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

Question

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.

score 35 · Accepted Answer · 2012-02-17T19:56:54.053

35

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.

edited Feb 17 '12 at 19:56

answered Feb 17 '12 at 12:12

2

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 05:31
@awllower Yes you've got it. – Feb 19 '12 at 00:50
Ah: it ought to be $zy=y$. Sorry for that mistake. – awllower Jul 07 '13 at 08:53

score 6 · Answer 2 · edited Jun 14 '14 at 14:18

6

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.

edited Jun 14 '14 at 14:18

user26857

52,094

answered Feb 17 '12 at 16:08

Math Gems

19,574

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 05:30

score -3 · Answer 3 · answered Jun 14 '14 at 12:35

-3

I have a more intuitive generator than the one given by user38268. Take $x+y$. We can of course that that $x-y\in\langle x+y\rangle$, as $x-y=(x+y)(x-y)$. Now since both $x+y$ and $x-y$ exist inside $\langle x+y\rangle$, it is easy to show $x,y\in\langle x+y\rangle$.

answered Jun 14 '14 at 12:35

freebird

130
2

4

Not at easy as you may think. Remember that 2x=0 for all x. – Ragib Zaman Jun 14 '14 at 12:52
@RagibZaman- Ah you're right. – freebird Jun 14 '14 at 12:56
1

Also, note that $x - y = x + y$, since a Boolean ring has characteristic $2$ – zcn Jun 14 '14 at 17:01

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

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