An algebraic number field is a subfield of the

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An algebraic number field is a subfield of the complex numbers that is a finite-dimensional vector space over the field of rational numbers. A complex number is said to be an algebraic integer if it is the root of a monic polyno- mial with integer coefficients. The set of algebraic integers contained within any algebraic number field constitute an integral domain embedded within that number field. A significant theorem of algebraic number theory, due to Richard Dedekind, guarantees that every non-zero proper ideal of the ring of algebraic integers in any algebraic number field can be factored uniquely as a product of prime ideals of that ring. The integral domain Z [ - 5] is the ring of integers of the algebraic number field Q ( - 5) that consists of all complex numbers that are of the form a + b - 5 for some rational numbers a and b . Therefore every non-zero ideal of this domain must factorize uniquely as a product of prime ideals. 2.9 Rings of Polynomials with Coefficients in a Unique Factorization Domain Let R be a unique factorization domain. We shall prove that the ring R [ x ] of polynomials with coefficients in R is also a unique factorization domain. We say that a polynomial f ( x ) with coefficients in the unique factorization domain R is primitive if the only elements of R that divide all the coefficients of f ( x ) are the units of R . Any non-zero element of R that is not a unit of R can be factored as a product of one or more prime elements of R , and is thus divisible by some prime element of R . It follows that a polynomial f ( x ) with coefficients in R is primitive if and only if there is no prime element of R that divides all the coefficients of f ( x ). 30
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Lemma 2.26 Let R be a unique factorization domain, and let f ( x ) be a non-zero polynomial with coefficients in R . Then there exists a non-zero element c of R and a primitive polynomial ˆ f ( x ) with coefficients in R such that f ( x ) = c ˆ f ( x ) . Proof If the polynomial f ( x ) is itself primitive, there is nothing to prove. Suppose that f ( x ) is not primitive. Let m be the largest positive integer with the property that all coefficients of the polynomial f ( x ) are divisible by some product of m prime elements of R . Such a positive integer must exist, because the number of factors in a product of prime elements of R dividing a non-zero coefficient of f ( x ) cannot exceed the number of factors in a representation of that coefficient as a product of prime elements of the ring R . Let c be a non-zero element of R dividing all the coefficients of f ( x ) that is a product of m prime elements of R . Then f ( x ) = c ˆ f ( x ) for some polynomial ˆ f ( x ) with coefficients in R . Moreover the polynomial ˆ f ( x ) is primitive, for if it were not primitive, then there would exist some prime element p of R dividing all the coefficients of ˆ f ( x ), and then cp would divide all the coefficients of f ( x ) and would be a product of more than m prime elements of R , contradicting the definition of m . The result follows.
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  • Fall '16
  • Jhon Smith
  • Algebra, Integers, Prime number, Integral domain, Ring theory, Principal ideal domain

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