I'm working with the following definition:
A integral domain $R$ is said to be an unique factorization domain if
$(i)$ every non-unit, non-zero element $r\in R$ has a decomposition $r = p_1\cdots p_n$ where each $p_i$ is a irreducible element of $R$.
$(ii)$ if there exists another decomposition $r = q_1\cdots q_m$ of irreducibles, then $n=m$ and there exists a bijection between the $p_i$ and $q_j$ where each $p_i$ is associated to a $q_j$ in the sense that $p_i = cq_j$ for some unit element $c$.
I want to prove that this definition of an UFD is equivalent to a definition consisting of the (i) above together with
$(ii)'$ if $a,b\in R$ and $p\in R$ is an irreducible such that $p|ab$, then $p|a$ or $p|b$.
I have already proved $(ii)\implies(ii)'$ but I'm stuck in some details on the other direction.
If we have two decompositions $r = p_1\cdots p_n = q_1\cdots q_m$ then applying $(ii)'$ we have that every $p_i$ is associated with some $q_j$, but I don't know how to show that $n=m$. Can someone help?
Best Answer
You can use the cancellation property for domains, namely that if $ab=ac$ for non-zero elements, then $a(b-c)=0$ and therefore $b-c=0$, i.e. $b=c$.
Then you can make some argument by reductio or induction, like, if $p_1\cdots p_n= q_1 \cdots q_m$ irreducible, then $p_1 = cq_k$ for some $k$, so $$cp_2\cdots p_n = q_1\cdots q_{k-1}q_{k+1} \cdots q_m$$ etc.