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

A ring is Boolean if $x²=x$ for any $x$ of $A$.
i) $2x=0$
ii) Every prime ideal of $A$ is maximal, whose 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.

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2 Answers

up vote 5 down vote accepted

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.

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And so $xz=x$ and $yz=y$, right? Thanks for providing the generator, though the mystery behind this generator is another matter then. –  awllower Feb 18 '12 at 5:31
@awllower Yes you've got it. –  fpqc Feb 19 '12 at 0:50
Ah: it ought to be $zy=y$. Sorry for that mistake. –  awllower Jul 7 '13 at 8:53
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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 algebrsa (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.

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Thanks for sharing the book, albeit I possess myself yet not enough time to understand the close relations said, but I am sure I will enjoy the book. –  awllower Feb 18 '12 at 5:30
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