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\isin 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\isin \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 \le i \le m\,. So each b_i\, is an associate of a prime a_j\, for some j\,, 1\le j\le 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\le i \le m\,, p_i\, is prime and therefore is an associate of a factor q_i\, for some j\,, 1\le j\le n\,. Cancelling pairs of associates (integral domains have the cancellation law) yeilds the fact m=n\,. \Box\,

Abstract Algebra

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