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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.

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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.

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  • $\begingroup$ okay . But, when $a$ is prime, does there always exist $b,c \in D$ such that $a=bc$. What if there do not exist any $b,c$ such that $a=bc$? If there do not exist any such pair, how can we be able to write $a=bc$? Thanks. $\endgroup$ – MathMan Jul 29 '14 at 10:21
  • $\begingroup$ The definition of an irreducible element $a$ says that : Whenever $b,c \in D$ with $bc=a$, then $b$ or $c$ is a unit. For this to happen, it should be possible that $a$ is expressed as a product of $b$ and $c$. Not every prime element $a$ might be capable of being expressed like this? $\endgroup$ – MathMan Jul 29 '14 at 10:30
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    $\begingroup$ Ohhh I get it , can't believe i couldnt see this.. thanks. $\endgroup$ – MathMan Jul 29 '14 at 10:35
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    $\begingroup$ @VHP The inference is much clearer if one uses an alternative equivalent definition (or characterization) of irreducible - see my answer. $\endgroup$ – Bill Dubuque Jul 29 '14 at 13:24
  • $\begingroup$ I can't believe it either. ;) $\endgroup$ – Martin Brandenburg Jul 29 '14 at 15:16
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Hint $\rm\ prime\Rightarrow\ irreducible$ is clearer when proved as follows.

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

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

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

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