Element in Integral Domain is Unit iff Principal Ideal is Whole Domain

Theorem

Let $\struct {D, +, \circ}$ be an integral domain.

Let $U_D$ be the group of units of $D$.

Let $\ideal x$ be the principal ideal of $D$ generated by $x$.

Let $x, y \in \struct {D, +, \circ}$.

Then:

$x \in U_D \iff \ideal x = D$

Proof

\(\ds x\)

\(\in\)

\(\ds U_D\)

\(\ds \leadsto \ \ \)

\(\ds \exists u \in U_D: \, \)

\(\ds u\)

\(\in\)

\(\ds \ideal x\)

Ideal of Unit is Whole Ring

\(\ds \leadsto \ \ \)

\(\ds \ideal x\)

\(=\)

\(\ds D\)

Ideal of Unit is Whole Ring

Conversely:

\(\ds \ideal x\)

\(=\)

\(\ds D\)

\(\ds \leadsto \ \ \)

\(\ds 1_D\)

\(\in\)

\(\ds \ideal x\)

\(\ds \leadsto \ \ \)

\(\ds \exists t \in D: \, \)

\(\ds 1\)

\(=\)

\(\ds t x\)

\(\ds \leadsto \ \ \)

\(\ds u\)

\(\in\)

\(\ds U_D\)

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 62.3$ Factorization in an integral domain: $\text{(ii)}$