Filomat 27:2 (2013), 215 226 DOI 10.2298/FIL1302215C Published b Facult of Sciences and Mathematics, Universit of Niš, Serbia Available at: http://www.pmf.ni.ac.rs/filomat Factoring bivariate polnomials with integer coefficients via Newton polgons Siniša Crvenković a, Ivan Pavkov b a Universit of Novi Sad, Facult of Sciences and Mathematics, Department of Mathematics and Informatics, Trg Dositeja Obradovića 4, 21000 Novi Sad b Higher School of Professional Business Studies, Vladimira Perića-Valtera 4, 21000 Novi Sad Abstract. It is known that the Newton polgon of a polnomial carries information on its irreducibilit. In this paper we shall give another proof of the main theorem and characterize the inner points of the polgon. From that proof it will be obvious that the inner points of the Newton polgon pla an important role in finding possible factorizations. We shall give a necessar and sufficient condition for the eistence of the integer polnomial factorization in integer factor-polnomials. Definition 0.1. The conve hull of a set S (denoted b conv(s)) is the smallest conve set that contains the set S. Definition 0.2. For two arbitrar sets A and B, being subsets of R 2, a 2-dimensional real Euclidean space, the set A + B = {a + b : a A, b B} is called the Minkowski sum of sets A and B. Definition 0.3. An arbitrar point from R 2 is called an integer point if both of its coordinates are integers. An arbitrar polgon in R 2 is called an integer polgon if all of its vertices are integer points. Definition 0.4. We sa that the integer polgon C is integrall decomposable if there eist integer polgons A and B that satisf C = A + B, A and B containing at least two points. Polgons A and B are called polgon summands of the polgon C. Otherwise, polgon C is integrall indecomposable, i.e. there is no non-trivial decomposition of polgon C. Definition 0.5. Consider an arbitrar polnomial in two variables with integer coefficients from Z[, ], the ring of polnomials in two variables over Z f (, ) = C e1 e 2 e 1 e 2. Consider an eponent vector (e 1, e 2 ) as a point in Z 2. The Newton polgon of the polnomial f (, ), denoted b P f, is defined as the conve hull in R 2 of all the points (e 1, e 2 ) with C e1 e 2 Z \ {0}. Definition 0.6. A polnomial over field F is called absolutel irreducible if it remains irreducible over ever algebraic etension of F. 2010 Mathematics Subject Classification. 12Y05 ) Kewords. bivariate polnomials, non-trivial factorization, Newton polgon Received: 10 Ma 2012; Accepted: 15 December 2012 Communicated b Miroslav Ćirić a Supported b the Ministr of Education and Science, Serbia, grant 174018 Email addresses: sima@dmi.uns.ac.rs (Siniša Crvenković), pavkov.ivan@gmail.com (Ivan Pavkov)
S. Crvenković, I. Pavkov / Filomat 27:2 (2013), 215 226 216 Definition 0.7. Let f (, ) Z[, ]. The non-etended lattice of nodes of the polnomial f (, ) consists of all the points (e 1, e 2 ) i, i = 1,..., k corresponding to the terms with non-zero coefficients. If the Newton polgon of f (, ) contains, in its inner area, some integer points different from (e 1, e 2 ) i, i = 1,..., k, these points, together with (e 1, e 2 ) i, i = 1,..., k, form an etended lattice of nodes. The following theorem is well known. Theorem 0.8. Let F be an arbitrar field. Let f (, ), (, ), h(, ) F[, ] with f (, ) 0 and f (, ) = (, )h(, ). Then P f = P + P h. Proof: It is known that the Minkowski sum of two polgons is a polgon and an verte of the resulting polgon is obtained as the sum of the vertices of the polgon summands. Let (α, β) P f be an arbitrar verte of the Newton polgon of polnomial f (, ). This implies that polnomial f contains monomial α β with non-zero coefficient. Due to the fact that f (, ) = (, )h(, ), we can conclude that polnomials (, ) and h(, ) contain monomials γ δ and α γ β δ in that order, both with non-zero coefficients. These monomials correspond to the points (γ, δ) P and (α γ, β δ) P h. It is obvious that (γ, δ) + (α γ, β δ) P + P h. Namel, we obtain (γ+α γ, δ+β δ) P +P h, i.e. (α, β) P +P h. In other words, the inclusion P f P +P h holds. We will show that the reverse inclusion P + P h P f also holds. Since a Newton polgon is the conve hull of its vertices, it is sufficient to show that an verte of a polgon P + P h lies in the polgon P f. Let v be a verte of the Newton polgon P + P h. Since v P + P h, we conclude that there eist points v P and v h P h such that v = v + v h. The assumption that v is a verte of the Newton polgon P + P h implies that such vectors v and v h are unique. Assume the opposite, v = v + v h = v + v h, v, v P, v h, v h P h, with v v and v h v h. Let v = (, ) and v = (a, b). As v = v + v h, it is obvious that v h = ( a, b). Let v = (c, d). Analogousl we obtain v h = ( c, d). Consider the point v + v h. It is clear that v + v h P + P h. We have the following v + v h = (a, b) + ( c, d) = ( + a c, + b d) = ( + (a c), + (b d)). Further, consider the point v + v h. It is clear that v + v h P + P h. We have the following v + v h = (c, d) + ( a, b) = ( a + c, b + d) = ( (a c), (b d)). Let us look for a midpoint of the line segment whose endpoints are v +v h and v +v h, v +v h, v +v h P +P h. ( + (a c) + (a c), 2 + (b d) + (b d) ) = (, ). 2 In other words, the center of the line segment whose endpoints are v + v h and v + v h, which both lie in the Newton polgon P + P h, is the point (, ). As a verte of the Newton polgon is not on an line segment connecting an other two points of the polgon, we conclude that the point (, ) is not the verte of the polgon P + P h, which is an obvious contradiction. Therefore, for an arbitrar verte v of the Newton polgon P + P h there eist unique vectors v P and v h P h such that v = v + v h. As v is a verte of the polgon P +P h, we can conclude that v and v h are also the vertices of the polgons P and P h. Since v and v h are unique, it is obvious that there eist unique terms of (, ) and h(, ) such that v is a resulting eponent vector in (, )h(, ). Therefore v P f. We have shown that all vertices of polgon P + P h lie in the polgon P f. Due to the fact that the Newton polgon is the conve hull of its vertices, we have P + P h P f. Remark 0.9. From the proof of Theorem 0.8 it is obvious that the monomials of the polnomial f (, ) that are obtained in, at least, two different was b multipling factor-polnomials (, ) and h(, ), certainl do not correspond to the vertices of the Newton polgon P f. Eample 0.10. Consider a polnomial: f (, ) = 2 + 2 + 2 Z[, ].
S. Crvenković, I. Pavkov / Filomat 27:2 (2013), 215 226 217 The factorization of polnomial f (, ) is given below f (, ) = ( + )( + ). The monomial with coefficient 2 is obtained in two different was b multipling factor-polnomials = ()() = ()(). Eamine the Newton polgon of the polnomial f (, ) shown in Figure 1. Figure 1 Clearl, P f = conv((1, 1), (0, 2), (2, 0)) is a line segment with endpoints (0, 2) and (2, 0). Indeed, the monomial of the polnomial f (, ) that is obtained in two different was b multipling factorpolnomials corresponds to the point (1, 1), which is not the verte of the polgon P f. Theorem 0.11. Let f (, ) be a non-zero bivariate polnomial over an arbitrar field F, non-divisible either b or b. If the Newton polgon of the polnomial is integrall indecomposable, then polnomial f (, ) is absolutel irreducible over F. Proof: Since polnomial f (, ) is non-divisible either b or b, it has no trivial decomposition. Assume the opposite, the polnomial f (, ) is not absolutel irreducible over field F, i.e. it does not remain irreducible over ever algebraic etension of F. This implies that over some algebraic field etension of F, f (, ) = (, )h(, ) holds, with (, ) and h(, ) having, at least, two non-zero terms. Due to the fact that (, ) and h(, ) have, at least, two non-zero terms, their corresponding Newton polgons have, at least, two points. Since f (, ) = (, )h(, ), from the previous theorem we get P f = P + P h, with P and P h having, at least, two points. This implies that polgon P f has non-trivial decomposition, which is a contradiction. Therefore, the polnomial f (, ) is absolutel irreducible over F. Definition 0.12. Let f (, ) be a polnomial in two variables over Z. Let P = {A 1, A 2,..., A n } be the lattice of nodes of the polnomial f (, ) possibl etended b some integer points that lie inside of the Newton polgon of the polnomial f (, ). Without loss of generalit, we can assume that after the construction of the Newton polgon of the polnomial f (, ), A 1, A 2,..., A k, k 2, become its vertices, and A k+1,..., A n do not. We sa that the grouping of the set P, G 1,..., G l, l 2, is a super-covering of P if: 1. Each group G i, i = 1,..., l, contains the same number of points not less than two, l 2. G i = P, i=1 3. Points A 1, A 2,..., A k appear in one and onl one of the sets G 1,..., G l, 4. Points A k+1,..., A n appear in at least one of the sets G 1,..., G l, 5. Conve polgons determined b G 1, G 2,..., G l are congruent and G 2,..., G l are obtained from G 1 b translation.
S. Crvenković, I. Pavkov / Filomat 27:2 (2013), 215 226 218 Definition 0.13. Let f (, ) be a bivariate polnomial with integer coefficients over Z. Let P = {A 1, A 2,..., A n } be the etended or non-etended lattice of nodes of the polnomial f (, ). Let G 1 = conv(a i1,1,..., A i1,k ),..., G l = conv(a il,1,..., A il,k ), l 2, with {i 1,1,..., i 1,k,..., i l,1,..., i l,k } = {1,..., n} be a super-covering of P b l congruent k gons. Due to the fact that the composition of two translations is also a translation, we conclude that, for an of the G p and G q, p q, p, q {1,..., l} there eists a translation τ p,q, such that τ p,q (G p ) = G q. For each polgon, we list vertices in such a wa that we firstl list the verte with smallest coordinate. If such verte is not unique, we choose the one having simultaneousl smallest coordinate. Then we list the other vertices in counterclockwise order. It is clear that τ p,q (A ip,w ) = A iq,w, for an of the p and q, p q, p, q {1,..., l} and each w = 1,..., k. Let us denote b coef(a i ) the coefficient of the monomial of the polnomial f (, ) corresponding to the eponent vector A i. Assume that polgons G 1, G 2,..., G l have no common node. We sa that the super-covering of P is suitable super-covering with respect to the coefficients of the polnomial f (, ) if coef(a i1,1 ) : coef(a i1,2 ) :... : coef(a i1,k ) =... = coef(a il,1 ) : coef(a il,2 ) :... : coef(a il,k ). Assume that polgons G 1,..., G l have common nodes. Each G i, i = 1,..., l, determines a polnomial p i (, ) such that f (, ) = p 1 (, ) +... + p l (, ), where polnomial summands p i (, ), i = 1,..., l, are ordered in the same wa as the vertices. For each node A c, that is common for s polgons, the coefficient of the monomial whose eponent vector is A c is partitioned into s summands such that ever summand belongs to one and onl one p i (, ), corresponding to the polgons having a common node A c, in a wa that the coefficients of p 1 (, ),...,p l (, ), p i (, ) = c i,1 α i,1 β i,1 +... + c i,k α i,k β i,k, are proportional, i.e. c 1,1 : c 1,2 :... : c 1,k =... = c l,1 : c l,2 :... : c l,k. If the above holds for ever common node, we sa that the super-covering is suitable. From Theorem 0.11 it follows that integral indecomposabilit of the corresponding Newton polgon of a polnomial, that is completel determined b its vertices, implies irreducibilit of a polnomial. Moreover, an polnomial having the same non-zero terms is also absolutel irreducible over that field. It remains irreducible if we add monomials whose eponent vectors lie in the inner area of the polgon. On the other hand, the Newton polgon of a polnomial does not carr the entire information on the eistence of the polnomial factorization. In what follows we are going to see that, when factoring a polnomial, it is important to consider the vertices that determine Newton polgon, as well as the points disregarded for vertices of the Newton polgon and integer points captured b the polgon. Remark 0.14. Note that it is sufficient to discuss possible factorization of polnomials non-divisible either b or b, and therefore, having no trivial factorization. For eample, multipling b and means, in the sense of Newton polgon, translation for vectors (1, 0) and (0, 1). Since we are not interested in the trivial factorizations, in the following we eamine polnomials non-divisible either b or b. So, if we intend to eamine possible non-trivial factorization of a polnomial divisible b, or both of them, firstl we etract a trivial factor α, β or α β and discuss possible factorizations of the non-trivial factor-polnomial. Let us formulate a necessar and sufficient condition for the eistence of the non-trivial integer factorization of the bivariate polnomial with integer coefficients. Theorem 0.15. Let f (, ) be a non-zero bivariate polnomial over Z. Polnomial f (, ) has non-trivial integer factorization if and onl if its lattice of nodes, possibl etended b some integer points captured b the Newton polgon of the polnomial f (, ), has suitable super-covering with respect to the coefficients of f (, ). Proof: ( ) Assume that f (, ) has non-trivial integer factorization, i.e. there eist integer polnomials (, ) and h(, ) both having, at least, two monomials with non-zero terms such that f (, ) = (, )h(, ). Let h(, ) = c 1 α 1 β 1 +... + c k α k β k, with ci 0, for at least, two i, i = 1,..., k. Let c p 0 and c q 0. It is clear that either (α p, β p ) (0, 0) or (α q, β q ) (0, 0). So, let us rewrite polnomial h(, ) in the following form h(, ) = c p α p β p + c q α q β q + c i α i β i, i I
S. Crvenković, I. Pavkov / Filomat 27:2 (2013), 215 226 219 where I {1,..., k} \ {p, q}, is a collection of all indices different from p and q, such that for each i I, c i 0. Clearl, if the polnomial h(, ) has no other non-zero terms ecept c p α p β p and c q α q β q, the set I is empt. We can rewrite the polnomial f(,) in the following wa f (, ) = (, ) ( c p α p β p + c q α q β q + c i α i i) β. Further, we obtain f (, ) = (, )c p α p β p + (, )c q α q β q + (, ) c i α i β i. Let us denote the Newton polgons of polnomials (, ) and f (, ) b P and P f. Due to the fact that the multiplication of the polnomial (, ) b monomials c p α p β p, c q α q β q, and eventuall others, corresponds to the translation of the polgon P for vectors (α p, β p ), (α q, β q ), and others, it is clear that the Newton polgon of f (, ) is the conve hull of polgon P translated b vectors (α p, β p ), (α q, β q ) and others. In other words, the lattice of nodes is covered b congruent polgons P. So, the super-covering of the lattice of nodes is obtained. From the representation above of the polnomial f (, ) it is obvious that, if monomials obtained in more than one wa eist, the are split into the polnomial summands (, )c p α p β p, (, )c q α q β q, and eventuall others, in the wa that appropriate coefficients of these polnomial summands are proportional as c p : c q :..., so it follows that such a covering is suitable super-covering. ( ) Assume that the lattice of nodes of f (, ), possibl etended b some integer points captured b the Newton polgon of the polnomial f (, ), has suitable super-covering with respect to the coefficients of f (, ). Further, we assume that this covering is reached b l congruent polgons G 1, G 2,..., G l. We group monomials of the polnomial f (, ) in the wa determined b polgons G 1, G 2,..., G l, l 2. We split monomials into polnomial summands that correspond to the nodes that are common for, at least, two polgons in order to obtain proportionalit of the coefficients of the polnomial summands. This is possible to achieve because of the fact that super-covering is suitable with respect to the coefficients of f (, ). It is obvious that such splitting of the monomials into polnomial summands defines a non-trivial integer factorization of the polnomial f (, ). Remark 0.16. Under the same conditions as in the previous theorem, the same holds if we consider the polnomial with rational coefficients. Eample 0.17. Consider a polnomial f (, ) over Z, i I f (, ) = 2 + 2 + 2 Z[, ]. The lattice of nodes of f (, ), shown in Figure 2, contains the vertices of the polgon (0, 2) and (2, 0), but also the point (1, 1) that is disregarded for the verte because of its collinearit with two consecutive vertices. i I Figure 2 The onl wa to achieve super-covering on this lattice is to cover it with two congruent line segments conv((2, 0), (1, 1)) and conv((1, 1), (0, 2)) having a common node (1, 1). This super-covering is suitable if the coefficient of the monomial corresponding to the comon node (1, 1), i.e., can be split in such a wa as to obtain proportionalit of the appropriate coefficients of polnomial summands induced b the super-covering. We use brackets to announce the grouping of the monomials f (, ) = ( 2 + a) + ((2 a) + 2 ).
S. Crvenković, I. Pavkov / Filomat 27:2 (2013), 215 226 220 Line segment conv((1, 1), (0, 2)) is obtained from the line segment conv((2, 0), (1, 1)) b the translation for the vector ( 1, 1). If we denote such translation b τ, it follows that τ((2, 0)) = (1, 1) and τ((1, 1)) = (0, 2). In the first brackets, the order of the monomials is such that firstl we list the monomial corresponding to the eponent vector (2, 0) and secondl the monomial corresponding to the eponent vector (1, 1). In the second brackets, firstl we list the monomial corresponding to the eponent vector τ((2, 0)) and secondl the monomial corresponding to the eponent vector τ((1, 1)). The super-covering is suitable if 1 : a = (2 a) : 1, i.e. a = 1. We have f (, ) = ( + ) + ( + ) = ( + )( + ). Eample 0.18. Consider a polnomial f (, ) over Z p(, ) = 2 2 Z[, ]. The eponent vectors corresponding to non-zero terms of the polnomial p(, ) are (0, 2) and (2, 0) and the Newton polgon of the polnomial p(, ) is a line segment with endpoints (0, 2) and (2, 0) as in the previous eample. It is not possible to reach a factorization of this polnomial if we onl consider the vertices of its Newton polgon. Nevertheless, this line segment contains an integer point (1, 1) that corresponds to the monomial with coefficient 0. Consider the etended lattice of nodes that consists of vertices (0, 2) and (2, 0) together with the point (1, 1). We reach the super-covering of the lattice in the same wa as in the previous eample. So let us rewrite polnomial p(, ) in the following wa p(, ) = 2 + 0 2. This super-covering is suitable if we can choose an appropriate wa to represent zero in order to obtain proportionalit of the appropriate coefficients of polnomial summands p(, ) = 2 + ( a + a) 2 = ( 2 a) + (a 2 ), a Z. As 1:(- a) = a:(-1), i.e. a 2 = 1, we conclude that suitable super-covering is reached for a = 1 and a = 1. Choosing a = 1, we get p(, ) = ( 2 ) + ( 2 ) = ( ) + ( ) = ( + )( ). Remark 0.19. Choosing a = 1 leads to the same factorization of the polnomial p(, ) = ( 2 + ) + ( 2 ) = ( + ) ( + ) = ( )( + ). Eample 0.20. Consider a polnomial p(, ) over Z p(, ) = 2 2 2 = 2 + 0 2 2 Z[, ]. As in Eample 0.17 and Eample 0.18, the etended lattice of nodes consists of points (0, 2), (2, 0) and (1, 1) and we cover it in the same wa. This super-covering is suitable if the monomial 0 can be split in such a wa as to obtain proportionalit of the appropriate coefficients, i.e. p(, ) = 2 a + a 2 2 = ( 2 a) + (a 2 2 ), a Z, with 1 : ( a) = a : ( 2), i.e. a 2 = 2. There is no a, a Z, such that a 2 = 2. Therefore, the polnomial f (, ) has no integer factorization. Since a = 2 R, p(, ) has factorization over the field of real numbers Eample 0.21. Consider a polnomial r(, ) over Z p(, ) = ( 2)( + 2). r(, ) = 2 4 2 = 2 + 0 4 2 Z[, ]
S. Crvenković, I. Pavkov / Filomat 27:2 (2013), 215 226 221 with the same lattice of nodes as in the previous eamples and we cover it in the same wa. This supercovering is suitable if the polnomial can be written in the following form r(, ) = 2 + 0 4 2 = ( 2 a) + (a 4 2 ), a Z with 1 : ( a) = a : ( 4), i.e. a = 2 or a = 2. Choosing a = 2, we get r(, ) = ( 2 2) + (2 4 2 ) = ( 2) + 2( 2) = ( + 2)( 2). Remark 0.22. We have stated earlier that, if we want to factor a bivariate polnomial, it is important to consider the vertices that determine the Newton polgon as well as the points disregarded as vertices of the Newton polgon, because of their collinearit with two consecutive vertices or their position inside the constructed polgon. In other words, all the points corresponding to an of the eponent vectors are important for finding the possible factorizations of a polnomial. In the previous eamples we have shown that integer points, that do not correspond to eponent vectors but lie inside the Newton polgon of a polnomial, ma also be important for its factorization. Remark 0.23. It would be wrong to conclude that if all the points that correspond to eponent vectors of a polnomial are vertices of its Newton polgon and the polgon has no other integer points in its inner area, the polnomial is irreducible. In other words, the inner points of the Newton polgon pla an important role in finding possible factorizations, but their absence does not necessar impl irreducibilit of the polnomial. This will be illustrated b the following eample. Eample 0.24. Consider a polnomial f (, ) over Z f (, ) = + + + 1 Z[, ]. The terms of the polnomial f (, ) with non-zero coefficients correspond to the following eponent vectors: (1, 1), (1, 0), (0, 1) i (0, 0). The Newton polgon of f (, ) is a square with integer vertices shown in Figure 3 having no other integer points in its inner area. Figure 3 P f can be integrall decomposed in a wa as shown in Figure 4, i.e. P f = conv((1, 1), (1, 0), (0, 1), (0, 0)) = conv((0, 0), (1, 0)) + conv((0, 0), (0, 1)). Figure 4
S. Crvenković, I. Pavkov / Filomat 27:2 (2013), 215 226 222 Since the Newton polgon of the polnomial f (, ) is integrall decomposable, the polnomial f (, ) can be either reducible or irreducible. We cover the lattice with two congruent line segments and factor f (, ) in the following wa f (, ) = ( + 1)( + 1). Note that the Newton polgon of f (, ) does not capture an integer point ecept its vertices. However, f (, ) has integer factorization. Due to the absence of the inner integer points, it is understandable that there is no monomial obtained in more than one wa from the factor-polnomials of f (, ). Eample 0.25. Consider a polnomial f (, ) over Z f (, ) = 2 2 + 2 + 2 + 1 Z[, ]. The terms of the polnomial f (, ) with non-zero coefficients correspond to the following eponent vectors: (2, 2), (2, 0), (0, 2) and (0, 0). The Newton polgon of the polnomial f (, ) is a square with integer vertices shown in the Figure 5. Figure 5 The polnomial f (, ) has following integer factorization f (, ) = ( 2 + 1)( 2 + 1). Remark 0.26. Note that the Newton polgon of the polnomial f (, ) contains the following integer points in its inner area: (0, 1), (1, 0), (1, 2), (2, 1) and (1, 1). These points are not of interest for polnomial factorization because super-covering can be reached on the non-etended lattice of nodes. Eample 0.27. Consider the following bivariate polnomials over Z f (, ) = 4 + 2 3 + 3 2 + 3 + 2 + + Z[, ], (, ) = 4 + 2 3 + 2 2 + 3 + 2 + + Z[, ], h(, ) = 4 + 2 3 + 5 2 + 3 + 2 + + Z[, ]. Since these polnomials have the same non-zero terms, their non-etended lattices of nodes are the same and Newton polgons are also the same, i.e. P f = P = P h. Let A = (1, 0), B = (0, 1), C = (1, 2), D = (0, 3), E = (1, 4), F = (2, 1) and G = (2, 3). The polgon P f is shown in Figure 6. E D G C B F A Figure 6
S. Crvenković, I. Pavkov / Filomat 27:2 (2013), 215 226 223 The onl point from the non-etended lattice of nodes that is not a verte of the Newton polgon is C and, if the polnomial is reducible, the monomial that corresponds to the point C = (1, 2), that is 3 2, would be obtained from the factor-polnomials in more than one wa. Since its coefficient in f (, ) is 3 and all the other coefficients are 1, intuitivel it is clear that, in order to obtain suitable super-covering with respect to coefficients of the polnomial f (, ), we have to cover the non-etended lattice of nodes b three congruent figures, such that each node belongs to one and onl one figure, ecept node C which should be common to all of these figures. Such covering b three congruent triangles ABC, CDE and FCG is shown in Figure 7. E D G C F B A Figure 7 In order to obtain factorization of the polnomial, we group monomials in such a wa that the vertices of each triangle are collected in the same brackets. Nevertheless, 2 appears in each of these three groups. This is appropriate because it is 2 + 2 + 2 = 3 2. We have f (, ) = ( + + 2 ) + ( 2 + 3 + 4 ) + ( 2 + 2 + 2 3 ). As triangles CDE and FCG are obtained from the triangle ABC b translation for vectors (0, 2) and (1, 1), we etract from the second brackets 2 and from the third f (, ) = ( + + 2 ) + 2 ( + + 2 ) + ( + + 2 ) = (1 + 2 + )( + + 2 ). Since the Newton polgons of the factor-polnomials 1 + 2 + and + + 2 have no super-covering, we conclude that these polnomials have no integer factorization. It is eas to prove that another possible suitable super-covering of this lattice of nodes with respect to coefficients of the polnomial f (, ) shown in Figure 8 leads to the same factorization. E D G C B F A Figure 8 Nevertheless, super-coverings of the lattice of nodes shown in Figure 7 and Figure 8 are not suitable with respect to the coefficients of (, ). Due to the fact that the coefficient of the monomial 2 is 2, suitable covering of the lattice is the covering b two congruent figures such that each node belongs to one and onl one figure, ecept the node (1, 2), which is common to these two figures. Such covering is given in Figure 9.
S. Crvenković, I. Pavkov / Filomat 27:2 (2013), 215 226 224 E D G C B F A Figure 9 Analogousl, we obtain integer factorization induced b this covering (, ) = 4 + 2 3 + 2 2 + 3 + 2 + + = ( + + 2 + 2 ) + ( 2 + 3 + 4 + 2 3 ) (, ) = ( + + 2 + 2 ) + 2 ( + + 2 + 2 ) = (1 + 2 )( + + 2 + 2 ). It is eas to prove that the polnomial 1 + 2 has no integer factorization. Further, we factor polnomial + + 2 + 2 with the lattice of nodes shown in Figure 10. Figure 10 We cover the lattice b two congruent line segments, as shown in Figure 11. Figure 11 This covering implies grouping of the monomials in the following wa + + 2 + 2 = ( + ) + ( 2 + 2 ). Since 1 : 1 = 1 : 1, the proportionalit of the coefficients of the polnomial summands is obtained. So, we conclude that this covering is suitable super-covering of the lattice of nodes of the polnomial ++ 2 + 2. Further, we obtain ( + ) + ( 2 + 2 ) = ( + ) + ( + ) = (1 + )( + ). So, the polnomial (, ) has integer factorization, that is (, ) = (1 + 2 )(1 + )( + ).
S. Crvenković, I. Pavkov / Filomat 27:2 (2013), 215 226 225 E D C G B F A Figure 12 Another suitable covering of the lattice of nodes with respect to the coefficients of the polnomial (, ) is shown in Figure 12 and it leads to the same integer factorization of (, ). Finall, consider the polnomial h(, ) and super-covering of its lattice of nodes shown in Figure 7. We group monomials in the wa induced b this covering. h(, ) = ( + + a 2 ) + (b 2 + 3 + 4 ) + ( 2 + c 2 + 2 3 ) Z[, ], a + b + c = 5. This covering is not suitable with respect to the coefficients of the polnomial h(, ) since: 1 : 1 : a = b : 1 : 1 = 1 : c : 1, implies: i.e., a = b = c = 1, a + b + c = 3. Completel analogousl, for each super-covering reached on its lattice of nodes, it can be shown that it is not suitable with respect to the coefficients of h(, ), so we conclude that h(, ) has no non-trivial integer factorization. Eample 0.28. Consider following polnomial over Z h(, ) = 9 4 + 6 2 3 + 10 2 + 6 3 + 2 2 + + 2 Z[, ]. Since this polnomial has the same non-zero terms as the polnomials f (, ) and (, ) in the previous eample, the lattice of nodes is the same too. Super-covering of the lattice shown in the Figure 7 is suitable super-covering with respect to the coefficients of the polnomial h(, ) if it can be written in the form of the sum of three polnomials determined b triangles ABC, CDE and FCG, while monomial 10 2 is split in order to satisf the proportionalit of appropriate coefficients if possible h(, ) = ( + 2 + 3 2 ) + (3 2 + 6 3 + 9 4 ) + (2 2 + 4 2 + 6 2 3 ). As 1 : 2 : 3 = 3 : 6 : 9 = 2 : 4 : 6, the proportionalit is satisfied. Further, we get h(, ) = ( + 2 + 3 2 ) + 3 2 ( + 2 + 3 2 ) + 2( + 2 + 3 2 ), i.e., h(, ) = (1 + 3 2 + 2)( + 2 + 3 2 ). It is eas to prove that both of the Newton polgons of the factor-polnomials of h(, ) have no supercovering.
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