Question

Prove that every principal ideal in a UFD is a product of prime ideals uniquely except for the order of the factors.

Short answer

It is proved that every principal ideal in a unique factorization domain is a product of prime ideals uniquely except for the order of the factors.

Step-by-step solution

Unit factorization domain

If $\mathit{R}$ is a unique factorization domain then every non-zero, non-unit element is a finite product of irreducible and also every irreducible element is a prime.

Prove the required statement

Let $a\in R$

According to the definition of unique factorization domain,

$a=p_1{p}_2\mathrm{...}{p}_m$

The ring $<Z,+,\cdot >$of integers is a unique factorization domain. So, it is an integral domain with unity.

If $n\in Z$be any non-zero, and non-unit element of $Z$then if $n>0$,

If $n<0$,

Let $n=(-m)$where $m>0$then $m$can be expressed as the product of primes in $Z$.

$\begin{array}{c}m=q_1{q}_2\cdots q_k\\ (-m)=(-q_1)\left(q_2\right)\cdots \left(q_k\right)\\ n=(-q_1)\left(q_2\right)\cdots \left(q_k\right)\end{array}$

Therefore, if every principal ideal in a unique factorization domain is a product of prime ideals uniquely except for the order of the factors.