Proving that an ideal in a PID is maximal if and only if it is generated by an irreducible

Solution 1:

Hint $ $ Recall for principal ideals: $\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$, $ $ i.e. $\,\color{#0a0}{(a)\supseteq (b)}\iff \color{#c00}{a\mid b},\,$ thus having no proper $\rm\color{#0a0}{containing}$ ideal (maximal) is the same as having no proper $\rm\color{#c00}{divisor}$ (irreducible), $ $ i.e.

$\qquad\quad\begin{eqnarray} (p)\,\text{ is maximal} &\iff&\!\!\ (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (d)\\ &\iff&\ p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ d\\ &\iff&\ p\ \ \text{ is irreducible}\\ \end{eqnarray}$

Solution 2:

Suppose that $R$ is a PID and that $\mathcal{I}$ is an an ideal. Then $\mathcal{I}$ is maximal iff for any $x$ generating $\mathcal{I}$, $x$ is irreducible.

Proof: $\Rightarrow$: Suppose $\mathcal{I}$ is maximal and that $\mathcal{I}$ is generated by $x$. Write $x = ab$ for some $a,b\in R$. Since $a|x$, $\mathcal{I}$ must be a subset of the ideal generated by $a$. Were this inclusion to be proper, by the maximality of $\mathcal{I}$, we would have $R$ being generated by $a$. This make $a$ a unit. By symmetry, $a$ or $b$ is a unit. We conclude that $x$ is irreducible.

$\Leftarrow$: Suppose that $x$ is irreducible. If $x$ is not a unit, there is a maximal ideal $\mathcal{I}$ of $R$ with $x\in\mathcal{I}$. Since $R$ is a PID, we can choose $y\in R$ so that $\mathcal{I}$ is generated by $y$. Since $x\in \mathcal{I}$, $y|x$. Write $x = ay$ for some $y\in R$. Since $x$ is irreducible $a$ is a unit. Hence $x$ and $y$ generate the same ideal; this ideal is maximal.

Solution 3:

Suppose $\langle x \rangle$ is maximal and $x = yz$. Then $\langle x \rangle \subseteq \langle y \rangle$. From here, you should be able to show that $y$ is either a unit or an associate of $x$, showing $x$ is irreducible.

Suppose instead $x$ is irreducible. Then $\langle x \rangle \subseteq \langle y \rangle$ would imply $x = yz$ for some $z$. From here, you should be able to show that $y$ is either a unit or an associate of $x$, showing $\langle x \rangle$ is maximal.