TITLE: Finitely generated ideal in Boolean ring; how do we motivate the generator?
QUESTION [8 upvotes]: This problem is Exercise 11.3 in Atiyah/Macdonald Commutative Algebra. They ask to prove every finitely generated ideal in a Boolean ring is in fact a principal ideal.
The question has been answered already on StackExchange: note that $(x,y) = (x+y+xy)$ and then use induction. However, is there any clear motivation for using $x+y+xy$ as the generator of this ideal? (apart from pulling it out of a hat and noticing it has nice properties)

REPLY [5 votes]: Boolean rings are the same as Boolean algebras.
1. A Boolean ring is a Boolean algebra
If $R$ is a Boolean ring, we can define a relation
$$
a\le b \quad\text{if and only if}\quad ab=a
$$
This is easily seen to be an order relation, since

$aa=a^2=a$ (the ring is Boolean),
if $ab=a$ and $ba=b$, then $a=b$ (a Boolean ring is commutative),
if $ab=a$ and $bc=b$, then $ac=(ab)c=a(bc)=ab=a$.

Any two elements in $R$ have a greatest lower bound under this order relation. Indeed, if $a,b\in R$, then $ab\le a$, because $(ab)a=a^2b=ab$ and, similarly, $ab\le b$. Next, if $c\le a$ and $c\le b$, we have $c\le ab$, since
$$
c(ab)=(ca)b=cb=c.
$$
Our purpose now is to find a least upper bound for $a$ and $b$. It should be an element $d\in R$ such that $a\le d$ and $b\le d$ and, moreover, if $a\le c$ and $b\le c$ for some $c\in R$, we should have $d\le c$.
An idea is to consider $d=a+b+x$; since we want $a\le d$, we must have
$$
a=a(a+b+x)=a^2+ab+ax=a+ab+ax
$$
so $ab+ax=0$. Similarly, from $b\le d$ we get $ab+bx=0$, which means $ax=bx$. Oh, we know a choice for $x$, namely $x=ab$. Now we prove the other condition: suppose $a\le c$ and $b\le c$; then
$$
c(a+b+ab)=ca+cb+cab=a+b+ab=d
$$
so $d\le c$ and we have proved that $a+b+ab$ is a greatest upper bound for $a$ and $b$.
What's the minimum for this order relation? Of course $0$, because $0a=0$ for all $a\in R$. Similarly from $a1=a$ we deduce that $1$ is the maximum.
What do we need now? A complement: for $a\in R$ we need $b$ such that
$$
ab=0\quad\text{and}\quad a+b+ab=1.
$$
But $ab=0$ implies $a+b=1$ or $b=1-a=1+a$ (in a Boolean ring we have $-x=x$).
We have a bounded lattice with complements. If it is distributive, then it is a Boolean algebra. Set $a\land b=ab$ and $a\lor b=a+b+ab$: we have to prove that
$$
a\land(b\lor c)=(a\lor b)\land(a\lor c).
$$
The left hand side is
$$
a(b+c+bc)=ab+ac+abc
$$
while the right hand side is
\begin{align}
(a+b+ab)(a+c+ac)
&=a^2+ab+a^2b+ac+bc+abc+a^2c+abc+a^2bc\\
&=a+ab+ab+ac+bc+abc+ac+abc+abc=ab+ac+abc
\end{align}
remembering that $2x=0$ in a Boolean ring.
2. A Boolean algebra is a Boolean ring
If $A$ is a Boolean algebra under the operations $\land$, $\lor$ and complement $'$, we define
\begin{gather}
ab=a\land b\\
a+b=(a\land b')\lor(a'\land b)
\end{gather}
It is just a tedious check to verify the ring axioms. We get a Boolean ring because $a^2=a\land a=a$. Moreover
$$
a+b+ab=a\lor b
$$
as it's readily verified.
Why finitely generated ideals are principal?
An ideal in a Boolean algebra $A$ is a non empty subset $I$ such that for $a,b\in I$ we have $a\lor b\in I$ and, if $a\in I$ and $b\le a$ (that is, $a\land b=b$ or, equivalently, $a\lor b=a$), then $b\in I$.
In particular $0\in I$ for any ideal. But it's also clear that an ideal in a Boolean algebra is also an ideal under the Boolean ring operations. Indeed, if $a,b\in I$, then
$$
a+b\le a\lor b
$$
(check it). Moreover, if $a\in I$ and $b\in A$, then $ab\le a$, so $ab\in I$.
Conversely, if $I$ is an ideal in a Boolean ring $R$, then it is also an ideal under the Boolean algebra operations (check it).
Thus the principal ideal generated by $a\in R$ is $aR=\{b\in R:b\le a\}$, so it is the Boolean algebra ideal generated by $a$. Most clearly, the Boolean algebra ideal generated by $a$ and $b$ is the ideal generated by $a\lor b=a+b+ab$.
Is the intuition clearer now?