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The only Boolean integral domain is Z/(2)

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.16

Prove that the only boolean ring that is an integral domain is $\mathbb{Z}/(2)$.

Solution: Let $B$ be a boolean ring which is an integral domain. If $a \in B$ is nonzero, then $a = a^2 = a^3$, and by the cancellation law, $a^2 = 1$. By Exercise 7.1.11, $a = 1$ or $a = -1$. Note also that $-1 = (-1)^2 = 1$, so that $B = \{ 0,1 \}$. Additively, $B \cong \mathbb{Z}/(2)$, and in fact $$0 \cdot 0 = 0 \cdot 1 = 1 \cdot 0 = 0$$ and $1 \cdot 1 = 1$, so that $B$ “is” $\mathbb{Z}/(2)$. (The book hasn’t introduced ring isomorphisms yet.)

Alternate proof: Suppose $a \in R$ is not 0 or 1. Then $a$ and $a-1$ are nonzero, but $$a(a-1) = a^2-a = a-a = 0,$$ so that $a$ is a zero divisor and we have a contradiction. Thus $R = \{0,1\}$.