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.