Boolean Ring has Proper Zero Divisor
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Theorem
Let $\left({R, +, \circ}\right)$ be a Boolean ring whose zero is $0_R$.
Suppose that $R$ has more than two elements.
Then $R$ has a proper zero divisor.
Proof
Since $R$ has more than two elements, there exist distinct non-zero elements $x, y \in R$.
Note that $x + y \ne 0_R$ since $x$ and $y$ are distinct (by Idempotent Ring has Characteristic Two).
If $x \circ y = 0$, $x$ is a proper zero divisor.
If $x \circ y \ne 0$, then:
\(\ds \left({x \circ y}\right) \circ \left({x + y}\right)\) | \(=\) | \(\ds x \circ y \circ x + x \circ y \circ y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x \circ y + x \circ y\) | $R$ is an idempotent ring, Idempotent Ring is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds 0_R\) | Idempotent Ring has Characteristic Two |
Hence $x \circ y$ is a proper zero divisor.
The result follows, from Proof by Cases.
$\blacksquare$
Sources
- 2008: Paul Halmos and Steven Givant: Introduction to Boolean Algebras ... (previous) ... (next): $\S 1$: Exercise $9$