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:


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

Also, as $D$ is an integral domain and hence a commutative ring, we have:


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

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