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

$\blacksquare$