**Every finitely generated ideal in a Boolean ring is principal.**
**Proof.** The proof is easy if you see the trick. Define the finitely generated ideal of a Boolean ring:

\begin{align*}
A = (a_1,a_2,\ldots,a_n).
\end{align*}

To understand what we will do, we will take the ideal

(a_1) of the Boolean ring. Our claim is that

(a_1) = (a_1,a_2,\ldots,a_n). This means we need to show for every

i \in \{2,\ldots,k\} that

a_i \in (a_1). Here is the trick we will apply:

\begin{align*}
a_1^2a_i - a_1a_i = 0 \in (a_1) \quad \text{since } a_1^2 = a_1.
\end{align*}

So we get

a_1(a_1a_i - a_i) = 0, which implies that

a_1 = 0 or

a_1a_i - a_i = 0. If

a_1 = 0, then we have the zero ideal, which we are not interested in. So let

a_1a_i - a_i = 0. Then this implies that

a_1a_i = a_i. But we already know that

a_1a_i \in (a_1), so

a_i \in (a_1), Since

a_i taken arbitrarily, we have that

(a_1) = (a_1,a_2,\ldots,a_n).
The same steps can also be done by taking an arbitrary element

x of the Boolean ring and letting

(x) be the ideal of the Boolean ring. This way, you can prove the general way by taking the same steps as above.