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In my abstract algebra book, it says that an element $p \in R$ where $R$ is a commutative ring is irreducible if

$\textbf{(i)}$: $p$ is not a unit $\textbf{(ii)}$: if $p=ab$ then either $a$ or $b$ is a unit of $R$.

However, in another book when $R$ becomes an integral domain, they give the definition that en element $p$ is irreducible if $a \mid p$ implies that either $a$ is a unit or an associate of $p$.

To me it seems like the statement " $a \mid p$ implies that either $a$ is a unit or an associate of $p$." is totally equivalent to $\textbf{(ii)}$. My question is, does $\textbf{(ii)}$ imply $\textbf{(i)}$ in an integral domain? So that in an integral domain we can shrink the definition of an irreducible element?

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The answer is no. If we take the ring of integers $\mathbb{Z}$ and $p=1$ then the condition $\textbf{(ii)}$ is satisfyied but $p$ is a unit.

  • So why does not the later book require that $p$ is a nonunit? I find this rather confusing. The books are Beachy and Blair: Abstract algebra and ireland and rosen: A classical introduction to modern number theory. –  Jun 26 '14 at 07:13
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    @Sodan I believe it is a mistake on the part of Ireland & Rosen. Cf. Corollary 1.3.2 to see an instance where they implicitly assume irreducible implies nonunit. – Karl Kroningfeld Jun 26 '14 at 08:59