**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.24**

Solution:

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.