Alg7.2

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Theorem: Let R\, be an integral domain. Suppose that existence of factorizations holds in R\,. Then R\, is a unique factorization domain if and only if every irreducible element is prime.

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 R\, is a unique factorization domain, then every irreducible element is prime.

Suppose an element k\in R\, is factored in two ways into irreducible elements, a_{1},...,a_{m}\, and b_{1},...,b_{n}\,.

Lemma 1: A prime factorization exists.

Let a,b\in {\mathbb  {Z}}\,, assume a>1\,, and by induction on a\, assume that there is a prime factorization for all b<a\,. Either a\, is prime or it can be written as a product of two proper divisors a=a_{1}\cdot a_{2}\,. Both a_{1}\, and a_{2}\, are less than a\, so they have prime factorizations by the induction hypothesis. The prime factorization of a\, is the result of putting the prime factorizations for a_{1}\, and a_{2}\, side by side.

Let a_{1},...,a_{m}\, be the prime factorization. Since a_{1}\, is prime and divides k\,, it divides a factor b_{i}\, for some i,1\leq i\leq m\,. So each b_{i}\, is an associate of a prime a_{j}\, for some j\,, 1\leq j\leq n\, and therefore each b_{j}\, is also prime. Cancelling each pair of associates leads to the fact m=n\,.

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 a\, is factored in two ways into irreducible elements, say a=p_{1}\cdot \cdot \cdot p_{m}=q_{1}\cdot \cdot \cdot q_{n}\,, then m=n\,, and with suitable ordering of the factors, p_{i}\, is an associate of q_{i}\, for each i\,.

If a\, is factored in two ways into irreducible elements, a=p_{1}\cdot \cdot \cdot p_{m}=q_{1}\cdot \cdot \cdot q_{n}\,, then for each i,1\leq i\leq m\,, p_{i}\, is prime and therefore is an associate of a factor q_{i}\, for some j\,, 1\leq j\leq n\,. Cancelling pairs of associates (integral domains have the cancellation law) yeilds the fact m=n\,. \Box \,

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