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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 ...
Multiplication is defined for ideals, and the rings in which they have unique factorization are called Dedekind domains. There is a version of unique factorization for ordinals, though it requires some additional conditions to ensure uniqueness. Any commutative Möbius monoid satisfies a unique factorization theorem and thus possesses ...
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.
Template: Commutative ring ... Download QR code ... rngs ⊃ rings ⊃ commutative rings ⊃ integral domains ⊃ integrally closed domains ⊃ GCD domains ⊃ unique ...
Nagata, Masayoshi (1958), "A general theory of algebraic geometry over Dedekind domains. II. Separably generated extensions and regular local rings", American Journal of Mathematics, 80 (2): 382–420, doi:10.2307/2372791, ISSN 0002-9327, JSTOR 2372791, MR 0094344
In particular if k is a field, the ring of integers, or a principal ideal domain, then the polynomial ring [, …,] is regular. In the case of a field, this is Hilbert's syzygy theorem. Any localization of a regular ring is regular as well. A regular ring is reduced [b] but need not be an integral domain. For example, the product of two regular ...
The converse is true for unique factorization domains [2] (or, more generally, GCD domains). Moreover, while an ideal generated by a prime element is a prime ideal , it is not true in general that an ideal generated by an irreducible element is an irreducible ideal .
The polynomial ring [,,, …] in infinitely many variables over a unique factorization domain is a Krull domain which is not noetherian. Let A {\displaystyle A} be a Noetherian domain with quotient field K {\displaystyle K} , and L {\displaystyle L} be a finite algebraic extension of K {\displaystyle K} .