Proof: In this case a prime element has the property that if it divides a quantity, then it must divide one of the factors of that quantity.
If two elements are associates then they divide into each other.
First I prove that if an integral domain is a unique factorization domain, then every irreducible element is prime.
Suppose an element is factored in two ways into irreducible elements, and .
Lemma 1: A prime factorization exists.
Let , assume , and by induction on assume that there is a prime factorization for all . Either is prime or it can be written as a product of two proper divisors . Both and are less than so they have prime factorizations by the induction hypothesis. The prime factorization of is the result of putting the prime factorizations for and side by side.
Let be the prime factorization. Since is prime and divides , it divides a factor for some . So each is an associate of a prime for some , and therefore each is also prime. Cancelling each pair of associates leads to the fact .
Second I prove that if in an integral domain every irreducible element is prime, then the integral domain is a unique factorization domain.
The properties of a UFD to be shown are:
- Existence of factorizations is true. (Given by theorem). - The irreducible factorization of an element is unique in the following sense: If is factored in two ways into irreducible elements, say , then , and with suitable ordering of the factors, is an associate of for each .
If is factored in two ways into irreducible elements, , then for each , is prime and therefore is an associate of a factor for some , . Cancelling pairs of associates (integral domains have the cancellation law) yeilds the fact .