Null Ring is Trivial Ring

Theorem
Let $R$ be the null ring.

That is, let:
 * $R := \left({\left\{{0_R}\right\}, +, \circ}\right)$

where ring addition and the ring product are defined as:


 * $0_R + 0_R = 0_R$
 * $0_R \circ 0_R = 0_R$

Then $R$ is a trivial ring and therefore a commutative 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:
 * $R := \left({\left\{{0_R}\right\}, +, \circ}\right)$

where ring addition and the ring product are defined as:


 * $0_R + 0_R = 0_R$
 * $0_R \circ 0_R = 0_R$

Taking the ring axioms in turn:

A: Addition forms a Group
$\left({\left\{{0_R}\right\}, +}\right)$ is a group.

This follows from the definition of the Trivial Group: the element $0_R$ is the identity for the operation $+$.

M0: Closure of Ring Product
$\left({\left\{{0_R}\right\}, \circ}\right)$ is closed:
 * $0_R \circ 0_R = 0_R$

by definition.

M1: Associativity of Ring Product

 * $0_R \circ \left({0_R \circ 0_R}\right) = 0_R \circ 0_R = \left({0_R \circ 0_R}\right) \circ 0_R$

Thus $\circ$ is associative.

D: Distributivity of Ring Product over Addition
$\circ$ distributes over $+$ in $\left({\left\{{0_R}\right\}, +, \circ}\right)$:

First we have:
 * $0_R \circ \left({0_R + 0_R}\right) = 0_R$

by definition.

Then we have:

Trivial Ring
The fact that the null ring is a trivial ring follows by definition from the fact that:
 * $\forall a \in R: a \circ a = 0_R$

as there is only one possible value for $a$, that is $0_R$.

The fact that $R$ is a commutative ring is trivial, or a consequence of Trivial Ring is Commutative Ring.