Search results
Results from the WOW.Com Content Network
A Noetherian integral domain is a UFD if and only if every height 1 prime ideal is principal (a proof is given at the end). Also, a Dedekind domain is a UFD if and only if its ideal class group is trivial. In this case, it is in fact a principal ideal domain. In general, for an integral domain A, the following conditions are equivalent: A is a UFD.
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 ...
If is an integral domain, then is an irreducible element of if and only if, for all ,, the equation = implies that the ideal generated by is equal to the ideal generated by or equal to the ideal generated by . This equivalence does not hold for general commutative rings, which is why the assumption of the ring having no nonzero zero divisors is ...
A Bézout domain (i.e., an integral domain where every finitely generated ideal is principal) is a GCD domain. Unlike principal ideal domains (where every ideal is principal), a Bézout domain need not be a unique factorization domain; for instance the ring of entire functions is a non-atomic Bézout domain, and there are many other
In mathematics, a noncommutative unique factorization domain is a noncommutative ring with the unique factorization property. ... Statistics; Cookie statement;
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 the case of K[X], it may be stated as: every non-constant polynomial can be expressed in a unique way as the product of a constant, and one or several irreducible monic polynomials; this decomposition is unique up to the order of the factors. In other terms K[X] is a unique factorization domain.
This lack of unique factorization is a major difficulty for solving Diophantine equations. For example, many wrong proofs of Fermat's Last Theorem (probably including Fermat's "truly marvelous proof of this, which this margin is too narrow to contain") were based on the implicit supposition of unique factorization.