Search results
Results from the WOW.Com Content Network
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 ...
1 Examples. 2 References. 3 Notes. ... In mathematics, a noncommutative unique factorization domain is a noncommutative ring with the unique factorization property.
There is a unique ring homomorphism φ from Z[α] to Z/nZ that maps α to m. For simplicity, we'll assume that Z[α] is a unique factorization domain; the algorithm can be modified to work when it isn't, but then there are some additional complications. Next, we set up two parallel factor bases, one in Z[α] and one in Z.
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 ...
Examples of regular rings include fields (of dimension zero) and Dedekind domains. If A is regular then so is A [ X ], with dimension one greater than that of A . In particular if k is a field, the ring of integers, or a principal ideal domain , then the polynomial ring k [ X 1 , … , X n ] {\displaystyle k[X_{1},\ldots ,X_{n}]} is regular.
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 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 ...
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.