Irreducibility of Polynomials and the Resultant 1 Introduction

Home , Generic polynomial, Irreducible polynomial, Monic polynomial, Resultant

International Journal of Algebra, Vol. 8, 2014, no. 14, 687 - 697 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ija.2014.4887

Irreducibility of Polynomials and the Resultant

Boukari Dahani

Universit´ede Ouagadougou 03 B.P. 7021 Ouagadougou, Burkina Faso

G´erardKientega

Département de mathématiqueset d’informatique Universitéde Ouagadougou 03 B.P. 7021 Ouagadougou, Burkina Faso

Copyright c 2014 Boukari Dahani and G´erardKientega. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract Let f and g be polynomials of degrees p and q respectively over a noetherian ring A (with p > q). If the polynomial g is monic then we know whether g divides f or not. Otherwise, we have no means to solve this problem. In this paper, our purpose is to give conditions on the coeﬃcients of f and g for divisibility of f by g. Using the result, we will obtain conditions of irreducibility of any polynomial. This result can be extended to polynomials over a commutative ring. We follow ideas of S. S. Woo.

Mathematics Subject Classiﬁcation: 11R09, 12D05, 13B25

Keywords: resultant, Fitting invariant, irreducible polynomial

1 Introduction

Given f and g two monic polynomials over a noetherian ring, S.S.Woo found in [4] a condition of divisibility of f by g by looking at the resultant matrix. Thus 688 Boukari Dahani and G´erardKientega she got a criterion of irreducibility for a monic polynomial and a generalization of the Eisenstein’s irreducibility criterion. In this paper we will generalize her results to not necessarily monic polynomials over a noetherian ring. In section 2, we give some preliminary results that will be needed in the others parts of the paper. In section 3, for a given polynomial f we give criteria of irreducibility and a condition using diophantine equations for the reducibility of f. In section 4, we generalize Eisenstein’s irreducibility criterion. Throughout all rings are commutative with a unit element denoted by 1.

2 Basic results

In this section we generalize some results obtained in the case of monic polynomials [4]. We ﬁrst deﬁne some notions to make things clear. In this part the ring A is not nesessarily noetherian. A zero divisor in A is a nonzero element a ∈ A such that there exists a nonzero element b ∈ A with ab = 0. An element c of A is nilpotent if there exists an integer n > 0 such that cn= 0; the smallest such integer is the nilpotency of c. p q Throughout for f(X) = apX + ··· + a0, g(X) = bqX + ··· + b0 two polynomials over A with respective degrees p and q, we set Y = apbqX, p−1 p q q−1 F(Y) = ap bqf(X), G(Y) = apbq g(X). Observe that F and G are monic polynomials in Y. We also suppose that ap and bq are not zero divisors. Given a strictly positive integer n, Sn will denote the A-submodule of A[Y] consisting of the polynomials of degree < n. Since ap and bq are not zero n−1 divisors then B = (Y , ··· , 1) is a basis of Sn. Choosing bases B1 (respectively B2) of Sq × Sp (respectively Sp+q) by q−1 q−2 p−1 B1 = {(Y , 0), (Y , 0), ··· ,(Y, 0), (1, 0),(0, Y ), ··· , (0, Y), (0, 1)} p+q−1 and B2 = {Y , ··· , 1} we observe that the resultant matrix of F and G which are elements of A[Y] that we denote by R(F, G), is the matrix of the A-linear map ϕ: Sq × Sp → Sp+q,(u, v) 7→ uF + vG with respect to B1 and B2. Proposition 2.1 The A − − module A[Y ]/(F ) is free of rank p.

Proof. Indeed, let H be an element of A[Y]. Since F is a monic polynomial of A[Y], then there exist Q and R two polynomials of A[Y] such that H = Q·F + R with R = 0 or deg R ≤ p − 1. Then in A[Y]/(F), we have H = R and B = ((Y )p−1, (Y )p−2, ··· , 1 ) generates A[Y]/(F). If p−1 p−1 p−1 p−1 αp−1 (Y ) + ··· + α0 1 = 0 then αp−1ap bq X + ··· + α0 = 0 and k−1 k−1 αk−1ap bq = 0 for all k ∈ {p, p − 1, ··· , 1}. Thence αk−1 = 0 for k = 1, ··· , p.

Lemma 2.2 A[Y ] = Sp+q + (F ). Irreducibility of polynomials and the resultant 689

Proof. Since (F) and Sp+q are submodules of A[Y] then (F) + Sp+q is a submodule of A[Y]. Let h be an element of A[Y]. Then there exist two polynomials Q and R of A[Y] such that h = QF + R with R = 0 or deg R < p. Then h ∈ (F) + Sp+q and A[Y] ⊂ (F) + Sp+q.

Lemma 2.3 Sp+q = Sp ⊕ FSq.

Proof. Let W be an element of Sp ∩ FSq. If W is a nonzero polynomial then there exists a polynomial u of degree at least one such that W = F · u and we have deg W = deg F + deg u; so W = 0. Moreover, it is obvious that Sp + FSq ⊂ Sp+q and if u is an element of Sp+q, then u = Q · F + R with R = 0 or deg R < deg F. Thus, Sp+q ⊂ FSq + Sp and Sp+q ⊂ FSq + Sp.

Proposition 2.4 We have FSq= Sp+q ∩ (F ).

Proof. Let w0 be an element of FSq. Then there exists an element w1 of Sq such that w0 = Fw1. Since F is a monic polynomial, then we have deg w0 = deg F + deg w1. Thus, deg w0 < p + q and w0 is an element of Sp+q. Then FSq ⊂ Sp+q ∩ (F). Now let h0 be an element of Sp+q ∩ (F). Then there exists an element h1 of A[Y] such that h0 = h1F and deg h0 < p + q. We have deg h0 = deg h1 + deg F. Then deg h1 < q and h0 is an element of FSq. Thus, Sp+q ∩ (F) ⊂ FSq and FSq = Sp+q ∩ (F). By the second isomorphism theo- ∼ rem, Lemma 2.2 and Proposition 2.4 we have an isomorphism δ: Sp+q/FSq = A[Y]/(F). Deﬁne the map ϕ’: Sp+q → Sp+q by ϕ’(uF + v) = uF + vG; then 0 ϕ |FSq = id. Now deﬁne the map θ: Sq × Sp → Sp+q by θ(u, v) = uF + v and the map ψ: A[Y]/(F) → A[Y]/(F), h 7→ h · G. Then we have the following result. Theorem 2.5 The following diagram where cl is the canonical surjection is commutative.

θ cl δ Sq × Sp / Sp+q / Sp+q/F Sq / A[Y ]/(F )

ϕ ϕ0 ϕ0 ψ id cl δ−1 Sp+q / Sp+q / Sp+q/F Sq o A[Y ]/(F )

Proof. Consider the diagram:

θ Sq × Sp / Sp+q (1)

ϕ ϕ0 id Sp+q / Sp+q. 690 Boukari Dahani and G´erardKientega

Let u be an element of Sq and v an element of Sp. We have ϕ0 o θ(u, v) = ϕ0(uF + v) = uF + vG and id o ϕ(u, v) = ϕ(u, v) = uF + vG. Then ϕ0 o θ = id o ϕ and diagram ( 1) is commutative. Now consider the diagram: cl Sp+q / Sp+q/F Sq (2)

ϕ0 ϕ0 cl Sp+q / Sp+q/F Sq.

0 Let w be an element of Sq. Since ϕ |FSq = id, we have 0 cl o ϕ (Fw) = cl(Fw) = 0 because Fw is an element of FSq = ker cl. Then, FSq ⊂ ker (cl o ϕ0). By the ﬁrst isomorphism theorem, there exists one and only 0 one morphism ϕ : Sp+q/F Sq → Sp+q/F Sq such that the following diagram is commutative:

ϕ0 cl Sp+q / Sp+q / Sp+q/F Sq . 4 cl Sp+q/F Sq

Then diagram ( 2) is commutative. To ﬁnish the proof, let us consider the following diagram: δ Sp+q/F Sq / A[Y ]/(F ) (3)

ϕ0 ψ δ−1 Sp+q/F Sq o A[Y ]/(F ).

0 0 Let h be an element of Sp+q. Set h = cl(h). We have ϕ o cl(h) = cl o ϕ (h). Then ϕ0(h) = ϕ0(h) = ϕ0(uF + v) with h = uF + v, where u and v are some uniquely deﬁned elements of Sq and Sp respectively. (The unicity is due to the 0 equality Sp+q = Sp ⊕ FSq.) Then we have: ϕ (h) = uF + vG + FSq = vG + FSq because uF belongs to FSq. Moreover, since δ(cl(h)) = h + (F) and GuF ∈ (F ) we have: δ−1 o ψ o δ(h) = δ−1 o ψ(h + (F)) = δ−1(GuF + Gv + (F)) = −1 δ (Gv + (F)) = Gv + FSq ( because deg Gv ≤ deg G + deg v < q + p). Hence, diagram ( 3) is commutative.

Proposition 2.6 Let M be the cokernel of

ψ : A[Y ]/(F ) → A[Y ]/(F ), h 7→ h · G.

If G is a factor of F then M ∼= A[Y]/(G). Irreducibility of polynomials and the resultant 691

Proof. Indeed, since M is the cokernel of ψ we have M ∼=(A[Y]/(F))/Im ψ. Since F is a multiple of G, we have (F) ⊂ (G). Let t be an element of Im ψ. There exists an element y of A[Y]/(F) such that ψ(y) = t. We have ψ(y) = yG = t. Since y·G is an element of (G), t is an element of (G)/(F). Thus, Im ψ ⊂ (G)/(F). Now let u be an element of (G)/(F). Then u is an element of (G), and there exists an element k of A[Y] such that u = kG. Now, ψ( k) = kG = u and ψ( k) is an element of Im ψ. Then we have (G)/(F) ⊂ Im ψ and Im ψ = (G)/(F). The ﬁrst isomorphism theorem gives A[Y]/(G) ∼= M.

Corollary 2.7 Let M be the cokernel of

ψ : A[Y ]/(F ) → A[Y ]/(F ), h 7→ h · G.

If G is a factor of F then M is also the cokernel of

ϕ : Sq × Sp → Sp+q, (u, v) 7→ uF + vG

Proof. Let Gv0 be an element of GSp. We have ϕ(0, v) = Gv0. Then 0 GSp ⊆ Im ϕ. Let h be an element of Im ϕ. Then there exists an element u of 0 0 0 0 Sq and an element v of Sp such that u F + v G = h. Now u F is an element of 0 GSp because (F) ⊆ (G) and v G is an element of GSp. Thus, h is an element of GSp and Im ϕ = GSp. By the isomorphism ∼ ∼ Sp+q/gSp = A[Y]/(G) and the previous proposition, M = A[Y]/(G). Then ∼ Sp+q/Im ϕ = M.

p q Lemma 2.8 Let f(X) = apX + ··· + a0, g(X) = bqX + ··· + b0 be two polynomials over A with respective degrees p and q (p > q). Then g is a factor p−q−1 p−q+1 of f if and only if F = GH and ap bq divides H.

Proof. If f(X) = g(X)·h(X) then p−1 p p−1−q p−q+1 q q−1 p−1−q p−q+1 ap bqf(X) = ap bq apbq g(X)h(X). Setting F(Y) = ap bq h(X), p−q−1 p−q+1 we have F(Y) = G(Y)·H(Y) and ap · bq divides H. p−q−1 p−q+1 Conversely, suppose that F(Y) = G(Y)·H(Y) and ap bq is a factor of H. Then there exists a polynomial h(X) over A[X], such that p−q−1 p−q+1 H(Y) = ap bq h(X). Therefore, we have p−1 p q q−1 p−q−1 p−q+1 ap bqf(X) = apbq g(X)ap bq h(X). Since ap and bq are not zero divisors, then f(X) = g(X)h(X).

3 Some irreducibility criteria

For an A–module M and an integer i, we denote by F itti(M) the i-th Fitting invariant of M. [1] 692 Boukari Dahani and G´erardKientega

p q Recall that for f(X) = apX + ··· + a0, g(X) = bqX + ··· + b0 two polynomials over A with respective degrees p and q, we set Y = apbqX, p−1 p q q−1 p−k−1 p−k F(Y) = ap bqf(X), G(Y) = apbq g(X). Also we set Ak = akap bq for j q−j−1 k = 0, 1, ··· , p − 1, and Bj = bjapbq for j = 0, 1, ··· , q − 1 will denote the respective coeﬃcients of the polynomials F and G. Recall that also Ap = Bq = 1. In this section we suppose that the ring A is noetherian.

p q Theorem 3.1 Let f(X) = apX + ··· + a0, g(X) = bqX + ··· + b0 be two polynomials over A with respective degrees p and q (p > q) which leading coeﬃcients are not zero divisors. Let M be the cokernel of the map:

ψ : A[Y ]/(F ) −→ A[Y ]/(F ), H 7→ H · G.

Then the following propositions are equivalent:

1. G(Y ) is a factor of F (Y ).

2. M is free of rank q.

3. M is projective of constant rank q.

4. The minors of ψ of size p − q generates A and the minors of size bigger than p − q of ψ vanish.

5. The minors of R(F, G) of size p generate A and the minors of R(F, G) of size bigger than p vanish.

Proof.1) ⇒ 2). By Proposition 2.6 we have M ∼= A[Y]/(G). Then M is free of rank q. 2) ⇒ 3). Obvious. 3) ⇒ 4). Since A is noetherian M is an A – module with ﬁnite presentation. Thus if M is projective of rank q then F ittq(M) = A and F ittq−1(M) = 0. 4) ⇐⇒ 5). Consider the following commutative diagram:

θ cl δ Sq × Sp / Sp+q / Sp+q/F Sq / A[Y ]/(F )

ϕ ϕ0 ϕ0 ψ id cl δ−1 Sp+q / Sp+q / Sp+q/F Sq o A[Y ]/(F ).

=∼ =∼ id M / M / M / M

0 0 Irreducibility of polynomials and the resultant 693

By Corollary 2.7, the ﬁrst and the last vertical rows of the diagram are ﬁnite free presentations of the same module M. Therefore by Fitting’s Lemma, F ittq(R(F,G)) = A and F ittq−1(R(F,G)) = 0 if and only if F ittq(ψ) = A and F ittq−1(ψ) = 0.

5) ⇒ 1). Indeed, consider the following determinant of size p + 1 that is zero:

0 Bq 0 ...... 0 ...... Bq . . . . . 0 . . .. .

. . .. . Ap ...... Dc(f, g) = ...... Ap−1 B0 ...... 0 B ...... 1 ...... B0 ...... 0

A 0 ... 0 B ...... B q 0 q F (Y ) Y p−1G(Y ) · ...... YG(Y ) G(Y ) Expanding this determinant according to the first column we obtain the result. The previous theorem gives a criterion to know whether a polynomial g which p−1 p leading coefficient bq is not a zero divisor and divides ap bqf and a method to p−1 p obtain the factorization ap bqf = gh. Example 3.2 Let f(X) = 5X3 − 4X2 − 1 and g(X) = 2X − 2 be two 3 2 polynomials over Z. We have F (Y ) = Y − 8Y − 200 and G(Y ) = Y − 10.  1 1 0 0   −8 −10 1 0  R(F, G) =   .  0 0 −10 1  −200 0 0 −10 satisfy the fifth condition of theorem 3.1. Then G(Y) is a factor of F(Y) and F(Y) = G(Y)·H(Y) with H(Y) = Y 2 + 2Y + 20 = 100X2 + 20X + 20 = 20(5X2 + X + 1). Then setting h(X) = 5X2 + X + 1, we have 10f(X) = g(X)·h(X). The previous Theorem and the Lemma 2.8 allow us to obtain new results on divisibility of polynomials.

p q Corollary 3.3 Let f(X) = apX + ··· + a0, g(X) = bqX + ··· + b0 be two polynomials over A with respective degrees p and q. g is a factor of f if and p−q−1 p−q+1 only if the minors of R(F,G) of size bigger than p vanish and ap bq divides H, where H is a polynomial deﬁned by F = GH. 694 Boukari Dahani and G´erardKientega

Example 3.4 Let f(X) = 6X3 + 3X2 − 2X − 1 and g(X) = 3X2 − 1 be 3 2 two polynomials over Z. We have F (Y ) = Y + 9Y − 108Y − 972 and G(Y ) = Y 2 − 108.

 1 0 1 0 0   9 1 0 1 0    R(F, G) =  −108 9 −108 0 1  .    −972 −108 0 −108 0  0 −972 0 0 −108 satisfy the ﬁfth condition of theorem 3.1. Then G(Y) is a factor of F (Y ) and F (Y ) = G(Y ) · H(Y )withH(Y ) = Y + 9 = 32(2X + 1). Then setting h(X) = 2X + 1 we have f(X) = g(X) · h(X).

The previous corollary gives also an irreducibility criterion for a polynomial over A, which leading coeﬃcient is not a zero divisor in A. Thus, let q q−1 gt(X) = tqX + tq−1X + ··· + t0 be a generic polynomial over A which p−1 p leading coeﬃcient tq is not a zero divisor. Set Ft(Y ) = ap tqf(X) and q q−1 Gt(Y ) = aptq g(X).

p Corollary 3.5 f is irreducible if and only if for all q ≤ , there exists a 2 p−q−1 p−q+1 nonzero minor of size > p in R(Ft, Gt) or if Ft = GtHt then ap tq divides Ht.

Now, we suppose that A is a noetherian domain and we will adapt the notations of [5, p.106]. Then we obtain conditions for f to be reducible.

p q Corollary 3.6 Let f(X) = apX + ··· + a0 and g(X) = bqX + ··· + b0 be two polynomials over A with respective degrees p and q (p > q), which leading coeﬃcients are not zero divisors. g divides f if and only if (B0,B1, ··· ,Bq−1) is a solution of the system of equations:  w(F,q)(T , ··· ,T ) = 0  1 0 q−1  (F,q) w2 (T0, ··· ,Tq−1) = 0  ·····················  (F,q)  wq (T0, ··· ,Tq−1) = 0

p−q−1 p−q+1 and ap bq divides H where H is deﬁned by F = GH.

Proof. We apply the ideas of S. S. Woo [5, p.105 − 106] to F and G. As a consequence of the above Corollary, we have the following result: Irreducibility of polynomials and the resultant 695

p Corollary 3.7 Let f(X) = apX +···+a0 be polynomial over A with degree p, which leading coeﬃcient is not a zero divisor. f is irreducible if and only if p for all q ≤ the system of equations: 2  w(Ft,q)(T , ··· ,T ) = 0  1 0 q−1  (Ft,q) w2 (T0, ··· ,Tq−1) = 0  ·····················  (Ft,q)  wq (T0, ··· ,Tq−1) = 0

p−q−1 p−q+1 has no solution or if Ft = GtHt then ap bq does not divide Ht.

4 Generalized Eisenstein criterion

In this section we will generalize the Eisenstein’s irreducibility criterion. Given p f(X) = apX + ··· + a0 a polynomial over a ring B and ϕ:B −→ C a ring p p X i homomorphism, ϕ(f(X)) = ϕ(ap)X + ··· + ϕ(a0) = then ϕ(ai)X is a i=0 polynomial over C. We still suppose that A is noetherian.

Lemma 4.1 Let ϕ:B −→ C be a ring homomorphism and let Pp i f(X) = i=0 aiX be a polynomial over B of degree p. If ϕ(ap) is a nonzero element of A and not a zero divisor in C and ϕ(f(X)) is irreducible over C then f(X) is irreducible.

This lemma is in fact, a generalization of those of S. Lang [3, p.185] and S. S. Woo [6, p.502].

Pp i Theorem 4.2 Let f(X) = i=0 aiX be a polynomial over a ring B and let ϕ:B −→ A be a ring homomorphism such that:

1. ϕ(ap) is a nonzero element of A and not a zero divisor in A.

2. ϕ(ai) = 0 for i = 0, ··· , p − 1.

3. If a0 = b0c0 then either ϕ(b0) or ϕ(c0) is of of nilpotency > p. Then f(X) is irreducible over A.

Pq j Proof. Suppose that f(X) = g(X)·h(X) with g(X) = j=0 bjX and Pr k h(X) = k=0 ckX . Then a0 = b0 · c0 and by the condition 3 of theorem, either ϕ(b0) or ϕ(c0) is of nilpotency > p. We may suppose that ϕ(b0) is of of nilpotency > p. (Otherwise reverse the role of g and h.) Since 696 Boukari Dahani and G´erardKientega

ϕ(f(X)) = ϕ(g(X))· ϕ(h(X)), then setting Z = ϕ(ap) · ϕ(bq) · X, p−1 p q q−1 P(Z) = ϕ(ap) ϕ(bq) ϕ(f(X)), Q(Z) = ϕ(ap) ϕ(bq) ϕ(g(X)), we have P(Z) = Q(Z)·S(Z) for some polynomial S of A[Z]. Then, the minors of q q−1 R(P, Q) of size bigger than p vanish. Now, setting Cj = ϕ(ap) ϕ(bq) ϕ(bj) for j = 0, ··· q − 1 the coeﬃcients of Q(Z) and using the condition 2, R(P, Q) is of the form:

1 0 ... 0 1 0 ...... 0 ...... 0 1 . Cq−1 1 ...... 0 .. 0 . C .. 0 q−1 ...... 1 . . . 1

R(P, Q) = .. . 0 . 0 C0 C1 Cq−1 . . . . 0 0 .. . 0 C .. . 0 ......

0 ... 0 0 0 ... 0 C 0

p pq p(q−1) p Then we have R(P, Q) = C0 = ϕ(ap) · ϕ(bq) ϕ(b0) which is nonzero because ϕ(b0) is of nilpotency > p and ϕ(ap) and ϕ(bq) are nonzero elements of A and not zero divisors in A. The following result may be viewed as a generalization of Eisenstein criterion. In fact, it is a result of S.S. Woo. [6, p. 505].

Pp i Corollary 4.3 Let ϕ:B −→ A be a ring homomorphism, f(X) = i=0 aiX a polynomial over B of degree p > 0 and k an integer such that 0 ≤ k < p. If:

1. ϕ(ap) is a nonzero element of A and not a zero divisor in A.

2. ϕ(ai) = 0 for all i ∈ { 0, 1, ··· , k }.

Then f(X) has no factor g(X) of degree ≥ p − k with ϕ(g(0)) of nilpotency > k + 1.

q X j Proof. Suppose that f(X) = g(X) · h(X) with g(X) = bjX , j=0 r X k h(X) = ckX such that q ≥ p − k and the nilpotency d of ϕ(g(0)) > k + k=0 1. Then we have F(Y) = G(Y)H(Y). Now, since F is monic then ϕ(Ap) = 1 p−i−1 p−i and ϕ (Ai) = ϕ(aiap bq ) = 0 for all i ∈ { 0, 1, ··· , k } because ϕ(ai) = 0. By [6, p.505], F has no factor of degree ≥ p − k with ϕ (G(0)) of index of nilpotence > k + 1. Now, we have Irreducibility of polynomials and the resultant 697

d q q−1 d ϕ (G(0)) = ϕ (b0apbq ) = 0. Then ϕ(B0) is of nilpotency d.

In the following result we show that the condition 2) of the above corollary can be replaced by a weaker one.

Corollary 4.4 Let ϕ:B −→ A be a ring homomorphism and let p X i f(X) = aiX be a polynomial over B of degree ≥ 1 and k an integer such i=1 that 0 ≤ k < p. If:

1. ϕ(ap) is a nonzero element of A and not a zero divisor in A.

2. ϕ(ai) is nilpotent for i = 0, ··· , p − 1. Then f(X) has no factor g(X) over B of degree ≥ p − k with g(0) a nonnilpotent element.

Proof. Assuming that f(X) has a factor g(X) over B of degree ≥ p − k with g(0) a nonnilpotent element, we obtain F = GH for some polynomial H of A[Y ] and F satisfy to the hypotheses of Theorem 4.6 of [6, p.507] and G(0) is nonnilpotent, which is impossible.

References

[1] D. Eisenbud, Commutative algebra with a view toward algebraic geometry, New York, Berlin, Springer-Verlag, 1995.

[2] M. Mignotte, D. Stef˜anescu, Polynomials. An Algorithmic approach, Springer-Verlag, 1999.

[3] S. Lang, Algebra, Springer GTM 211, 2002.

[4] S.S.Woo, Dividing polynomials using the resultant matrix, Comm. Algebra 35, no. 11, (2007), 3263-3272 .

[5] S.S.Woo, Irreducibility of polynomials and diophantine equations, Math. Soc. 47, no. 1, (2010), 101-112.

[6] S.S.Woo, Some remarks on Eisenstein’s criterion, Math. Soc. 23, no. 4, (2008), 499-509.

[7] V.V. Prasolov, Polynomials, Springer-Verlag, 2010.

Received: August 8, 2014