Prove that every finitely generated ideal in a Boolean ring is principal

The Mathematician — Thu, 06 Apr 2023 13:00:49 +0000

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

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.

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Prove that every finitely generated ideal in a Boolean ring is principal