,a direct consequence of (16). Thus, we can assume that there is k < l such that k ≥ and α k > . We can also assume that we have already proved that(17) c ( i − α − re = α rα k c ( i + s − k ) α − e k . Now, using (C1) we obtain that(18) [( α k − 1) + 1] c ( i + s − l )( α − e l − e k )+ e k = [( α l − 1) + 1] c ( i + s − k )( α − e l − e k )+ e l and so(19) α rα l c ( i + s − l ) α − e l = α rα k c ( i + s − k ) α − e k . This completes the proof. (cid:3) Lemma 3.3. Assume that r ∈ N + , h is a homogeneous polynomial that is no power of any otherpolynomial of degree < deg h, polynomials G i , . . . , G i + s − , defined as in Lemma 3.1, satisfy thecondition (C1) and with polynomial G i + s , defined similarly as in Lemma 3.1, satisfy the followingcondition:(C2) for all ≤ l, k ≤ s and α = ( α , . . . , α s ) ∈ Z × N s if α k > , then c ( i + l − α − re = α rα k c ( i + l + s − k ) α − e k If polynomial G i − is such that [ h r , G i − ] + [ F s , G i ] + . . . + [ F , G i + s − ] = 0 , then(a) G i − , G i , . . . , G i + s − satisfy the condition (C1),(b) G i − , G i , . . . , G i + s − satisfy the condition (C2),all with the same r and i replaced by i − . Proof. We start with the proof of (b). Take any ≤ l, k ≤ s and α = ( α , . . . , α s ) ∈ Z × N s with α k > . Assume that l ≥ . Then, by (C2) we have c (( i − l − α − re = c ( i +( l − − α − re = α rα k c ( i +( l − s − k ) α − e k = α rα k c (( i − l + s − k ) α − e k . (20)For l = 1 we can use property (P1) with l replaced by k. Now, we prove (a). Take any ≤ k, l ≤ s and α = ( α , . . . , α s ) ∈ Z × N s . Let us notice that ( α l + 1) c (( i − s − k ) α + e l = ( α l + 1) c ( i +( s − k ) − α + e l = ( α l + 1) α r ( α l +1) c ( i +( s − k )+ s − l ) α + re . (21)Indeed, if k = s, then it is a consequence of (P1): ( α l + 1) c ( i +( s − k ) − α + e l = ( α l + 1) c ( i − α + e l = ( α l + 1) c ( i − α + e l + re ) − re (22) = ( α l + 1) α r ( α l +1) c ( i + s − l ) α + re = ( α l + 1) α r ( α l +1) c ( i +( s − k )+ s − l ) α + re and if k < s it follows from (C2): ( α l + 1) c ( i +( s − k ) − α + e l = ( α l + 1) c ( i +( s − k ) − α + e l + re ) − re = ( α l + 1) α r ( α l +1) c ( i +( s − k )+ s − l ) α + re . (23)Similarly, we have ( α k + 1) c (( i − s − l ) α + e k = ( α k + 1) c ( i +( s − l ) − α + e k = ( α k + 1) α r ( α k +1) c ( i +( s − l )+ s − k ) α + re . (24) Daria Holik, Marek Karaś By (21) and (24), we have ( α l + 1) c (( i − s − k ) α + e l = ( α k + 1) c (( i − s − l ) α + e k . (25) (cid:3) The following lemma is obvious. Lemma 3.4. Assume that h is a homogeneous polynomial such that deg h divides N. Let G N = α N h N deg h , G N +1 = . . . = G N + s = 0 . Then(a) polynomials G N , G N +1 , . . . , G N + s − satisfy the condition (C1),(b) for any r ∈ N + , polynomials G N , G N +1 , . . . , G N + s satisfy the condition (C2),all with i replaced by N. Formulas for G j The main result of this section is the following formula for polynomials G j . Theorem 4.1. Let F = F + · · · + F s + F d and G = G + · · · + G N be elements of C [ x , . . . , x n ] , where F i and G j are homogeneous polynomials of degree i and j, respectively. Assume that (26) F d = h d and G N = a N h N for some homogeneous polynomial h of degree 1 and a N ∈ C \ { } . If deg[ F, G ] < d + i for i ∈ { , . . . , N } , then there are a i , . . . , a N − ∈ C such that for j = i, i + 1 , . . . , N, we have (27) G j = X α =( α ,...,α s ) ∈ Z × N s α + α +2 α + ··· + sα s = jα + d ( α + ··· + α s ) ≤ N c ( j ) α F α · · · F α s s h α , where for α = ( α , . . . , α s ) ∈ Z × N s with α + α +2 α + · · · + sα s = j and α + d ( α + · · · + α s ) ≤ N, we have (28) c ( j ) α = Q α + ··· + α s k =1 ( α + dk ) (cid:0)Q α i =1 di (cid:1) · · · (cid:0)Q α s i s =1 di s (cid:1) a α + d ( α + ··· + α s ) . Proof. In what follows, we assume that c ( j )( α ,...,α s ) = 0 if α + α + 2 α + · · · + sα s = j or α + d ( α + · · · + α s ) > N. Let us notice that the above formula holds for G N . Indeed, if ( α , . . . , α s ) ∈ N s \ { (0 , . . . , } and α + α + 2 α + · · · + sα s = N, then α + d ( α + · · · + α s ) > N. Hence, in the sum (27), for G N , there is only one summand a N h N . Let us also notice that the above formula holds for thepolynomials G N +1 = . . . = G N + s = 0 . By Lemma 3.4, we know that the polynomials G N , . . . , G N + s − satisfy the condition (C1),and that the polynomials G N , . . . , G N + s satisfy the condition (C2). Assume that i ≤ N − . Then, deg[ F, G ] < d + N − and so [ h d , G N − ] + [ F s , G N ] + · · · + [ F , G N + s − ] = 0 . Since G N , . . . , G N + s − satisfy the condition (C1), we see from Lemma 3.2 that the polynomial G N − is of the form G N − = X α =( α ,...,α s ) ∈ Z × N s α + α +2 α + ··· + sα s = N − c ( N − α F α · · · F α s s h α and satisfy (P1). But, since G N = a N h N and G N +1 = . . . = G N + s − = 0 , we see, from (P1),that if ( α , . . . , α s ) ∈ N s \ { (0 , . . . , , (0 , . . . , , } , then c ( N − α ,...,α s ) = 0 . For α s = 1 , by (P1), wehave c ( N − N − d, ,..., , = Nd · c ( N )( N, ,..., = Nd · a N = ( N − d )+ d · d · a ( N − d )+ d (0+ ··· +0+1) . ependence of Homogeneous Components 9 Setting a N − = c ( N − N − , ,..., , we obtain that G N − satisfy formulas (27) and (28). By Lemma 3.3,we also know that the polynomials G N − , . . . , G N + s − satify the condition (C1), and that thepolynomials G N − , . . . , G N + s − satify the condition (C2). Thus, we can use the induction.Now, assume that for some l ∈ { i + 1 , i + 2 , . . . , N } , polynomials G l , . . . , G l + s are given bythe formulas (27) and (28). Assume also that G l , . . . , G l + s − satisfy the condition (C1) and G l , . . . , G l + s satisfy the condition (C2). Since deg[ F, G ] < d + i ≤ d + ( l − , it follows that [ h d , G l − ] + [ F s , G l ] + · · · + [ F , G l + s − ] = 0 . Hence, by Lemma 3.2, the polynomial G l − is ofthe form(29) G l − = X α =( α ,...,α s ) ∈ Z × N s α + α +2 α + ··· + sα s = l − c ( l − α F α · · · F α s s h α and satisfy (P1). Moreover, by Lemma 3.3, the polynomials G l − , . . . , G l + s − satisfy the con-dition (C1) and G l − , . . . , G l + s − satisfy the condition (C2). Since G l − satisfy (P1), we havethat G l − = a l − h l − + X α =( α ,...,α s ) ∈ Z × N + × N s − α + α +2 α + ··· + sα s = l − α + ddα c ( l + s − α + d,α − ,α ,...,α s ) F α · · · F α s s h α (30) + X α =( α ,...,α s ) ∈ Z ×{ }× N + × N s − α + α +2 α + ··· + sα s = l − α + ddα c ( l + s − α + d, ,α − ,α ,...,α s ) F α · · · F α s s h α ... + X α =( α ,...,α s ) ∈ Z ×{ (0 ,..., }× N + α + α +2 α + ··· + sα s = l − α + ddα s c ( l + s − s )( α + d, ,..., ,α s − F α s s h α . Notice that c ( l + s − α + d,α − ,α ,...,α s ) can be different from zero only if ( α + d )+ d ( α − α + · · · + α s ) ≤ N (because there is no components with degree > N in the polynomial G ), that is only if α + d ( α + α + · · · + α s ) ≤ N. The same holds for c ( l + s − α + d, ,α − ,α ...,α s ) , . . . , c ( l )( α + d, ,..., ,α s − . Hence, in (29), we can add the restriction α + d ( α + · · · + α s ) ≤ N. Assume that α > , α + α + 2 α + · · · + sα s = l − and α + d ( α + · · · + α s ) ≤ N. Then,by (30) and the above assumptions, we have c ( l − α ,...,α s ) = α + ddα c ( l + s − α + d,α − ,α ,...,α s ) (31) = α + ddα Q α − ··· + α s k =1 (( α + d ) + dk ) (cid:16)Q α − i =1 di (cid:17) · · · (cid:0)Q α s i s =1 di s (cid:1) a ( α + d )+ d (( α − α + ··· + α s ) = Q α + ··· + α s k =1 ( α + dk ) (cid:0)Q α i =1 di (cid:1) · · · (cid:0)Q α s i s =1 di s (cid:1) a α + d ( α + α + ··· + α s ) . Now, assume that α = 0 , α > , α + 2 α + · · · + sα s = l − and α + d ( α + · · · + α s ) ≤ N. Similarly, as before, we have c ( l + s − α ,...,α s ) = α + ddα c ( l +1)( α + d, ,α − ,α ,...,α s ) (32) = α + ddα Q α − α + ··· + α s k =1 (( α + d ) + dk ) (cid:16)Q α − i =1 di (cid:17) · · · (cid:0)Q α s i s =1 di s (cid:1) a ( α + d )+ d (( α − α + ··· + α s ) = Q α + α + ··· + α s k =1 ( α + dk ) (cid:0)Q α i =1 di (cid:1) · · · (cid:0)Q α s i s =1 di s (cid:1) a α + d ( α + α + ··· + α s ) The similar calculation shows that formula (28) holds also in the cases α = . . . = α q = 0 , and α q +1 > , with α + α + 2 α + · · · + sα s = l − and α + d ( α + · · · + α s ) ≤ N, for all q = 3 , . . . , s − . Since, by Lemma 3.3 we can use induction, the result follows. (cid:3) Using similar arguments as above one can prove the following result in which we consider thecase deg h ≥ interested one is deg h > . Of course, the following theorem is a generalizationof the above one. Theorem 4.2. Let F = F + · · · + F s + F d and G = G + · · · + G N be elements of C [ x , . . . , x n ] , where F i and G j are homogeneous polynomials of degree i and j, respectively. Assume that h ∈ C [ x , . . . , x n ] is a homogeneous polynomial that is no power of any other polynomial ofdegree < deg h, such that deg h | gcd( d, N ) and (33) F d = h d/ deg h and G N = a N h N/ deg h for some a N ∈ C \ { } . In what follows, we put r = d deg h and t = deg h. If deg[ F, G ] < d + i for i ∈ { , . . . , N } , then there are a i , . . . , a N − ∈ C such that for j = i, i + 1 , . . . , N, we have (34) G j = X α =( α ,...,α s ) ∈ Z × N s tα + α +2 α + ··· + sα s = jtα + d ( α + ··· + α s ) ≤ N c ( j ) α F α · · · F α s s h α , where for α = ( α , . . . , α s ) ∈ Z × N s with tα + α +2 α + · · · + sα s = j and tα + d ( α + · · · + α s ) ≤ N, we have (35) c ( j ) α = Q α + ··· + α s k =1 ( α + rk ) (cid:0)Q α i =1 ri (cid:1) · · · (cid:0)Q α s i s =1 ri s (cid:1) a tα + d ( α + ··· + α s ) . Moreover, we have a j = 0 when deg h ∤ j. Divisibility of F s In this section, we show that, if deg[ F, G ] is small enough then F s is divisible by h (seeTheorem 5.1) or even by suitable power of h (see Theorem 5.5). This will be established in thecase deg h = 1 and in the case deg h > with the additional assumption that h is a square-freepolynomial (see Theorem 5.8). In the case deg h > but without assumption that h is square-free,we obtain that F s is divisible by sqrf h, where sqrf h is defined as the product of all irreduciblefactors of the polynomial h but taken without multiplicity, for example sqrf (cid:0) x ( x − (cid:1) = x ( x − . A polynomial sqrf h for a given h ∈ C [ x , . . . , x n ] is defined only up to a constantfactor u ∈ C \ { } , but for our purpose it is not a disadvantage. If the polynomial h is itselfa square-free polynomial, that is with irreducible factors with multiplicity one, then we put ependence of Homogeneous Components 11 sqrf h = h. For a general definition of square-free elements of any commutative rings we refer thereader to [3].5.1. The case deg h = 1 . In this subsection we consider the case deg h = 1 . First of all, let usnotice that if deg h ≤ and h is no power of any other polynomial with degree < deg h, thenthe polynomial h is square-free. Theorem 5.1. Let F and G be as in Theorem 4.1. If N d ) and deg[ F, G ] < d + i with i < sd N = d − d N, then F s is divisible by h. In other words, there is a homogeneous polynomial ˜ F s − of degree s − such that F s = ˜ F s − h. In the proof of the above theorem we use the following Lemma 5.2. For all integers ≤ j < N, the minimal value (36) min { α | ( α , . . . , α s ) ∈ Z × N s , α + α + 2 α + · · · + sα s = j, α + d ( α + · · · + α s ) ≤ N } is equal to dj − sN and is attained at only one point ( dj − sN, , . . . , , N − j ) . Proof. Let Z j,N,d := { ( α , . . . , α s ) ∈ Z × N s | α + α +2 α + · · · + sα s = j, α + d ( α + · · · + α s ) ≤ N } . Notice that for all ( α , . . . , α s ) ∈ Z j,N,d , we have sα + ( s − α + · · · + 2 α s − + α s ≤ N − j (in particular, α s ≤ N − j ). Thus, sα s ≤ s ( N − j ) − s ( sα + ( s − α + · · · + 2 α s − ) and α + 2 α + · · · + sα s ≤ s ( N − j ) − [ s − α − [ s ( s − − α − · · · − [ s · − ( s − α s − . Since s − , s ( s − − , . . . , s · − ( s − are strictly positive numbers, it follows that for ( α , . . . , α s ) ∈ Z j,N,d if ( α , . . . , α s − ) = (0 , . . . , ∈ N s − , then α > dj − sN. Of course, if ( α , , . . . , , α s ) ∈ Z j,N,d with α s < N − j, then also α = j − sα s > j − s ( N − j ) = dj − sN. (cid:3) Now we can prove Theorem 5.1. Proof of Theorem 5.1. Since deg[ F, G ] < d + i, we see by Theorem 4.1 that(37) G i = X α =( α ,...,α s ) ∈ Z × N s α + α +2 α + ··· + sα s = iα + d ( α + ··· + α s ) ≤ N c ( i ) α F α · · · F α s s h α . By Lemma 5.2, all summands of (37), except c ( i )( di − sN, ,..., ,N − i ) F N − is h di − sN , involves h u with u > di − sN. Let us also notice that di − sN < . Thus, multiplying both sides of (37) by h sN − di , we obtain that h divides c ( i )( di − sN, ,..., ,N − i ) F N − is . Since N 0( mod d ) , we see that − sN d ) and so c ( i )( di − sN, ,..., ,N − i ) = Q N − ik =1 ( di − sN + dk ) Q N − iis =1 di s a N = 0 . Thus, we see that h | F N − is and so h | F s because h is a square-free polynomial. (cid:3) By Theorems 4.1 and 5.1, we obtain the following Corollary 5.3. Let F and G be as in Theorem 4.1. If N d ) and deg[ F, G ] < d + i with i < sd N, then for j = i, i + 1 , . . . , N, we have (38) G j = X α =( α ,...,α s ) ∈ Z × N s α + α +2 α + ··· + sα s = jα + d ( α + ··· + α s ) ≤ N c ( j ) α F α · · · F α s − s − ˜ F α s s − h α + α s , where a i , . . . , a N − ∈ C and c ( j ) α are as in Theorem 4.1, and ˜ F s − is as in Theorem 5.1. The following lemma we use in the proof of Theorem 5.5 below. Lemma 5.4. For all k ∈ N such that d > k and for all integers ≤ j < N, the minimal value (39) min { α + kα s | ( α , . . . , α s ) ∈ Z j,N,d } is equal to ( d − k ) j − ( s − k ) N and is attained at only one point ( dj − sN, , . . . , , N − j ) . Proof. As in the proof of Lemma 5.2, we have ( s − k ) α s ≤ ( s − k )( N − j ) − ( s − k ) ( sα + ( s − α + · · · + 2 α s − ) . From this we see that α + 2 α + · · · + ( s − α s − + ( s − k ) α s ≤ ( s − k )( N − j ) − [( s − k ) s − α − [( s − k )( s − − α − · · · − [( s − k ) · − ( s − α s − and so α + kα s = j − [ α + 2 α + · · · + ( s − α s − + ( s − k ) α s ] (40) ≥ j − ( s − k )( N − j ) + [( s − s − α +[( s − k )( s − − α + · · · + [( s − k ) · − ( s − α s − Since ( s − k ) s − > ( s − k )( s − − > . . . > ( s − k ) · − ( s − 1) = s − (2 k − > , it follows that for ( α , . . . , α s ) ∈ Z j,N,d if ( α , . . . , α s − ) = (0 , . . . , ∈ N s − , then α + kα s > ( d − j − ( s − N. Of course, if ( α , , . . . , , α s ) ∈ Z j,N,d with α s < N − j, then also α + kα s = j − ( s − k ) α s >j − ( s − k )( N − j ) = ( d − k ) j − ( s − k ) N. (cid:3) By the above lemma, one can use, several times, the similar arguments as in the proof ofTheorem 5.1 to show the following result (see also proof of Theorem 5.8 in the next subsection). Theorem 5.5. Let F, G and h be as in Theorem 4.1. If N d ) and deg[ F, G ] < d + i with i < s − kd − k N with k < d, then F s is divisible by h k +1 . In other words, there is a homogeneouspolynomial ˜ F s − k − of degree s − k − such that F s = ˜ F s − k − h k +1 . By Theorems 4.1 and 5.5, we obtain the following Corollary 5.6. Let F and G be as in Theorem 4.1. If N d ) and deg[ F, G ] < d + i with i < s − kd − k N with k < d, then we have (41) G j = X α =( α ,...,α s ) ∈ Z × N s α + α +2 α + ··· + sα s = jα + d ( α + ··· + α s ) ≤ N c ( j ) α F α · · · F α s − s − ˜ F α s s − k − h α +( k +1) α s , where a i , . . . , a N − ∈ C and c ( j ) α are as in Theorem 4.1 and ˜ F s − k − is as in Theorem 5.5. The case deg h > . The results given in this subsection are also true for deg h = 1 , butof course are interested in the case deg h > . We start with the following lemma. Lemma 5.7. Let d, t and N be as in Theorem 4.2. For all k ∈ N such that d > kt and for allintegers ≤ j < N, the minimal value (42) min { α + kα s | ( α , . . . , α s ) ∈ Z ( t ) j,N,d } is equal to t [( d − kt ) j − ( s − kt ) N ] and is attained at only one point ( t ( dj − sN ) , , . . . , , N − j ) , where Z ( t ) j,N,d = { ( α , . . . , α s ) ∈ Z × N s : tα + α + 2 α + · · · + sα s = j, tα + d ( α + · · · + α s ) ≤ N } Proof. As in the proof of Lemma 5.2, we have ( s − kt ) α s ≤ ( s − kt )( N − j ) − ( s − kt ) ( sα + ( s − α + · · · + 2 α s − ) . From this we see that α + 2 α + · · · + ( s − α s − + ( s − kt ) α s ≤ ( s − kt )( N − j ) − [( s − kt ) s − α − [( s − kt )( s − − α − · · · − [( s − kt ) · − ( s − α s − . ependence of Homogeneous Components 13 Since α = t [ j − ( α + 2 α + · · · + sα s )] , it follows that α + kα s = t [ j − ( α + 2 α + · · · + sα s )] + ktα s t (43) = t [ j − ( α + 2 α + · · · + ( s − α s − + ( s − kt ) α s )] ≥ t { j + ( s − kt )( j − N ) + [( s − kt ) s − α + [( s − kt )( s − − α + · · · + [( s − kt )2 − ( s − α s − } . Since ( s − kt ) s − > ( s − kt )( s − − > . . . > ( s − kt ) · − ( s − 1) = s − (2 kt − > , it follows that for ( α , . . . , α s ) ∈ Z ( t ) j,N,d if ( α , . . . , α s − ) = (0 , . . . , ∈ N s − , then α + kα s > t { ( d − kt ) j − ( s − kt ) N } . Of course, if ( α , , . . . , , α s ) ∈ Z ( t ) j,N,d with α s < N − j, then also α + kα s = t { j − ( s − kt ) α s } > t { j − ( s − k )( N − j } ) = t { ( d − kt ) j − ( s − kt ) N } . (cid:3) Now, we can use the above lemma and Theorem 4.2 to prove the following result. Theorem 5.8. Let F, G and h be as in Theorem 4.2. Assume also that the polynomial h issquare-free. If N d ) and deg[ F, G ] < d + i with i < s − kd − k N and kt < d, then F s is divisible by h k +1 . In other words, there is a homogeneous polynomial ˜ F s − t ( k +1) of degree s − t ( k + 1) such that F s = ˜ F s − t ( k +1) h k +1 . Proof. Since deg[ F, G ] < d + i, we see by Theorem 4.2 that(44) G i = X α =( α ,...,α s ) ∈ Z × N s tα + α +2 α + ··· + sα s = itα + d ( α + ··· + α s ) ≤ N c ( i ) α F α · · · F α s s h α . By Lemma 5.7 used for k = 0 , all summands of (44), except c ( i )( t ( di − sN ) , ,..., ,N − i ) F N − is h t ( di − sN ) , involves h u with u > t ( di − sN ) . Let us also notice that t ( di − sN ) < . Thus, multiplyingboth sides of (44) by h t ( sN − di ) , we obtain that h divides c ( i )( t ( di − sN ) , ,..., ,N − i ) F N − is . Since N d ) , we see that − sN 0( mod d ) and so c ( i )( t ( di − sN ) , ,..., ,N − i ) = Q N − ik =1 ( 1 t ( di − sN )+ rk ) Q N − iis =1 ri s a N = Q N − ik =1 ( di − sN + dk ) Q N − iis =1 di s a N = 0 . Thus, we see that h | F N − is and so h | F s because h is a square-free poly-nomial.Now, we can (44) rewrite in the form(45) G i = X α =( α ,...,α s ) ∈ Z × N s tα + α +2 α + ··· + sα s = itα + d ( α + ··· + α s ) ≤ N c ( i ) α F α · · · F α s − s − (cid:0) F s h (cid:1) α s h α + α s , where F s h is a homogeneous polynomial of degree s − t. Using Lemma 5.7 for k = 1 , we seethat all summands of (45), except c ( i )( t ( di − sN ) , ,..., ,N − i ) (cid:0) F s h (cid:1) N − i h t [( d − i − ( s − N ] , involves h u with u > t [( d − i − ( s − N ] . Let us also notice that t [( d − i − ( s − N ] < . Thus, multiplyingboth sides of (45) by h t [( d − i − ( s − N ] , we obtain that h divides c ( i )( t ( di − sN ) , ,..., ,N − i ) (cid:0) F s h (cid:1) N − i . Since c ( i )( t ( di − sN ) , ,..., ,N − i ) = 0 , it follows as above that h divides F s h . Now, we can (45) rewrite in the form(46) G i = X α =( α ,...,α s ) ∈ Z × N s tα + α +2 α + ··· + sα s = itα + d ( α + ··· + α s ) ≤ N c ( i ) α F α · · · F α s − s − (cid:0) F s h (cid:1) α s h α +2 α s , where F s h is a homogeneous polynomial of degree s − t. Thus, we can repeat the above argumentsonce more again.Repeating this arguments several times we obtain the thesis. (cid:3) Corollary 5.9. Let F and G be as in Theorem 4.2. Moreover, let h be a square-free polynomial.If N d ) and deg[ F, G ] < d + i with i < s − kd − k N and kt < d, then we have (47) G j = X α =( α ,...,α s ) ∈ Z × N s tα + α +2 α + ··· + sα s = jtα + d ( α + ··· + α s ) ≤ N c ( j ) α F α · · · F α s − s − ˜ F α s s − t ( k +1) h α + t ( k +1) α s , where a i , . . . , a N − ∈ C and c ( j ) α are as in Theorem 4.2 and ˜ F s − t ( k +1) is as in Theorem 5.8. Dependence between F s − and F s In this section we consider the situation when d = gcd( d, N ) = gcd(deg F, deg G ) is such that d = 2 d . In such a situation N must be an odd multiplicity of d and so must be of the form N = d (2 k + 1) for some nonnegative integer k. Since we always assume that N ≥ d, we see thatthe number k maybe considered as positive integer. In this setup we will show that if deg[ F, G ] is small enough that there is a relation between F s − and F s . We start this with the following numerical lemma which we will use in both considered cases,namely the case deg h = 1 and the case deg h > . Lemma 6.1. For a given positive integers j, N, d, t such that j < N, d ≤ N and t | gcd( d, N ) letus consider the following sets Z j,N,d = ( ( α , . . . , α s ) ∈ Z × N s | α + s X l =1 lα l = j, α + d s X l =1 α l ≤ N ) , (48) Z ( t ) j,N,d = ( ( α , . . . , α s ) ∈ Z × N s | tα + s X l =1 lα l = j, tα + d s X l =1 α l ≤ N ) , (49) ∆ j,N,d = ( ( α , . . . , α s ) ∈ R s + | s X l =1 ( d − l ) α l ≤ N − j ) . (50) Then, the following assertions hold:(a) The mapping ϕ : ∆ j,N,d ∩ N s ∋ ( α , . . . , α s ) ( j − P sl =1 lα l , α , . . . , α s ) ∈ Z j,N,d isa bijection.(b) The mapping ψ : Z j,N,d ∩ ( t Z × N s ) ∋ ( ˜ α , α , . . . , α s ) ( ˜ α t , α , . . . , α s ) ∈ Z ( t ) j,N,d isa bijection.(c) The set ∆ j,N,d is the simplex with the following set of vertices: (0 , . . . , , N − js e , N − js − e , . . . , N − j e s − and ( N − j ) e s , where e = (1 , , . . . , , . . . , e s = (0 , . . . , , isthe standard basis of R s . In other words vertices of ∆ j,N,d others than (0 , . . . , areof the form N − jd − l e l for l = 1 , . . . , s. ependence of Homogeneous Components 15 (d) For any ( β , . . . , β s ) ∈ N s and for the function (a linear function actually) f β ,...,β s :∆ j,N,d ∋ ( α , . . . , α s ) j − P sl =1 ( l − β l ) α l ∈ R we have f β ,...,β s (0 , . . . , 0) = j and f β ,...,β s ( N − jd − l e l ) = j − l − β l d − l ( N − j ) . (e) If β s < d then the minimal value of f ,..., ,β s on the set ∆ j,N,d equals to j − ( s − β s )( N − j ) and is attained exactly at one point (0 , . . . , , N − j ) . (f ) If β s = d then the minimal value of f ,..., ,β s on the set ∆ j,N,d equals to j − ( s − β s )( N − j ) and is attained exactly on the line segment with the following endpoints (0 , . . . , , N − j , and (0 , . . . , , N − j ) . Proof. The assertions (a)-(d) are easy to check. To prove assertions (e) and (f) let us noticethat, since the set ∆ j,N,d is a convex set and the function f ,..., ,β s is linear, it follows that weonly need to examine the values of the function on the set of vertices of ∆ j,N,d . Since we areinterested in minimal value of f ,..., ,β s and N − j > , we see that what is interested for us isthe maximal value of the fraction l − β l d − l for l = 1 , . . . , s. Since we have β = . . . = β s − = 0 , thevalues of l − β l d − l for l = 1 , . . . , s − are the following: d − < d − < . . . < d − = s − . To concludethe proof, we only need to notice that s − < s − β s in the case β s < d and s − = s − β s in thecase β s = d. (cid:3) Before we go to the subsection devoted to the case deg h = 1 we give the one more lemma,which is a consequence of the above numerical one. Lemma 6.2. Let d ∈ N \ { , } and set d = 2 d , N = d (2 k + 1) for some k ∈ N + . Then, thefollowing statements hold(a) min { α + d α s | ( α , . . . , α s ) ∈ Z j,N,d } is equal to ( d − d ) j − ( s − d ) N = d j − ( d − N and is attained exactly at the points of the following set { ( dj − sN + dl, , . . . , , l, N − j − l ) ∈ Z × N s | ≤ l ≤ N − j } , (b) Let t be an integer such that t divides d and set r = dt and ˜ r = d t (in particular r = 2˜ r ). Then, min { α + ˜ rα s | ( α , . . . , α s ) ∈ Z ( t ) j,N,d } is equal to ˜ rj − ( d − r (2 k +1) and is attained exactly at the points of the following set { ( rj − s ˜ r (2 k + 1) + rl, , . . . , , l, N − j − l ) ∈ Z × N s | ≤ l ≤ N − j } . Proof. Consider the set ∆ j,N,d = (cid:8) ( α , . . . , α s ) ∈ R s + | P sl =1 ( d − l ) α l ≤ N − j (cid:9) and the func-tion f ,...,d ( α , . . . , α s ) = j − P sl =1 ( l − δ ls d ) α l , where δ ls is the Kronecker delta. Let us noticethat from Lemma 6.1 (a), we have Z j,N,d = ϕ j (∆ j,N,d ∩ N s ) . Notice also that f ,...,d ( α , . . . , α s ) = α + d α s , where ( α , α , . . . , α s ) = ϕ j ( α , . . . , α s ) for any ( α , . . . , α s ) ∈ ∆ j,N,d ∩ N s . Now usingthe above considerations and Lemma 6.1 (c) and (f), we obtain the assertion (a).In order to prove (b) we apply Lemma 6.1 (b) and already proved Lemma 6.2 (a). In-deed, by Lemma 6.1 (b) we have that ( α , . . . , α s ) ∈ Z ( t ) j,N,d if and only if ( ˜ α , . . . , α s ) =( tα , . . . , α s ) ∈ Z j,N,d ∩ ( t Z × N s ) . On the other hand minimizing α + ˜ rα s is equivalent tominimizing t ( α + ˜ rα s ) = ˜ α + d α s . Thus searching the points at which α + ˜ rα s achieve min-imal values on the set Z ( t ) j,N,d is the same as searching the points at which ˜ α + d α s achieveminimal values on the set Z j,N,d ∩ ( t Z × N s ) and so we can use Lemma 6.2 (a). It should bementioned that from Lemma 6.2 (a) we get information about the points at which the minimumis achieved and what is its value. This completes the proof. (cid:3) To conclude this part of the section, we make the following observation. Remark 6.3. One can notice that Lemma 6.1 (e) can be used, instead of Lemmas 5.2 and 5.7,in order to obtain alternative proofs of Theorems 5.1 and 5.8. The case deg h = 1 . In this subsection we consider the case deg h = 1 . Thus, we have F = F + · · · + F s − + h d and G = G + · · · + G N − + a N h N with d = 2 d for some d ∈ N \ { , } ,a N ∈ C \ { } and N = d (2 k + 1) for some positive integer k. In the proof of the main result of this subsection we will use the following lemma. Lemma 6.4. Assume that d = 2 d for some d ∈ N \ { , } , N = 2 d k + d for some k ∈ N + , and that deg[ F, G ] < d + N − k − . Assume also that F d = h d and G N = a N h N for somehomogeneous polynomial h of degree one and a N ∈ C \ { } . Then, for all l ∈ { , , . . . , k + 1 } , we have (51) c ( N − k − − d k − d +2 d l, ,..., ,l, k +2 − l ) = ( − l (cid:18) k + 1 l (cid:19) c ( N − k − − d k − d , ,..., , , k +2) . Proof. Since deg[ F, G ] < d + N − k − , we see that the polynomial G N − k − is given bythe formulas (27) and (28) with d = 2 d and s = 2 d − . One can check, that c ( N − k − − d k − d +2 d ( l +1) , ,..., ,l +1 , k +2 − l +1)) (52) = Q k +2 − ( l +1) i =1 ( − d k − d + 2 d ( l + 1) + 2 d i ) Q l +1 i s − =1 d i s − · Q k +2 − l +1) i s =1 d i s a N and c ( N − k − − d k − d +2 d l, ,..., ,l, k +2 − l ) = Q k +2 − li =1 ( − d k − d + 2 d l + 2 d i ) Q li s − =1 d i s − · Q k +2 − li s =1 d i s a N , (53)for any l = 0 , , . . . , k. From the above, one can see that c ( N − k − − d k − d +2 d ( l +1) , ,..., ,l +1 , k +2 − l +1)) (54) = 2 d (2 k + 1 − l )2 d (2 k + 2 − l )2 d ( l + 1)( − d k − d + 2 d l ) c ( N − k − − d k − d +2 d l, ,..., ,l, k +2 − l ) = ( k + 1 − l )( − l + 1 c ( N − k − − d k − d +2 d l, ,..., ,l, k +2 − l ) . In particular, for l = 0 , we obtain that c ( N − k − − d k − d , ,..., , , k ) = ( k + 1)( − c ( N − k − − d k − d , ,..., , , k +2) (55) = ( − (cid:18) k + 11 (cid:19) c ( N − k − − d k − d , ,..., , , k +2) . Similarly, one can see that c ( N − k − − d k + d , ,..., , , k − = ( k + 1 − − c ( N − k − − d k − d , ,..., , , k ) = ( k + 1) k ( − · c ( N − k − − d k − d , ,..., , , k +2) = ( − (cid:18) k + 12 (cid:19) c ( N − k − − d k − d , ,..., , , k +2) . Continuing the above calculation inductively, we obtain the result. (cid:3) Now we can prove the following theorem. Theorem 6.5. Assume that d = 2 d for some d ∈ N \ { , } , N = 2 d k + d for some k ∈ N + , and that deg[ F, G ] < d + N − k − . Assume also that F d = h d and G N = a N h N for ependence of Homogeneous Components 17 some homogeneous polynomial h of degree one and a N ∈ C \ { } . Then, there is a homogeneouspolynomial ˆ F s − of degree s − such that (56) F s − = ( ˜ F d − + h ˆ F s − ) . Proof. First of all notice that s − ( d − d − ( d − = d d +1 > d − d and so s − ( d − d − ( d − N = d d +1 N > d − d N = d − d (2 d k + d ) = N − k − . By the above and the assumption that deg[ F, G ] < d + N − k − , we can use Theorem 5.5 to obtain that F s is divisible by h d . Thus, there exists a homogeneouspolynomial ˜ F d − of degree d − such that F s = h d ˜ F d − . Now, using the above considerations,assumption that deg[ F, G ] < d + N − k − and Theorem 4.1 with d = 2 d , s = 2 d − and j = N − k − we obtian that(57) G N − k − = X α =( α ,...,α s ) ∈ Z × N s α + α +2 α + ··· + sα s = jα + d ( α + ··· + α s ) ≤ N c ( N − k − α F α · · · F α s − s − ˜ F α s d − h α + d α s . Notice that, for N = 2 d k + d and j = N − k − , the set from Lemma 6.2 (a) is equal tothe following one { ( − d k − d + 2 d l, , . . . , , l, k + 2 − l ) ∈ Z × N s | l = 0 , , . . . , k + 1 } . Thus, in the polynomial G N − k − all summand other than included in the following sum(58) k +1 X l =0 c ( N − k − − d k − d +2 d l, ,..., ,l, k +2 − l ) F ls − ˜ F k +2 − ld − h − d involves h u with u > − d . By Lemma 6.4, the above sum can be written as follows c ( N − k − − d k − d , ,..., , k +2) h − d k +1 X l =0 (cid:18) k + 1 l (cid:19) ( − l F ls − ˜ F k +2 − ld − (59) = c ( N − k − − d k − d , ,..., , k +2) h − d ( ˜ F d − − F s − ) k +1 . Now, we see by multiplying both sides of (57) by h d that h divides c ( N − k − − d k − d , ,..., , k +2) ( ˜ F d − − F s − ) k +1 . But, since c ( N − k − − d k − d , ,..., , k +2) = Q k +2 i =1 ( − d k − d +2 d i ) Q k +2 is =1 d i s a d k + d = 0 , it follows that h | ( ˜ F d − − F s − ) k +1 , and since h is not a power of any other polynomial (in this case this meansthat h is square-free), h | ( ˜ F d − − F s − ) . Thus there is a homogeneous polynomial ˆ F s − of degree s − such that − h ˆ F s − = ˜ F d − − F s − . Solving the last equality with respect to F s − we obtainthe thesis. (cid:3) The case deg h > . As the title of subsection suggests, we consider here the situation of deg h > . Thus, we have F = F + · · · + F s − + h dt and G = G + · · · + G N − + a N h Nt where t = deg h and a N ∈ C \ { } . As before, we assume that d = 2 d for some d ∈ N \ { , } and N = d (2 k + 1) for some positive integer k. Since t must divide d and N and gcd( d, N ) = d , one can see that we must have, in this situation, that d is divisible by t. In the proof of the main result of this subsection we will use the following lemma. Lemma 6.6. Let d, N and deg[ F, G ] be as in Lemma 6.4. Moreover, suppose that F d = h dt and G N = a N h Nt for some square-free, homogeneous polynomial h of degree t , where t ≥ ,a N ∈ C \ { } and r = dt , ˜ r = d t . Then, for all l ∈ { , , . . . , k + 1 } , we have (60) c ( N − k − − k − l )˜ r, ,..., ,l, k +2 − l ) = ( − l (cid:18) k + 1 l (cid:19) c ( N − k − − k − r, ,..., , , k +2) . Proof. According with the assumption that deg[ F, G ] < d + N − k − , the polynomial G N − k − is given by the formulas (34) and (35) with d = 2 d , s = 2 d − , so c ( N − k − − k − l +1))˜ r, ,..., ,l +1 , k +2 − l +1)) (61) = c ( N − k − (cid:16) − d k − d d l +1) t , ,..., ,l +1 , k +2 − l +1) (cid:17) = Q k +2 − ( l +1) i =1 − d k − d +2 d ( l +1)+2 d it Q l +1 i s − =1 2 d i s − t · Q k +2 − l +1) i s =1 2 d i s t a N = Q k +2 − ( l +1) i =1 ( − d k − d + 2 d ( l + 1) + 2 d i ) Q l +1 i s − =1 d i s − · Q k +2 − l +1) i s =1 d i s a N and c ( N − k − − k − l )˜ r, ,..., ,l, k +2 − l ) = Q k +2 − li =1 ( − d k − d + 2 d l + 2 d i ) Q li s − =1 d i s − · Q k +2 − li s =1 d i s a N , (62)for any l = 0 , , . . . , k. Further part of this proof is similar to the proof of Lemma 6.4. (cid:3) Theorem 6.7. Assume that d = 2 d for some d ∈ N \ { , } , N = 2 d k + d for some k ∈ N + , and that deg[ F, G ] < d + N − k − . Assume also that F d = h dt = h r and G N = a N h Nt for somesquare-free, homogeneous polynomial h of degree t , where t ≥ , a N ∈ C \ { } and r = dt , ˜ r = d t .Then, there is a homogeneous polynomial ˆ F s − t − of degree s − t − such that (63) F s − = ( ˜ F d − + h ˆ F s − t − ) . Proof. Using the assumption that deg[ F, G ] < d + N − k − , and the following observationthat s − ( d t − d − ( d t − = d (2 t − d (2 t − t > d − d and so s − ( d t − d − ( d t − N > d − d N = N − k − we can applyTheorem 5.8 to obtain the divisibility of F s by h ˜ r . It means that, there exists a homogeneouspolynomial ˜ F d − of degree d − such that F s = h ˜ r ˜ F d − . Referring to the above calculationsand Theorem 4.2, the polynomial G N − k − takes the form(64) G N − k − = X α =( α ,...,α s ) ∈ Z × N s tα + α +2 α + ··· + sα s = jtα + d ( α + ··· + α s ) ≤ N c ( N − k − α F α · · · F α s − s − ˜ F α s d − h α +˜ rα s . According to Lemma 6.2 (b) for j = N − k − , the set of summands with minimal possiblepower of h , in the above sum, corresponds to α belonging to { (( − k − l )˜ r, , . . . , , l, k +2 − l ) ∈ Z × N s | l = 0 , , . . . , k + 1 } , so all components under than occurring the presenting sum(65) k +1 X l =0 c ( N − k − − k − l )˜ r, ,..., ,l, k +2 − l ) F ls − ˜ F k +2 − ld − h − ˜ r involves h u with u > − ˜ r. Applying Lemma 6.6, we can rewrite the above sum in the followingway c ( N − k − − k − r, ,..., , k +2) h − ˜ r k +1 X l =0 (cid:18) k + 1 l (cid:19) ( − l F ls − ˜ F k +2 − ld − (66) = c ( N − k − − k − r, ,..., , k +2) h − ˜ r ( ˜ F d − − F s − ) k +1 . ependence of Homogeneous Components 19 The rest of the proof is analogous to the proof of the Theorem 6.5 using the assumption that h is square-free. (cid:3) References [1] V. Drensky, J.-T. Yu, Degree estimate for commutators. J. Algebra 322, no. 7, 2321–2334 (2009)[2] E. Edo, T. Kanehira, M. Karaś, S. 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