Divisor of Unit is Unit

Theorem
Let $$\left({D, +, \circ}\right)$$ be an integral domain whose unity is $$1_D$$.

Let $$\left({U_D, \circ}\right)$$ be the group of units of $$\left({D, +, \circ}\right)$$.

Then:


 * $$x \in D, u \in U_D: x \backslash u \implies x \in U_D$$

That is, if $$x$$ is a divisor of a unit, $$x$$ must itself be a unit.

Proof
Let $$x \in D, u \in U_D$$ such that $$x \backslash u$$.

Then by definition, $$\exists t \in D: u = t \circ x$$.

Thus:
 * $$1_D = u^{-1} \circ u = u^{-1} \circ t \circ x$$.

It follows from the definition of a unit that $$x$$ is a unit, as it has an inverse, namely $$u^{-1} \circ t$$.