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)
On
It's actually pretty nice to reason through! This is all possible thanks to two lattice theoretic properties of Boolean rings. It's good to know about these two properties, and it turns out they're pretty instructive in this proof.
One thing we need is the involutory map $(x)\mapsto (1-x)$ establishing an inclusion reversing bijection on the principal ideals of the Boolean ring $R$, which I will call $L$. So $L((a)):=(1-a)$, and $(a)\subseteq (b)$ implies $L((b))\subseteq L((a))$. (Hint: if $(a)\subseteq(b)$, take $a=br$ and do something with $1-b=(1-b)(1-a)+(1-b)a$.)
The other thing afforded us in Boolean rings that we need is that $(a)\cap (b)=(ab)$. The $\supseteq$ is obvious, and I leave the $\subseteq$ as an exercise.
Using these two things, we can cook up the formula $a+b+ab$.
Let's seek the least upper bound of $(a)$ and $(b)$. A natural candidate is to find the greatest lower bound of $L((a))$ and $L((b))$, and then map it back above $(a)$ and $(b)$ using the inclusion reversing map.
Explicitly, since we know the greatest lower bound of $(1-a)$ and $(1-b)$ is $((1-a)(1-b))$ we have $(1-a)\supseteq ((1-a)(1-b))$, and after applying $L$ we have $(a)\subseteq (1-(1-a)(1-b))$, and similarly $(b)\subseteq (1-(1-a)(1-b))$.
But look at $1-(1-a)(1-b)=a+b-ab$. We've just shown that $(a+b-ab)$ is an upper bound of $(a)$ and $(b)$. Finally, it's obviously contained in $(a,b)$, so it is actually equal.
Here's a picture of that proof:
Remember that all of the subtraction signs may as well be addition because the characteristic is 2.
So in hindsight, the "mysterious" character $a+b+ab$ is just a product of the interplay of this involution $a\mapsto 1-a$ with the way principal ideals intersect. We can retrieve it via a Galois connection on the principal ideals of the ring.
On
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
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.
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.
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?
We may assume that $R$ is finite (otherwise consider the subring generated by $x,y$). Then $R=(P(S),\Delta,\cap)$ for some finite set $S$. We have $x \in (p)$ iff $x \subseteq p$. Hence, we have $(x,y) \subseteq (p)$ iff $x \subseteq p$ and $y \subseteq p$, i.e. $x \cup y \subseteq p$. We want a minimal such $p$. Hence, $p = x \cup y$ is the only candidate. Now observe that $x \cup y = x \Delta y \Delta (x \cap y)$.
Alternatively, observe that the subring generated by $x,y$ is the $\mathbb{F}_2$-vector space generated by $1,x,y,xy$. We may assume that it is a basis (since this is the case in the universal example). If $p = \lambda_0 + \lambda_1 x + \lambda_2 y + \lambda_3 xy$ is a generator of $(x,y)$, then $p \bmod x = \lambda_0 + \lambda_2 y$ is a generator of $(y)$, which implies $\lambda_2=1$ and $\lambda_0=0$. In the same way we get $\lambda_1=1$. But $x+y$ doesn't work, so $p=x+y+xy$ is the only choice.