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. 

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.
