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

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?

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.

• 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. – MathMan Jul 29 '14 at 10:21
• 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? – MathMan Jul 29 '14 at 10:30
• Ohhh I get it , can't believe i couldnt see this.. thanks. – MathMan Jul 29 '14 at 10:35
• @VHP The inference is much clearer if one uses an alternative equivalent definition (or characterization) of irreducible - see my answer. – Bill Dubuque Jul 29 '14 at 13:24
• I can't believe it either. ;) – Martin Brandenburg Jul 29 '14 at 15:16

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

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

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

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