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Let R be a domain and $a$ a nonzero nonunit in R. Show that $a$ is irreducible if and only if the principal ideal (a) is maximal in the set {($b$) where $b$ a nonzero nonunit in R}.

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If $(a)$ is not maximal in the given set, there is $b$, a nonzero nonunit, such that $(a)\subsetneq (b)$, which means $a=rb$ for some $r\in R$. So $a$ is reducible.

The converse is equally straight forward. I leave it to you.

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  • $\begingroup$ why is r necessarily not a unit in this case? $\endgroup$ – davidh Jan 27 at 8:29
  • $\begingroup$ If $r$ is a unit, we get $b=r^{-1}a$, which implies $(b)\subset (a)$, contradicting our assumption. $\endgroup$ – Chris Custer Jan 27 at 12:09

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