I'm reading a proof from C. Musili's Rings and Modules that every PID is a factorisation domain.
The author defines a factorisation domain as a commutative integral domain $R$ with a unit such that every non-zero $x \in R$ can be written as a unit times a finite product of irreducible elements. I will write down an outline of the proof here and my queries at the end. Those points that I am not sure about I will put in bold.
Let $R$ be a PID, and $\Omega$ the set of all non-zero elements of $R$ that cannot be written as a product of irreducible elements in $R$. We want to show that $\Omega = \emptyset$. So for a contradiction suppose that $\Omega \neq \emptyset$. Consider the non-empty family of principal ideals
$\mathcal{F} = \{(x) \subseteq R : x \in \Omega\}$
that is also a poset with respect to set inclusion. Now given any chain in $\mathcal{F}$ we can check that it has an upper bound in $\mathcal{F}$ so that by Zorn's Lemma $\mathcal{F}$ has a maximal element, say $(a)$.
Now $a$ cannot be a unit or irreducible for $a \in \Omega$. So write $a = bc$ for non-units $b$ and $c$. Now $(a) \subsetneqq (b)$ for otherwise $a$ and $b$ would be associates contradicting the fact that $a$ is irreducible. Since $(a)$ is maximal in $\mathcal{F}$, this means that $b$ and $c$ can be factored into irreducibles so that $a \notin \Omega$, a contradiction. Hence $\Omega = \emptyset$.
(1) For the first sentence in bold, should $\Omega$ not be the set of all non-zero elements in $R$ that cannot be written as a unit times an infinite number of irreducibles?
(2) If $\Omega$ is like what I have written above, then $a$ cannot be a unit for then
$a = a \cdot 1 = $ a unit $\times$ an irreducible element.
However if $\Omega$ is as the author has defined it to be, why must $a$ not be unit?
(3) If $a$ is not irreducible, why can we always write $a = bc$ for non-units $b$ and $c$? Does such a decomposition always exist?
Thanks.