# Ring Product with Zero

## Theorem

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.

Then:

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

That is, the zero is a zero element for the ring product, thereby justifying its name.

## Proof

Because $\struct {R, +, \circ}$ is a ring, $\struct {R, +}$ is a group.

Since $0_R$ is the identity in $\struct {R, +}$, we have $0_R + 0_R = 0_R$.

From the Cancellation Laws, all group elements are cancellable, so every element of $\struct {R, +}$ is cancellable for $+$.

Thus:

 $\ds x \circ \paren {0_R + 0_R}$ $=$ $\ds x \circ 0_R$ Definition of Ring Zero $\ds \leadsto \ \$ $\ds \paren {x \circ 0_R} + \paren {x \circ 0_R}$ $=$ $\ds x \circ 0_R$ Ring Axiom $\text D$: Distributivity of Product over Addition $\ds \leadsto \ \$ $\ds \paren {x \circ 0_R} + \paren {x \circ 0_R}$ $=$ $\ds \paren {x \circ 0_R} + 0_R$ Definition of Ring Zero $\ds \leadsto \ \$ $\ds x \circ 0_R$ $=$ $\ds 0_R$ Cancellation Laws

Next:

 $\ds \paren {0_R + 0_R} \circ x$ $=$ $\ds 0_R \circ x$ Definition of Ring Zero $\ds \leadsto \ \$ $\ds \paren {0_R \circ x} + \paren {0_R \circ x}$ $=$ $\ds 0_R \circ x$ Ring Axiom $\text D$: Distributivity of Product over Addition $\ds \leadsto \ \$ $\ds \paren {0_R \circ x} + \paren {0_R \circ x}$ $=$ $\ds 0_R + \paren {0_R \circ x}$ Definition of Ring Zero $\ds \leadsto \ \$ $\ds 0_R \circ x$ $=$ $\ds 0_R$ Cancellation Laws

$\blacksquare$