Null Ring is Commutative Ring
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Theorem
Let $R$ be the null ring.
That is, let:
- $R := \struct {\set {0_R}, +, \circ}$
where ring addition and ring product are defined as:
| \(\ds 0_R + 0_R\) | \(=\) | \(\ds 0_R\) | ||||||||||||
| \(\ds 0_R \circ 0_R\) | \(=\) | \(\ds 0_R\) |
Then $R$ is a commutative ring.
Proof
From Null Ring is Trivial Ring, we have that $R$ is a trivial ring.
The result follows from Trivial Ring is Commutative Ring.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Example $21.4$