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Irreducibility of Polynomials and the Resultant 1 Introduction 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 D´epartement de math´ematiqueset d'informatique Universit´ede 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 poly- nomials [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 2 A such that there exists a nonzero element b 2 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 = f(Y , 0), (Y , 0), ··· ,(Y, 0), (1, 0),(0, Y ), ··· , (0, Y), (0, 1)g p+q−1 and B2 = fY , ··· , 1g 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 ': 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 2 fp; p − 1; ··· ; 1g. 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 2 (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 '': Sp+q ! Sp+q by ''(uF + v) = uF + vG; then 0 ' jFSq = id. Now define the map θ: Sq × Sp ! Sp+q by θ(u, v) = uF + v and the map : 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 ) ' '0 '0 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) ' '0 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 '0 o θ(u, v) = '0(uF + v) = uF + vG and id o '(u, v) = '(u, v) = uF + vG. Then '0 o θ = id o ' and diagram ( 1) is commutative. Now consider the diagram: cl Sp+q / Sp+q=F Sq (2) '0 '0 cl Sp+q / Sp+q=F Sq: 0 Let w be an element of Sq. Since ' jFSq = id, we have 0 cl o ' (Fw) = cl(Fw) = 0 because Fw is an element of FSq = ker cl. Then, FSq ⊂ ker (cl o '0). By the first isomorphism theorem, there exists one and only 0 one morphism ' : Sp+q=F Sq ! Sp+q=F Sq such that the following diagram is commutative: '0 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) '0 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 ' o cl(h) = cl o ' (h). Then '0(h) = '0(h) = '0(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: ' (h) = uF + vG + FSq = vG + FSq because uF belongs to FSq. Moreover, since δ(cl(h)) = h + (F) and GuF 2 (F ) we have: δ−1 o o δ(h) = δ−1 o (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 : 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 we have M ∼=(A[Y]/(F))/Im . Since F is a multiple of G, we have (F) ⊂ (G). Let t be an element of Im . There exists an element y of A[Y]/(F) such that (y) = t. We have (y) = yG = t. Since y·G is an element of (G), t is an element of (G)/(F). Thus, Im ⊂ (G)/(F).
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