Trivial Ring is Commutative Ring

Theorem
Let $\left({R, +, \circ}\right)$ be a trivial ring.

Then $\left({R, +, \circ}\right)$ is a commutative ring.

Proof
First we need to show that a trivial ring is actually a ring in the first place.

Taking the ring axioms in turn:

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

This follows from the definition.

M0: Closure of Ring Product
$\left({R, \circ}\right)$ is closed:

From Ring Product with Zero, we have $x \circ y = 0_R \in R$.

M1: Associativity of Ring Product
$\circ$ is associative on $\left({R, +, \circ}\right)$:


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

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


 * $x \circ \left({y + z}\right) = 0_R$ by definition.

Then:

... and the same for $\left({y + z}\right) \circ x$.

Commutative
From the definition of trivial ring:
 * $\forall x, y \in R: x \circ y = 0_R = y \circ x$

Hence its commutativity.