Selçuk J. Appl. Math. Selçuk Journal of
Vol. 6. No.2. pp. 3-7, 2005 Applied Mathematics
Irreducible Polynomials by the Polytope Method over Integral Do-mains
Fatih Koyuncu
Department of Mathematics, Mugla University, 48000 Mugla,Turkey; e-mail:fatih@ mu.edu.tr
Received : January 7, 2005
Summary. Ostrowski realized that there is a relation between the factorization of a multivariate polynomial over any …eld and the integral decomposition of the Newton polytope of this polynomial. Here, we extended this result to the multivariate polynomials over any integral domain. Moreover, we observed that this result is also valid for some special kind of multivariate polynomials over any commutative ring.
Key words: Polytopes, integral indecomposability, multivariate polynomials.
1. Introduction
Let Rndenote the n-dimensional Euclidean space and S be a subset of Rn: The smallest convex set containing S; denoted by conv(S), is called the convex hull of S: If S = fa1; a2; :::; ang is a …nite set then we shall write conv(a1; :::; an)
instead of conv(S). It can be shown easily that
conv(S) = ( k X i=1 ixi: xi2 S; i 0; k X i=1 i= 1 ) :
The principle operation for convex sets in Rn is de…ned as follows. De…nition 1 For any two sets A and B in Rn; the sum
A + B = fa + b : a 2 A; b 2 Bg is called Minkowski sum, or vector addition of A and B:
A point in Rn is called integral if its coordinates are integers. A polytope in Rn is called integral if all of its vertices are integral. An integral polytope C is called integrally decomposable if there exist integral polytopes A and B such that C = A + B where both A and B have at least two points. Otherwise, C is called integrally indecomposable.
De…nition 2 Let F be any …eld and consider any multivariate polynomial f (x1; x2; :::; xn) = X ce1e2:::enx e1 1 x e2 2 :::xenn 2 F [x1; :::; xn]:
We can think an exponent vector (e1; e2; :::; en) of f as a point in Rn: The
Newton polytope of f; denoted by Pf; is de…ned as the convex hull in Rn of all
the points (e1; :::; en) with ce1e2:::en 6= 0:
A polynomial over a …eld F is called absolutely irreducible if it remains irre-ducible over every algebraic extension of F .
Using Newton polytopes of multivariate polynomials, we can determine in…nite families of absolutely irreducible polynomials over an arbitrary …eld F by the following result due to Ostrowski [4], c.f. [1].
Lemma 1 Let f; g; h 2 F [x1; :::; xn] with f = gh: Then Pf = Pg+ Ph:
s a direct result of Lemma 1, we have the following corollary which is an irreducibility criterion for multivariate polynomials over arbitrary …elds. Corollary 1 Let F be any …eld and f a nonzero polynomial in F [x1; :::; xn] not
divisible by any xi: If the Newton polytope Pf of f is integrally indecomposable
then f is absolutely irreducible over F:
When Pf is integrally decomposable, depending on the given …eld, f may be
re-ducible or irrere-ducible. For example, the polynomial f = x9+y9+z9has the New-ton polytope Pf = conv((9; 0; 0); (0; 9; 0); (0; 0; 9)) = conv((6; 0; 0); (0; 6; 0); (0; 0; 6))+
conv((3; 0; 0); (0; 3; 0); (0; 0; 3)): But, while f = x9+ y9+ z9= (x + y + z)9over
F3; it is irreducible over F2; F5; F7; F11; where Fmrepresents the …nite …eld with
m elements. Moreover, the polynomial g = x6+ y6+ 1 has the decomposable
Newton polytope Pg = conv((0; 0); (6; 0); (0; 6)) = conv((0; 0); (3; 0); (0; 3)) +
conv((0; 0); (3; 0); (0; 3)) where as, g = (x3+ y3+ 1)2over F
2; g = (x2+ y2+ 1)3
over F3and g is irreducible over F5; F7; F11, F13; F41; F103:
In [1], [2] and [3], in…nitely many integrally indecomposable polytopes in Rn
are presented and then, being associated to these polytopes, in…nite families of absolutely irreducible polynomials are determined over any …eld F:
2. Main Result
We have observed that Lemma 1 is also true if we consider integral domains instead of …elds.
Lemma 2 Let R be an integral domain and f; g; h 2 R[x1; x2; :::; xn] with f = gh: Then Pf = Pg+ Ph: Proof. Let f (x1; x2; :::; xn) = X ce1e2:::enx e1 1 xe22:::xenn; g(x1; x2; :::; xn) = X c0e0 1e02:::e0nx e01 1 x e02 2 :::x e0n n ; h(x1; x2; :::; xn) = X c00e00 1e002:::e00nx e00 1 1 x e00 2 2 :::x e00 n n : Then, we have (1) f = X (e0 1;e02;:::;e0n) X (e00 1;e002;:::;e00n) c0e0 1e02:::e0nc 00 e00 1e002:::e00nx e01+e001 1 x e02+e002 2 :::x e0n+e00n n :
In this expanded product, let us assume that there are r terms containing xe1 1 x e2 2 :::xenn; and write (2) S = (d1+ ::: + dr)xe11x e2 2 :::x en n = ce1e2:::enx e1 1 x e2 2 :::x en n :
Note that S may drop out if r 2; but not if r = 1:
We will prove that the set f(e1; e2; :::; en)g of exponents of the polynomial f and
the set f(e0
1+ e001; e02+ e002; :::; e0n+ e00n)g of exponents of gh determine the same
polytope. From the relations (1) and (2), we see that every element of the set f(e1; e2; :::; en)g is contained in the set f(e01+e001; e02+e002; :::; e0n+e00n)g. Therefore,
we have Pf Pg+ Ph:
To prove the other inclusion Pg+ Ph Pf; we must show that every member of
the set f(e01+ e001; e02+ e200; :::; e0n+ e00n)g; which may be a vertex of or in Pf+ Pg; is
a member of the set f(e1; e2; :::; en)g: Note that this fact is not necessarily true,
in general, since d1+ ::: + dr may be equal to zero if r 2: But, this follows
from the fact that each vertex of Pf+ Pg is uniquely determined,
Let v be any vertex of Pg+ Ph: We will show that there are unique vertices v1
of Pg and v2 of Ph such that v = v1+ v2: Assume that there is another pair of
vertices v0
12 Pgand v022 Ph such that
(3) v = v1+ v2= v10 + v02: Then, we have v =1 2(v1+ v 0 2) + 1 2(v 0 1+ v2):
Since v1+ v20; v10 + v22 Pg+ Ph and v is a vertex of Pg+ Ph; we must have
Now, using (3) and (4) we get
v2 v02= v02 v2;
i.e. 2(v2 v20) = 0: Hence, we have v2= v02 and v1 = v10: Since v is a vertex of
Pg+ Ph; v1 and v2 must be vertices of Pg and Ph respectively.
Consequently, if the point v = (e1; e2; :::; en) is a vertex of Pg+ Ph; there is only
one term c0 e0 1e02:::e0nx e01 1 x e02 2 :::x e0n
n of g(x1; x2; :::; xn) and just one term c00e00 1e002:::e00nx e001 1 x e002 2 :::x e00n n of h(x1; x2; :::; xn) such that (e1; e2; :::; en) = (e01+ e001; e02+ e002; :::; e0n+ e00n): Since, by assumption, c0e0 1e02:::e0n6= 0 and c 00 e00
1e002:::e00n6= 0, we must have
c0e0
1e02:::e0n c
00 e00
1e002:::e00n 6= 0;
i.e. (e01+e001; e02+e002; :::; e0n+e00n) is a member of the set f(e1; e2; :::; en)g: Therefore,
there is a unique term in the expansion of g h that has v as its exponent vector. Thus, v 2 Pf: Hence, we have Pf Pg+ Ph:
As a result of Lemma 2, we have the following irreducibility criterion for multivariate polynomials over arbitrary integral domains.
Corollary 2 Let R be any integral domain and f a nonzero multivariate polynomial in R[x1; x2; :::; xn] not divisible by any xi: If the Newton polytope
Pf of f is integrally indecomposable then f is irreducible over any extension of
R:
Proof. Since f is not divisible by any xi; it has no factor having only one
term. Let f be reducible over some extension of R: This means that we have f = gh, where both g and h have at least two nonzero terms. Then, the Newton polytopes of g and h have at least two points. By Lemma 2, we have Pf = Pg+ Ph; which is a contradiction.
Remark 1In…nite families of absolutely irreducible polynomials are presented over any …eld F in [1], [2] and [3]. By Corollary 2, these examples are also valid over integral domains. So, for the examples of absolutely irreducible multivariate polynomials over integral domains, see examples there.
We have the following special case of Lemma 2
Proposition 1 Let R be a commutative ring and f; g; h 2 R[x1; x2; :::; xn] with
f = gh: If the coe¢ cients of the terms forming Pf in f and the coe¢ cients of
the terms forming Pg in g are not zero divisors in R then Pf= Pg+ Ph:
Finally, we give the following result of Proposition 1.
Corollary 3 Let R be a commutative ring and f a nonzero multivariate polyno-mial in R[x ; x ; :::; x ] not divisible by any x: If the coe¢ cients of the terms,
forming the Newton polytope Pf of f , are not zero divisors in R and Pf is
integrally indecomposable then f is irreducible over any extension of R: For example, the polynomial
f = 7x88+ 5y63+ 2x7+ 4y6+ 1 +Xcijxiyj2 Z8[x; y];
with (i; j) 2 Pf = conv((0; 0); (88; 0); (0; 63)); is irreducible over any extension
of Z8: Because, Pf is an integrally indecomposable triangle in R2 while 88 and
63 are relatively prime integers.
References
1. Gao S.(2001): Absolute irreducibility of polynomials via Newton polytopes, Journal of Algebra 237 , No.2 501-520.
2. Gao S.(2001): Decomposition of Polytopes and Polynomials, Discrete and Compu-tational Geometry 26 , no. 1, 89-104.
3. Koyuncu F. and Özbudak F.: A Geometric Approach to Absolute Irreducibility of Polynomials, submitted.
4. Ostrowski A.M. (1975): On multiplication and factorization of polynomials, I. Lexicographic orderings and extreme aggregates of terms, Aequationes Math. 13 , 201-228.
