Unique Factorization Domain - Non-examples

Non-examples

• The quadratic integer ring of all complex numbers of the form, where a and b are integers. Then 6 factors as both (2)(3) and as . These truly are different factorizations, because the only units in this ring are 1 and −1; thus, none of 2, 3, and are associate. It is not hard to show that all four factors are irreducible as well, though this may not be obvious. See also algebraic integer.

• Most factor rings of a polynomial ring are not UFDs. Here are two examples:

• Let be any UFD. Then is not a UFD. The proof is in two parts.

First we must show, and are all irreducible. Grade by degree. Assume for a contradiction that has a factorization into two non-zero non-units. Since it is degree one, the two factors must be a degree one element and a degree zero element . This gives . In, then, the degree one element must be an element of the ideal, but the non-zero elements of that ideal are degree two and higher. Consequently, must be zero in . That implies that, so is a unit, which is a contradiction., and are irreducible by the same argument.

Next, the element equals the element because of the relation . That means that and are two different factorizations of the same element into irreducibles, so is not a UFD.

• The ring of holomorphic functions in a single complex variable is not a UFD, since there exist holomorphic functions with an infinity of zeros, and thus an infinity of irreducible factors, while a UFD factorization must be finite, e.g.:

.

• A Noetherian domain is not necessarily a UFD. Although any non-zero non-unit in a Noetherian domain is the product of irreducible elements, this product is not necessarily unique.