In an Integral Domain, every prime is an irreducible. Flaw in the Proof?

Question

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.

Answer 1

You can always write $a=a \cdot 1$. This has nothing to do with finiteness conditions. But if you want to prove that $a$ is irreducible, you have to write $a=bc$ and show that $b$ or $c$ is a unit (see the definition of "irreducible"). This proof is possible for example when $a$ is a prime, as you have shown.

Answer 2

Hint $\rm\ prime\Rightarrow\ irreducible$ is clearer when proved as follows.

$\qquad\! \rm nonunit\ p\ne 0\,\rm\ is\ \ {\bf irred}\ \ iff\ \ p = ab\,\Rightarrow\, p\mid a\ \ or\ \ p\mid b$

$\qquad\!\! \rm nonunit\ p\ne 0\,\rm\ is\,\ {\bf prime}\,\ iff\ \ p\,\mid\, ab\,\Rightarrow\, p\mid a\ \ or\ \ p\mid b$

$\rm So\ \ prime\Rightarrow\ irred\ by\ p = ab\,\Rightarrow\,p\mid ab\,\Rightarrow\ p\mid a\ \ or\ \ p\mid b$