In a Boolean ring, all finitely generated ideals are principal

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

We begin with a lemma.

Lemma: Let R be a ring. Suppose that for all A \subseteq R with |A| = 2, (A) is principal. Then for all finite nonempty A \subseteq R, (A) is principal. Proof: We proceed by induction on |A|. For the base case, If |A| is 1 or 2, then (A) is principal. For the inductive step, suppose that for some n \geq 2, for all A \subseteq R with |A| \leq n, (A) is principal. Let A \subseteq R have cardinality n+1 and write A = \{x\} \cup A^\prime, where |A^\prime| = n. Now (A) = (x) + (A^\prime), and by the induction hypothesis, (A^\prime) = (y) for some y \in R. Now (A) = (x,y) = (z) for some z \in R by hypothesis. \square

Now let R be a Boolean ring, and let a,b \in R. We claim that (a,b) = (a+ab+b). It is clear that a+ab+b \in (a,b) since R has a 1 and is commutative. Moreover, note that (a+ab+b)(ab+a+1) = b and (a+ab+b)(ab+b+1) = a, so that (a,b) \subseteq (a+ab+b). Thus every subset of order 2 generates a principal ideal in R, and by the lemma, every finitely generated ideal in R is principal.

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