Exercise A0.V.1.12

Proposition

Let $R$ be an integral domain and let $a \in R$ be nonzero. Then $a$ is irreducible if and only if $(a)$ is maximal among proper principal ideals of $R.$

Proof

Suppose $a$ is irreducible and let $(b)$ be an ideal such that $(a) \subseteq (b).$ This implies that there exists $r \in R$ such that $a = rb.$ Since $a$ is irreducible, either $r$ is unit, in which case $(a) = (b),$ or $b$ is unit, in which case $(b) = R.$ Thus $a$ is maximal among proper principal ideals of $R.$

Conversely, suppose $(a)$ is maximal among proper principal ideals of $R,$ and let $a = bc$ for some $b, c \in R.$ In this case $(a) \subseteq (b)$ and $(a) \subseteq (c).$ By maximality of $(a)$ we know that $(b)$ must be either equal to $(a)$ or to $R,$ in which case $b$ is a unit. The same holds for $(c).$

If either $b$ or $c$ is a unit, we’re done. It cannot be that both of them are units, since that would make $a$ into a unit. The only remaining case is when neither of them is a unit, that is $(a) = (b) = (c).$ Hence there exists a unit $r \in R$ such that $a = r c.$ Since $a = bc = r c$ we can use cancellativity to infer $b = r,$ so $b$ is a unit. Our use of cancellativity is justified, since $a$ is non-zero, and since $a = bc,$ neither $b$ nor $c$ can be zero.

$\blacksquare$