Element in Integral Domain is Unit iff Principal Ideal is Whole Domain
Jump to navigation
Jump to search
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: $\text{(ii)}$