Unit of Integral Domain divides all Elements

Theorem

Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.

Let $\struct {U_D, \circ}$ be the group of units of $\struct {D, +, \circ}$.

Then:

$\forall x \in D: \forall u \in U_D: u \divides x$

That is, every unit of $D$ is a divisor of every element of $D$.

Proof

\(\ds \forall x \in D, u \in U_D: \, \)

\(\ds x\)

\(=\)

\(\ds u \circ \paren {u^{-1} \circ x}\)

Definition of Unit of Ring

\(\ds \leadsto \ \ \)

\(\ds u\)

\(\divides\)

\(\ds x\)

Definition of Divisor of Ring Element

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 62.1$ Factorization in an integral domain: $\text{(ii)}$