"the prime factorization of an element, if it exists, is always unique"

Question

Why exactly? I'm seeing this in proofs that a ring is a UFD, but it's not jumping out from the ideal definition of primality.

Answer 1

Note: My proof uses the assumption that $R$ is an Integral Domain (in particular is commutative).

Suppose $a=p_1 p_2...p_n$ for primes $p_i \in R$. We show this prime factorization is unique.

Suppose $q_1 q_2 ...q_m$ is another prime factorization of $a$. $q_1$ divides $p_i$ for some $1\leq i \leq n$. To see this, note that $q_1$ divides $p_1 (p_2 ... p_n)$ implies $q_1$ divides $p_1$ or $(p_2 ... p_n)$. If is divides $p_1$ we are done, if not it divides $(p_2 ... p_n)$, etc.

We have $q_1|p_i$, i.e., $p_i=xq_1$ for some $x\in R$. If $p_i$ divides $x$ then you will get $yq_1=1$ for some $y\in R$ implying $q_1$ is a unit, contradiction. (Note, you need $R$ to be an integral domain for this argument to work). So $p_i|q_1$.

Thus $p_i$ and $q_1$ are associates. Equivalently $q_1=up_i$ for some unit $u\in R$. Now you can factor $p_i$ to get $p_i(p_1p_2 ...p_{i-1}p_{i+1}... p_n-uq_2q_3...q_n)=0$. Again using the assumption that $R$ is an integral domain, you have $p_1p_2 ...p_{i-1}p_{i+1}... p_n=uq_2q_3...q_m$. Rinse and repeat.

This way you cross off associates on each side of $p_1...p_n=q_1...q_m$ in pairs. In the end you must have $n=m$, else you get a unit with a prime factor.