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

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

Then $R$ has no zero divisors iff $\left({R^*, \circ}\right)$ is a semigroup.

Proof

 * Let $\left({R, +, \circ}\right)$ be a non-null ring with no zero divisors.

The set $R^* = R - \left\{{0_R}\right\} \ne \varnothing$ as $\left({R, +, \circ}\right)$ is non-null.

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

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

From Restriction of Operation Associativity, ring product is associative on $\left({R^*, +, \circ}\right)$, as it is associative on $\left({R, +, \circ}\right)$.

Thus $\left({R^*, \circ}\right)$ is a semigroup.


 * Now suppose $\left({R^*, \circ}\right)$ is 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.