Trivial Ring is Commutative Ring

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Theorem

Let $\struct {R, +, \circ}$ be a trivial ring.

Then $\struct {R, +, \circ}$ 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:


Ring Axiom $\text A$: Addition forms an Abelian Group

$\struct {R, +}$ is an abelian group:

This follows from the definition.

$\Box$


Ring Axiom $\text M0$: Closure under Product

$\struct {R, \circ}$ is closed:

From Ring Product with Zero:

$x \circ y = 0_R \in R$

$\Box$


Ring Axiom $\text M1$: Associativity of Product

$\circ$ is associative on $\struct {R, +, \circ}$:

$x \circ \paren {y \circ z} = 0_R = \paren {x \circ y} \circ z$

$\Box$


Ring Axiom $\text D$: Distributivity of Product over Addition

$\circ$ distributes over $+$ in $\struct {R, +, \circ}$:

$x \circ \paren {y + z} = 0_R$

by definition.


Then:

\(\ds x \circ y + x \circ z\) \(=\) \(\ds 0_R + 0_R\)
\(\ds \) \(=\) \(\ds 0_R\) Definition of Ring Zero


and the same for $\paren {y + z} \circ x$.

$\Box$


Commutativity

From the definition of trivial ring:

$\forall x, y \in R: x \circ y = 0_R = y \circ x$

Hence its commutativity.

$\blacksquare$


Sources

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