On Certain Divisibility Property of Polynomials over Integral Domains

Received: February 14, 2011 Accepted: February 28, 2011 doi:10.5539/jmr.v3n3p28 Abstract An integral domain R is a degree-domain if for given two polynomials f (x) and g(x) in R[x] such that for all k ∈ R (g(k) 0 ⇒ g(k)| f (k)), then f (x) = 0 or deg f ≥ deg g. We prove that the ring of integers OL is a degree-domain, where Q ⊆ L is a finite Galois extension. Then we study degree-domains that are also unique factorization domains to determine divisibility of polynomials using polynomial evaluations.


Introduction
All the rings are assumed to be commutative with identity.
Definition 1.An integral domain R is a degree-domain if given two polynomials g(x), f (x) ∈ R [x] such that for all k ∈ R, (g(k) 0 ⇒ g(k)| f (k)) then f (x) = 0 or deg f ≥ deg g.
Note that fields cannot be degree-domains.
In Section 2, we prove that Z is a degree-domain.We also present an example of an integral domain that is neither a field nor a degree-domain.In Section 3, we show that the ring of integers O L is also a degree-domain, where Q ⊆ L is a finite Galois extension of fields.In Section 4, we study divisibility of polynomials over degree-domains that are also unique factorization domains.The obtained results allow us to determine non-trivial conditions on polynomials f (x) and g(x) with integer coefficients such that the following statement holds: ( * ) Note that the statement ( * ) does not always hold: let p be a prime number and consider the polynomials g(x) = p and f (x) = x p − x, it follows (from Fermat's Little Theorem) that g(n)| f (n) for all integers n but clearly g(x) f (x) in Z[x].
For all n ≥ 0 consider p n (x) and q n (x) defined as follows.
In (Jones J.P. & Matiyasevich Y.V., 1991, Equation (2.14)) it is proved that if a ≥ 2 then p n (a)|q 2n (a).Can we say that p n (x)|q 2n (x) as polynomials?Consider the particular case n = 4: The above calculations show that p 4 (x)|q 8 (x).Using the results obtained in Section 4, we prove that indeed p n (x)|q 2n (x) in Z[x], and hence the polynomials in ( * * ) provide non-trivial examples where the statement ( * ) holds.
This paper is based on results from the second author M.Sc.thesis (Vélez-Marulanda, J.A., 2005), which was based on results from the first author Ph.D. thesis (Cáceres, L.F., 1998).The latter advised the former in the writing of his thesis.

Rings that are degree-domains
Lemma 2. The integral domain Z is a degree-domain.
Proof.Let g(x) and f (x) be polynomials with integer coefficients such that f (x) 0 and deg g > deg f .If follows that lim . This argument proves Lemma 2 by contradiction.
Example 3. Let Q be the set consisting of prime numbers p such that p = 2 or p ≡ 1 mod 4. Consider the domain Z [W] where W = {1/p : p ∈ Q}.Note that the non-integer elements in Z[W] are of the form c/d where c and d are relatively prime and p is a prime factor of d if and only if p ∈ Q.Moreover, an element c/d is a unit in Z[W] if and only if any prime factor of c is an element of Q.To see this, assume that (c/d)(u/t) = 1 for some u/t in Z[W] and let p be a prime factor of c.Since cu = dt and c and d are assumed to be relatively prime then p is a prime factor of t.Since u/t is an element of Z[W] with u and t relatively prime, then with a and b relatively prime.Look at g(x) = x 2 + 1 as a polynomial with coefficients in Z[W] (note in particular that g(r) 0 for all r ∈ Z So assume that a 2 + b 2 ≡ 0 mod p for some odd prime p.The condition that a and b are relatively prime implies that a or b , say a, is relatively prime to p. Let a satisfying aa ≡ 1 mod p.It follows that 1 + (ba ) 2 ≡ (aa ) 2 + (ba ) 2 ≡ 0 mod p, which implies that (ba ) 2 ≡ −1 mod p making −1 a quadratic residue of p. Therefore p ≡ 1 mod 4 (see (Burton, D.M., 2002, Theorem 9.2)), and hence p ∈ Q.This argument together with the observation above shows that g , which contradicts our choice of t.Therefore f (x, y) = 0 or deg y f ≥ deg y g.
We have the following direct consequence of Proposition 4.
Corollary 5. Let R be an integral domain.Then the ring of polynomials R[x] is a degree-domain.
Proof.Assume that R[x] is not a degree-domain.Then there exist two polynomials f (x, y) and g(x, y) in R [x, y] such that for all p(x) ∈ R[x] (g(x, p(x)) 0 ⇒ g(x, p(x))| f (x, p(x))) but f (x, y) 0 and deg y g > deg y f .Note that the latter implies that g(x, y) 0. Let t ∈ Z + sufficiently large such that g(x, x t ) 0. Using the assumptions on the polynomials f (x, y) and g(x, y), we obtain that g(x, x t )| f (x, x t ).It follows from Proposition 4 that f (x, y) = 0 or deg y g ≤ deg y f , which is a contradiction.

Ring of Integers
We review the definition of integral elements.
Let R be a subring of a ring L.An element α ∈ L is integral over R if there exists a monic polynomial f (x) ∈ R[x] such that f (α) = 0.In particular, when R = Z, the element α is said to be an algebraic integer in L. It is well-known that the set C consisting of all the elements that are integral over R is a ring which is called the integral closure of R in L. In particular, if R = Z and L is a field containing Z, the integral closure of Z in L is called the ring of integers of L, and we denote this ring by O L .For example, let d be a square-free integer and consider Q( We say that an integral domain R is integrally closed if R is equal to its integral closure in its field of fractions.In particular, Z is integrally closed. Proposition 6.Let R be an integral domain and K be its field of fractions.Assume that K ⊆ L is a finite Galois extension of fields and let C be the integral closure of R in L. Then σ(C) = C for all σ ∈ Gal(L/K), where Gal(L/K) denotes the Galois group of the extension For a proof of Proposition 6, see (Lorenzini, D., 1996, Chapter 1, Proposition 2.19 (iv)).
Let R, K, L and C as in the hypotheses of Proposition 6.For all p(x) Lemma 7. Let R, K, L and C be as in the hypotheses of Proposition 6 with R integrally closed and let p(x) Proof.Statement (i) follows directly from the fact that R ⊆ K. Assume that p(x) = p σ (x) for all σ ∈ Gal(L/K).Then the coefficients of p(x) are fixed by any element of Gal(L/K).Since R is integrally closed, the second conclusion in Proposition 6 implies that p Let R, K, L and C be as in the hypotheses of Proposition 6.For all p(x) ) and let τ be a fixed element in Gal(L/K).Then τ • σ ∈ Gal(L/K) and (p τ ) σ (x) = p τ•σ (x) for all σ ∈ Gal(L/K).Note that τ induces a permutation of the finite group Gal(L/K).Then ), which implies by Lemma 7(ii) that q(x) ∈ R[x] proving (i).Note that (ii) is a direct consequence of (i) and the statement (iii) follows from Lemma 7(i).
Proposition 9. Let R, K, L and C as in the hypotheses of Proposition 6 with R integrally closed.If R is a degree-domain then the ring C is also a degree-domain.
Proof.Let f (x) and g(x) be two polynomials in Since Z is integrally closed, the following result follows directly from Proposition 9.
Corollary 10.Let Q ⊆ L be a finite Galois extension.Then the ring of integers O L is a degree-domain.

Degree-domains that are unique factorization domains
Let R be a unique factorization domain.For all p(x) ∈ R[x] we denote by C(p) the content of p(x), i.e., the greatest common divisor of the coefficients of p(x).Remember that p(x) ∈ R[x] is said to be primitive if C(p) is a unit.Gauss Lemma states that the product of two primitive polynomials over a unique factorization domain is also primitive.This implies that if f (x), g(x) and h(x) are polynomials in R[x] with g(x) primitive and m f (x) = h(x)g(x) for some m ∈ R, there exists a polynomial q(x) ∈ R[x] such that h(x) = mq(x).
Proposition 10.Let R be a unique factorization domain and let K be its field of fractions.The following properties are equivalent.
(i) R is a degree-domain.
Hence, assume that deg r < deg g and denote α = a s n β.Thus for all k ∈ R with g(k) 0 we have both g(k)|α f (k) and g(k)|g(k)q(k), which implies g(k)|r(k).Using the hypothesis for the polynomials g(x) and r(x) we obtain r(x) = 0 or deg r ≥ deg g.Therefore r(x) = 0 and hence α f (x) = g(x)q(x).It follows that Example.Let n ≥ 1 and consider the polynomials p n (x), q n (x) in ( * * ).For all a ≥ 2 we know that p n (a)|q 2n (a); looking at the proof of this in (Jones J.P. & Matiyasevich Y.V., 1991, Equation (2.14)), we see that it can be extended to any a ∈ Z with |a| ≥ 2. Observe also that the polynomials p n (x) are primitive.Applying Corollary 11 to the polynomials g(x) = p n (x) and f (x) = q 2n (x), we obtain that p n (x)|q 2n (x) in Z[x].
to be the polynomial σ∈Gal(L/K) p σ (x).Note that deg N L/K (p) = |Gal(L/K)| deg p and N L/K (p)(x) = 0 if and only if p(x) = 0. Lemma 8. Let R, K, L and C as in the hypotheses of Proposition 6 with R integrally closed and let p(x) ∈ C[x].
Let b be an arbitrary element of R such that G(b) 0. It follows that g(b) 0, and hence g(b)| f (b).Therefore σ(g(b))|σ( f (b)) for all σ ∈ Gal(L/K).Using properties of divisibility together with Lemma 8(ii) and Lemma 8(iii) we obtain that G is a constant polynomial and deg f < deg g = 0 then f (x) = 0. Suppose that deg g ≥ 1 and let h(x) be a primitive polynomial in R[x] such that g(x) = C(g)h(x).By hypothesis, for all k ∈ R we have (h(k) 0 ⇒ h(k)| f (k)), which implies h(x)| f (x) in R[x].It follows that f (x) = 0 or deg f ≥ deg h = deg g.In both cases for g(x) we obtain that f (x) = 0 or deg g ≤ deg f .Therefore R is a degree-domain.The following result, which is a consequence of Proposition 10 together with Lemma 2 provides non-trivial conditions on polynomials f (x) and g(x) such that statement ( * ) holds.Corollary 11.Let f (x), g(x) ∈ Z[x] with g(x) a non-constant and primitive such that for all k ∈ Z, (g(k) 0 ⇒ g(k)| f (k)).Then g(x)| f (x) in Z[x].