In an Integral Domain, every prime is an irreducible. The proof is as follows :
Let $D$ be the integral domain, then, if $a \in D$, it's possible to express a = $bc$ where $b,c \in D ...(1)$.
Then if $a$ is a prime $\implies a | mn \implies a|m $ or $ a|n$.
Since, $D$ possesses the unity, $a.1 = a \implies a|a \implies a|bc \implies a|b$ or $a|c$.
$\implies b = at \implies b = bct $ (from $(1) ) \implies ct=1 \implies c$ is a unit.
Hence, $a$ is irreducible.
The basis of this proof is the one shown in the highlights which says that if $D$ be the integral domain, then, if $a \in D$, it's possible to express a = $bc$ where $b,c \in D$.
This is surely the case in a finite integral domain. But, this may not be always possible in an infinite integral domain? How do we explain this reasoning in an infinite integral domain?
Thank you for your help.