Null Ring is Trivial Ring
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Theorem
Let $R$ be the null ring.
Then $R$ is a trivial ring.
Proof
We have that $R$ is the null ring.
That is, by definition it has a single element, which can be denoted $0_R$, such that:
- $R := \struct {\set {0_R}, +, \circ}$
where ring addition and the ring product are defined as:
\(\ds 0_R + 0_R\) | \(=\) | \(\ds 0_R\) | ||||||||||||
\(\ds 0_R \circ 0_R\) | \(=\) | \(\ds 0_R\) |
Consider the operation $+$.
By definition, the algebraic structure $\struct {\set {0_R}, +}$ is a trivial group.
Then:
- $\forall a, b \in R: a \circ b = 0_R$
Thus by definition, $R$ is a trivial ring.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Example $21.5$