The only Boolean integral domain is ZZ/(2)

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


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 §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\}.

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