Bosch Linear Algebra p. 179
This is a part of the proof where where we want to Show that in every principle ideal Domain every element which is not a unit can be factorized with primeelements.
I just want to understand this part which is the beginning of a contradiction proof.
According to our definitions
In a integral Domain with an element $p\neq 0$ such that $p$ is also not a unit we call
$p$ irreducible if from a a equation $p=ab$ follows that either $a$ or $b$ is an unit. Otherwise we call $p$ reducible. Hence we have established a dichotomy for Elements which are in $R$ but not $0$ and also not a unit.
We call $p$ a primelement if we know $p|ab\Rightarrow p|a \vee p|b$ which is also equivalent to $ab\in(p)\Rightarrow a\in(p)\vee b\in(p)$
We already know that every primeelement is also irreducible and in a principal ídeal Domain also the Converse.
With that said we want to prrof from our assumptions above that $a$ is reducible. We want to Show that there exists $q,w$ such that $a=qw$ and $q$ or $w$ are not Units.
I have started a contradiction proof. Assuming that $a$ is irreducible
If we look at a product $a=qr$ then $q$ or $r$ must be an unit. If both are Units we get a contadiction because then $a$ would also be a unit. So let $q$ be the unit and $r$ not a unit. We can evoke a contradiction if we could Show that $r$ is a unit nevertheless. But I don't know how.
Edit: I think I got it now
If we write $a$ as a producit of two element so that $a=qw$ for example. Then we can say with our assumption that not $q$ and $w$ can be irreducible. Therefore let $q$ not be reducible without loss. Bacause $q$ is not irreducible $q$ is either reducible or it is a unit or it is a Zero. If it is Zero then $a$ would be Zero too, contradiction. If it is reducible there one can write $q=et$ without loss let $e$ then be a unit then we can write $a=e(tw)$ and $tw$ is not $0$ because otherwise we would get a new contradiction therefore $a$ is irreducible, if $q$ is a unit $a$ also would be irreducible. Therefore in every case $a$ is reducible.
Well, let $a$ be reducible. Write $a=a_1b_1$. Then $a_1$ and $b_1$ cannot be units. Write $a_1=a_2b_2$ similarly. This gives you a sequence of elements $a,a_1,a_2,\ldots$ such that $a_1\mid a$, $a_2\mid a_1$, and so on, or an ascending chain of ideals $$(a) \subset (a_1)\subset (a_2)\subset \ldots.$$ This series becomes eventually constant in a Noetherian ring such as an integral domain where every finitely generated ideal is principal.
This is what is required in the 2nd part of your proof.