Unit Ideal iff Radical is Unit Ideal
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Theorem
Let $A$ be a commutative ring with unity.
Let $\mathfrak a$ be an ideal of $A$.
Then:
- $ \mathfrak a = A \iff \map \Rad {\mathfrak a} = A$
where:
- $A$ is called the unit ideal of $A$
- $\map \Rad {\mathfrak a}$ denotes the radical of $\mathfrak a$
Proof
\(\ds \mathfrak a\) | \(=\) | \(\ds A\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds 1\) | \(\in\) | \(\ds A\) | Ideal is Unit Ideal iff Includes Unity | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists n \in \Z_{>0}: \, \) | \(\ds 1^n\) | \(\in\) | \(\ds A\) | since $ 1^n = 1$ by Definition of Unity of Ring | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds 1\) | \(\in\) | \(\ds \map \Rad {\mathfrak a}\) | Definition 1 of Radical of Ideal of Ring | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \map \Rad {\mathfrak a}\) | \(=\) | \(\ds A\) | Ideal is Unit Ideal iff Includes Unity |
$\blacksquare$