Finitely generated ideals in a Boolean ring are principal, why?

The classical book on commutative algebra Introduction to Commutative Algebra, by Atiyah and Macdonald, has the following as exercise I.11.

A ring is Boolean if $x^2=x$ for any $x$ of $A$. In a Boolean ring $A$, show that
i) $2x=0$ for all $x\in A$;
ii) Every prime ideal of $A$ is maximal, and its residue field consists of two elements;
iii) Every finitely generated ideal of $A$ is principal.

I can but solve the first two of them; for the last one, I cannot even perceive the situation here, for it will not, generally speaking, be a Dedekind domain, whereof my knowledge is somewhat more. I have hitherto tried to find one element for each ideal of two generators which can stand the stead of them, and then proceed by induction. But even the first step fails to produce any result.
Thanks for any hints and inspirations in advance.

Solution 1:

Massive hint: If your ideal $I$ is generated by two elements $x$ and $y$, consider the element $ z = x + y + xy$. What is your ideal $I$ generated by now? Recall if you want to show that $(a,b) = (c)$, you need to show that $a$ and $b$ are in the right hand side, and also that $c$ is in the left hand side.

Now complete the problem by induction.

Solution 2:

Hint: a proof of (iii) is much simpler when translated into the language of Boolean algebras, using the well-known equivalence between Boolean rings and Boolean algebras (see e.g. Chapter IV of Burris and Sankappanavar.) In particular, this allows one to very naturally discover the generator $z$ in $(x,y) = (z),\:$ vs. pulling $z$ out of hat. Analogous remarks hold for many Boolean ring proofs.