Idempotent Elements of Ring with No Proper Zero Divisors
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Theorem
Let $\struct {R, +, \circ}$ be a non-null ring with no (proper) zero divisors.
Let $x \in R$.
Then:
- $x \circ x = x \iff x \in \set {0_R, 1_R}$
That is, the only elements of $\struct {R, \circ}$ that are idempotent are zero and unity.
Proof
We have $0_R \circ 0_R = 0_R$, so that sorts out one element.
Let $R^*$ be the ring $R$ without the zero: $R^* = R \setminus \set {0_R}$.
By Ring Element is Zero Divisor iff not Cancellable, all elements of $R^*$ that are not zero divisors are cancellable.
Therefore all elements of $R^*$ are cancellable.
Suppose $x \circ x = x$ where $x \ne 0_R$.
Then $x \circ x = x = x \circ 1_R$.
As $x$ is cancellable, the result follows.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.3$