On Wikipedia, UFD is defined as an integral domain in which every element can be uniquely factored as product of primes (irreducibles), up to multiplication by units and arrangement.
My question is about why primes or irreducibles can be interchangeably placed in definition. I mean, in other words, are following two definitions equivalent?
(1) $R$ is an integral domain in which every element can be uniquely factored as product of primes (up to unit multiplication and permutation).
(2) $R$ is an integral domain in which every element can be uniquely factored as product of irreducibles (up to unit multiplication and permutation).
(from comments below, my question actually boils down to following:)
If in a domain, every element can be uniquely factored into product of primes, then certainly it can be decomposed into irreducibles; how can we ensure uniqueness into irreducible factorization?
Best Answer
Assume property (1); since any prime is irreducible, you have property (2).
Assume $R$ satisfies property (2). We want to see that any irreducible is prime, so also property (1) is satisfied.
Let $a$ be irreducible and suppose $a\mid bc$, so $bc=ad$ for some $d\ne0$ (the case when $bc=0$ is trivial). Decompose $b$, $c$ and $d$ into a product of irreducibles: \begin{align} b&=b_0b_1b_2\dots b_m && \text{$b_0$ invertible, $b_i$ irreducible for $1\le i\le m$} \\ c&=c_0c_1c_2\dots c_n && \text{$c_0$ invertible, $c_i$ irreducible for $1\le i\le n$} \\ d&=d_0d_1d_2\dots d_p && \text{$d_0$ invertible, $d_i$ irreducible for $1\le i\le p$} \end{align} By assumption (2), we deduce that $a=ub_i$, for some $i$ and some invertible $u$, or $a=vc_j$, for some $j$ and some invertible $v$.
In the former case, $a\mid b$; in the latter case, $a\mid c$. Therefore $a$ is prime.