Ideal of Ring is Contained in Radical
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Theorem
Let $A$ be a commutative ring with unity.
Let $\mathfrak a \subseteq A$ be an ideal.
Then $\mathfrak a$ is contained in its radical:
- $\mathfrak a \subseteq \operatorname{Rad} \left({\mathfrak a}\right)$
Proof
Let $a \in \mathfrak a$.
By definition of power:
- $a^1 = a$
Thus:
- $a \in \operatorname{Rad} \left({\mathfrak a}\right)$
$\blacksquare$