Modern Algebra: Groups, Rings, Modules, Fields

Pick $x\in R$ with $x\ne 0$ and $x\notin R^{\times }\text{.}$ If $x$ is irreducible, there is nothing to prove. Otherwise, we have $x=x_1{x}_2$ for non-units $x_1,x_2\text{.}$ If both $x_1,x_2$ are irreducible, the proof is complete. Otherwise, one or both of them factors non-trivially. We may express this conveniently by saying that $x_1=x_3{x}_4$ and $x_2=x_5{x}_6$ such that either $x_3$ and $x_4$ are both non-units or $x_5$ and $x_6$ are both non-units. (E.g., if $x_2$ is irreducible, we could set $x_5=x_2,x_6=1\text{.}$) Continuing in the this manner, we form a binary tree with $x$ at the top, $x_1,x_2$ one level down, $x_3,x_4,x_5,x_6$ one level below that, etc.

We halt the process of building the tree if at some stage all the leaves of the tree are irreducible elements, at which point we will have proven that $x$ factors in to a product of the irreducible elements given by these leaves.

We need to rule out the possibility that the process never terminates. If it never terminates, we will have built an infinite binary tree with the property that some route downward through the tree consists of an infinite list of irreducible elements $y_1,y_2,y_3,\dots$ such that $x=y_1{z}_1$ for a non-unit $z_1$ and, for each $i\ge 1\text{,}$ $y_i=y_{i+1}{z}_{i+1}$ for a non-unit $z_{i+1}\text{.}$ Since $R$ is an integral domain, we have $(x)⊊(y_1)$ and $(y_i)⊊(y_{i+1})$ for all $i\ge 1\text{.}$ (E.g., if $(x)=(y_1)$ then $y_1=xv$ and hence $x=xv{z}_1\text{,}$ so that $v{z}_1=1\text{,}$ contrary to $z_1$ being a non-unit.)