# Alg7.2

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.
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 $m=n\,$. $\Box \,$