Null Ring is Trivial Ring

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 $21.5$