Proposition

A ring $A$ is Boolean if $x^2 = x$ for all $x \in A$. In a Boolean ring $A$, show that

$2x = 0$ for all $x \in A$;
every prime ideal $p$ is maximal, and $A/p$ is a field with two elements;
every finitely generated ideal in $A$ is principal.

Solution

1

$(x + x)^2 = 4x^2 = 4x$, and $(x + x)^2 = x + x = 2x$. Thus $2x = 0$.

2

Let $p$ be a prime ideal. Let $x \in A \setminus p$. $x(1 - x) = x - x^2 = x - x = 0$. Thus $x(1 - x) \in p$ and $p$ is a prime ideal not containing $x$. Therefore, $1 - x \in p$. This implies that $p + \ev{x}$ contains $1 = (1 - x) + x$. Therefore, $p + \ev{x}$ is a maximal ideal, so there exists no proper ideal of $A$ containing $p$ properly. Thus $p$ is a maximal ideal.

3

Let $\ev{x_1, x_2} \subset A$ be given. Then $x_1 + x_2 + x_1x_2 \in \ev{x_1, x_2}$. On the other hand,

$x_1(x_1 + x_2 + x_1x_2) = x_1^2 + x_1x_2 + x_1^2x_2 = x_1 + x_1x_2 + x_1x_2 = x_1$.
$x_2(x_1 + x_2 + x_1x_2) = x_2x_1 + x_2^2 + x_1x_2^2 = x_2 + 2x_1x_2 = x_2$.

Therefore, $\ev{x_1, x_2} \subset \ev{x_1 + x_2 + x_1x_2}$. Hence, $\ev{x_1, x_2} = \ev{x_1 + x_2 + x_1x_2}$.

It is clear that this can be extended to an ideal generated by $n$ elements for any finite $n$ by using mathematical induction.