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Formally, a unique factorization domain is defined to be an integral domain R in which every non-zero element x of R which is not a unit can be written as a finite product of irreducible elements p i of R: x = p 1 p 2 ⋅⋅⋅ p n with n ≥ 1. and this representation is unique in the following sense: If q 1, ..., q m are irreducible elements ...
As the positive integers less than s have been supposed to have a unique prime factorization, must occur in the factorization of either or Q. The latter case is impossible, as Q , being smaller than s , must have a unique prime factorization, and p 1 {\displaystyle p_{1}} differs from every q j . {\displaystyle q_{j}.}
The class number of a number field is by definition the order of the ideal class group of its ring of integers. Thus, a number field has class number 1 if and only if its ring of integers is a principal ideal domain (and thus a unique factorization domain). The fundamental theorem of arithmetic says that Q has class number 1.
In the case of coefficients in a unique factorization domain R, "rational numbers" must be replaced by "field of fractions of R". This implies that, if R is either a field, the ring of integers, or a unique factorization domain, then every polynomial ring (in one or several indeterminates) over R is a unique factorization domain. Another ...
The irreducible elements are the terminal elements of a factorization process; that is, they are the factors that cannot be further factorized. If the irreducible factors of every non-zero non-unit element are uniquely defined, up to the multiplication by a unit, then the integral domain is called a unique factorization domain , but this does ...
In number theory, the fundamental theorem of ideal theory in number fields states that every nonzero proper ideal in the ring of integers of a number field admits unique factorization into a product of nonzero prime ideals. In other words, every ring of integers of a number field is a Dedekind domain.
The unique factorization property means that a non-zero non-unit r can be represented as a product of prime elements r = p 1 p 2 ⋯ p n {\displaystyle r=p_{1}p_{2}\cdots p_{n}} Then r is square-free if and only if the primes p i are pairwise non-associated (i.e. that it doesn't have two of the same prime as factors, which would make it ...
The integers and the polynomials over a field share the property of unique factorization, that is, every nonzero element may be factored into a product of an invertible element (a unit, ±1 in the case of integers) and a product of irreducible elements (prime numbers, in the case of integers), and this factorization is unique up to rearranging ...