I am interested in verifying the existence aspect of the theorem asserting that every Principal Ideal Domain is a Unique Factorization Domain. In the first paragraph, I (think that I) have provided an explanation for an arbitrary nonzero element $r$ that is distinct from any unit in a Principal Ideal Domain $R$ being divisible by a prime (irreducible) $s_{N}$. Is that correct? In the second paragraph, I (think that I) have provided an explanation for $r$ being equal to an associate of a finite number of primes (irreducibles). Is that correct?
It seems like this is a straight forward argument, but I have not found anything like it on the Internet. Due to the pandemic, I have not been able to look at any textbooks in Abstract Algebra in college libraries. I have the textbook by Dummit and Foote, but the explanation in there is not clear.
If anyone would like to comment on this post, please tell me whether my argument is correct or flawed, and if it is flawed, please tell me what edits could be made to correct it. I don't want alternative arguments ... if the argument that I have can be easily corrected.
Existence
$R$ is a Principal Ideal Domain, and $r$ is a nonzero element which is not a unit. If $r$ were not divisible by a prime, it would itself be a reducible: there would elements $s_{1}$ and $t_{2}$, neither of which are units, such that $r = s_{1}t_{1}$. $\langle r \rangle \subsetneq \langle s_{1} \rangle \subsetneq R$. If $s_{1}$ were reducible, there would be elements $s_{2}$ and $t_{2}$, neither of which are units, such that $s_{1} = s_{2}t_{2}$. $\langle s_{1} \rangle \subsetneq \langle s_{2} \rangle \subsetneq R$. So, there would be a chain \begin{equation*} \langle r \rangle \subsetneq \langle s_{1} \rangle \subsetneq \langle s_{2} \rangle \subsetneq \cdots \subsetneq \langle s_{n} \rangle \subsetneq R. \end{equation*} If this chain were infinite, $I = \cup_{\scriptscriptstyle{n\in\mathbb{N}}} \langle s_{n} \rangle$ would be an ideal containing the chain. Since $R$ is a Principal Ideal Domain, there would be some $a \in I$ such that $I = \langle a \rangle$. There would be a natural number $N$ such that $a \in \langle s_{{}_{\scriptstyle{N}}} \rangle$. So, $\langle s_{{}_{\scriptstyle{N}}} \rangle = \langle s_{{}_{\scriptstyle{N+1}}} \rangle = \langle s_{{}_{\scriptstyle{N+2}}} \rangle \ldots$, and the chain terminates. This would imply $s_{{}_{\scriptstyle{N}}}$ is prime. $s_{{}_{\scriptstyle{N}}}$ divides $r$, though. This is a contradiction.
So, $r$ is divisible by some primes irreducibles $p_{1}$, ... $p_{k}$. If $r/(p_{1} \cdots p_{k})$ is not a unit, it is divisible by another prime $p_{k+1}$. The number of primes dividing $r$ must be finite, though, because the chain \begin{equation*} \langle r \rangle \subsetneq \langle r/p_{1} \rangle \subsetneq \langle r/(p_{1}p_{2}) \rangle \subsetneq \langle r/(p_{1}p_{2}p_{3}) \rangle \cdots \end{equation*} is an ascending chain in the Principal Ideal Domain $R$.
