# Ring Less Zero is Semigroup for Product iff No Proper Zero Divisors

## Theorem

Let $\struct {R, +, \circ}$ be a non-null ring.

Then $R$ has no zero divisors if and only if $\struct {R^*, \circ}$ is a semigroup.

## Proof

### Necessary Condition

Let $\struct {R, +, \circ}$ be a non-null ring with no zero divisors.

The set $R^* = R \setminus \set {0_R} \ne \O$ as $\struct {R, +, \circ}$ is non-null.

All elements of $R$ are not zero divisors, and therefore are cancellable.

$\struct {R, +, \circ}$ is closed under $\circ$, from the fact that there are no zero divisors, and also that $\struct {R, \circ}$ is also closed.

From Restriction of Associative Operation is Associative, ring product is associative on $\struct {R^*, +, \circ}$, as it is associative on $\struct {R, +, \circ}$.

Thus $\struct {R^*, \circ}$ is a semigroup.

$\Box$

### Sufficient Condition

Let $\struct {R^*, \circ}$ be a semigroup.

Then:

- $\neg \exists x, y \in R^*: x \circ y \notin R^*$

Thus:

- $\neg \exists x, y \in R^*: x \circ y = 0_R$

and the result follows.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 21$: Theorem $21.2$: Corollary - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 55.2$ Special types of ring and ring elements