Proving that an ideal in a PID is maximal if and only if it is generated by an irreducible 
Let $R$ be a PID (Principal Ideal Domain) and $x$ is an element R. Prove that the ideal $\langle x\rangle$ is maximal if and only if $x$ is irreducible.

Ok, so I know what an irreducible is. I'm thinking that this problem is asking us to set up a proof by contradiction but I can't see how. No one in my study group has any clue. 
 A: 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.
A: 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.
A: 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.
$\qquad\quad\begin{eqnarray} (p)\,\text{ is maximal} 
&\iff&\!\!\ (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (d)\\
&\iff&\  p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ d\\
&\iff&\  p\ \ \text{ is irreducible}\\
\end{eqnarray}$
A: Proof by contradiction is a perfectly good idea. 
First, suppose $x = y z$, and neither $y$ nor $z$ are units. Can you find an ideal containing $\left< x \right>$?
Second, suppose $\left< x \right>$ is not maximal, so there is an ideal containing it. Can you find a factor of $x$?
