Idempotent Ring is Commutative

Theorem

Let $\struct {R, +, \circ}$ be an idempotent ring.

Denote with $0_R$ the zero of $R$.

Then $\struct {R, +, \circ}$ is a commutative ring.

Proof

Let $x, y \in R$.

Then:

\(\ds x + y\)

\(=\)

\(\ds \paren {x + y}^2\)

Definition of Idempotent Ring

\(\ds \)

\(=\)

\(\ds x^2 + x \circ y + y \circ x + y^2\)

Binomial Theorem: Ring Theory

\(\ds \)

\(=\)

\(\ds x + x \circ y + y \circ x + y\)

Definition of Idempotent Ring

Subtracting $x + y$ from both sides yields:

\(\ds x \circ y + y \circ x\)

\(=\)

\(\ds 0_R\)

\(\ds \)

\(=\)

\(\ds y \circ x + y \circ x\)

Idempotent Ring has Characteristic Two

Finally, subtracting $y \circ x$ from both sides, we obtain:

$x \circ y = y \circ x$

and conclude $R$ is a commutative ring.

$\blacksquare$

Sources

• 2008: Paul Halmos and Steven Givant: Introduction to Boolean Algebras ... (previous) ... (next): $\S 1$
• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $9$: Rings: Exercise $1$