MA3412S2_Hil2014.pdf

# The following result is a generalization of gausss

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The following result is a generalization of Gauss’s Lemma concerning products of primitive polynomials with integer coefficients. Lemma 2.27 Let R be a unique factorization domain, and let f ( x ) and g ( x ) be polynomials with coefficients in R . If f ( x ) and g ( x ) are both primitive then so is their product f ( x ) g ( x ) . Proof Let p be a prime element of R , and let R p = R/ ( p ). Then ( p ) is a prime ideal of R , and therefore the quotient ring R p is an integral domain (Lemma 2.14). Let ν p : R R p be the quotient homomorphism defined such that ν p ( a ) = a + ( p ) for all a R . Then ν p induces a ring homomorphism ν p * : R [ x ] R p [ x ], where ν p * m X k =0 a k x k ! = m X k =0 ν p ( a k ) x k for all a 0 , a 1 , . . . , a m R . Let f = ν p ( f ) and g = ν p ( g ). Then f ( x ) and g ( x ) are polynomials with coefficients in the quotient ring R p whose coefficients are the images of the corresponding coefficients of f ( x ) and g ( x ) under the quotient homomorphism ν p : R R p . Moreover f ( x ) g ( x ) is a polynomial in R p [ x ] whose coefficients are the images of the corresponding coefficients of f ( x ) g ( x ) under the quotient homomorphism. 31

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Now f ( x ) is a primitive polynomial, and therefore the prime element p of R does not divide all the coefficients of f ( x ). It follows that the poly- nomial f ( x ) is a non-zero polynomial in R p [ x ]. Similarly g ( x ) is a non-zero polynomial in R p [ x ]. Now the coefficient ring R p is an integral domain. It follows that f ( x ) g ( x ) is a non-zero polynomial whose leading coefficient is the product of the leading coefficients of f ( x ) and g ( x ). Therefore f ( x ) g ( x ) has at least one coefficient that is not divisible by the prime element p of R . We have thus shown that there cannot exist any prime element of R that divides all the coefficients of f ( x ) g ( x ). It follows that f ( x ) g ( x ) is a primitive polynomial, as required. Lemma 2.28 Let f ( x ) and g ( x ) be polynomials with coefficients in a unique factorization domain R . Suppose that the polynomial f ( x ) is primitive and that there exists some non-zero element c of R such that f ( x ) divides cg ( x ) in the polynomial ring R [ x ] . Then f ( x ) divides g ( x ) in the polynomial ring R [ x ] . Proof The result follows immediately in the case where c is a unit of the coefficient ring R . Suppose that the primitive polynomial f ( x ) divides pg ( x ), where p is a prime element of R . Then pg ( x ) = f ( x ) h ( x ) for some non-zero polynomial h ( x ) with coefficients in R . Moreover there exists a primitive polynomial k ( x ) and a non-zero element b of R such that h ( x ) = bk ( x ) (Lemma 2.26). Then pg ( x ) = bf ( x ) k ( x ). But f ( x ) k ( x ), being a product of primitive polynomials, is itself a primitive polynomial (Lemma 2.27). It follows that at least one coefficient of f ( x ) k ( x ) is not divisible by the prime element p of R , and therefore p must divide b . Let b = pa . Then g ( x ) = af ( x ) k ( x ), and thus f ( x ) divides g ( x ) in the polynomial ring R [ x ].
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• Fall '16
• Jhon Smith
• Algebra, Integers, Prime number, Integral domain, Ring theory, Principal ideal domain

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