Finitely generated ideal in boolean ring

Question

A boolean ring is a commutative ring where $x^{2} = x$ for every $x$.

Why in such a ring a finitely generated ideal is principal ?

Answer 1

You'll want to show this by induction on the number of generators of your ideal.

I'll show this for two generators. If $I=(x,y)$, then certainly $(x+y+xy) \subseteq I$.

Conversely, $x(x+y+xy) = x^2 +xy + x^2y = x+2xy = x \in (x+y+xy)$ (Assuming you've shown that $2x = 0\; \forall \; x$ Similarly, $(x+y+xy)y = y \in (x+y+xy)$ so $(x,y) \subseteq (x+y+xy)$ and we have equality.