$\langle r\rangle$ maximal $\iff r$ irreducible

Question

Let $R$ be a PID, and $r\in R- \{0\}$. Prove that $\langle r\rangle$ maximal $\iff r$ irreducible.

"$\Leftarrow$"Easy.

"$\Rightarrow$"If $J=\langle r \rangle$ then we will prove that $r$ is irreducible. If $r=ab$, we want to prove that $a\in U(R)$ or $b \in U(R)$.

If we take the ideal which is generated by $\langle a\rangle$ then (because $J$ is maximal)$$\langle a\rangle \subseteq\langle r \rangle \iff r\mid a \iff a=kr, k\in R \Longrightarrow r=krb\iff r(1-kb)=0_R \iff kb=1_R$$ so $b\in U(R)$. Same way if we work with $\langle b \rangle$.

Is this proof right?

Answer 1

The argument is not completely correct.

If I understand correctly you start with "(because $J$ is maximal) $\langle a\rangle \subseteq\langle r \rangle $", but it is not true that if you chose some maximal ideal then every other ideal is contained in it.

Instead argue like this if $r= ab$ then $r \in \langle a \rangle$ and thus $\langle r \rangle \subset \langle a \rangle$. Since $\langle r \rangle$ is maximal it follows that $\langle r \rangle = \langle a \rangle$ or that $ \langle a \rangle =R$.

Then continue from there.

Answer 2

Hint $ $ Note that for principal ideals: $\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$, $ $ i.e. $(a)\supset (b)\iff a\mid b,\,$ thus

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