MATH
MA3412S2_Hil2014.pdf

# Proof suppose that the irreducible primitive

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Proof Suppose that the irreducible primitive polynomial f ( x ) does not di- vide g ( x ) in R [ x ]. We must prove that f ( x ) then divides h ( x ) in R [ x ]. Let I the ideal of R [ x ] generated by the polynomials f ( x ) and g ( x ), and let m be the smallest non-negative integer with the property that the ideal I contains a non-zero polynomial of degree m . Then there exists a primitive polynomial 34

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k ( x ) of degree m and a non-zero element a of R such that ak ( x ) I . It then follows from Lemma 2.31 that there exist non-zero elements c and d of R such that cf ( x ) and dg ( x ) are divisible by k ( x ). It then follows from Lemma 2.28 that f ( x ) and g ( x ) are both divisible by the primitive polynomial k ( x ) in R [ x ]. But f ( x ) is an irreducible element of R [ x ] (Lemma 2.29) and no asso- ciate of f ( x ) in R [ x ] can divide g ( x ). It follows that k ( x ) must be a unit of the ring R [ x ], and therefore k ( x ) is a polynomial of degree zero whose coeffi- cient u is a unit of the ring R . The polynomial of degree zero with coefficient au then belongs to the ideal I generated by f ( x ) and g ( x ). It follows that there exist polynomials q ( x ) and r ( x ) such that q ( x ) f ( x ) + r ( x ) g ( x ) = au . Then q ( x ) f ( x ) h ( x ) + r ( x ) g ( x ) h ( x ) = auh ( x ) . It follows that f ( x ) divides auh ( x ) in R [ x ], because f ( x ) divides g ( x ) h ( x ) in R [ x ]. It then follows from Lemma 2.28 that f ( x ) divides h ( x ) in R [ x ], as required. Proposition 2.33 Let R be a unique factorization domain. Then the ring R [ x ] of polynomials with coefficients in R is also a unique factorization do- main. Proof An integral domain is a unique factorization domain if and only if every non-zero element of the domain that is not a unit can be factored as the product of one or more prime elements of the domain. We have shown that any non-zero polynomial in R [ x ] that is not a unit can be factored as a product of irreducible elements of R [ x ], and that the irreducible elements of R [ x ] are the polynomials of degree zero whose coefficients are prime elements of R and the primitive irreducible polynomials of positive degree in R [ x ] (Lemma 2.29). Moreover the polynomials of degree zero whose coefficients are prime are prime elements of R [ x ] (Lemma 2.30), and the irreducible primitive polynomials are also prime elements of R [ x ]. (Lemma 2.32). Therefore every element of R [ x ] that is not a unit can be factored as a product of one or more prime elements of R [ x ], and thus R [ x ] is a unique factorization domain. Example It follows from Proposition 2.33 that the ring Z [ x ] of polynomials with integer coefficients is a unique factorization domain. This integral do- main is not a principal ideal domain. Indeed let p be a prime number, and let I p = { f ( x ) Z [ x ] : p divides f (0) } . Then I p is a prime ideal of Z [ x ], for if f ( x ) and g ( x ) are polynomials with integer coefficients, and if fg I p then p | f (0) g (0). But then either p | f (0) or 35
else p | g (0), and thus either f I p or g I p . The prime ideal I p is generated by the constant polynomial p and the polynomial x . However this prime ideal is not a principal ideal, because the only common divisors of p and x in Z [ x ] are the constant polynomials with values ± 1.

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• Fall '16
• Jhon Smith
• Algebra, Integers, Prime number, Integral domain, Ring theory, Principal ideal domain

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