Ideal of Ring is Contained in Radical

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:
 * $\mathfrak a \subseteq \sqrt {\mathfrak a}$

Proof
Let $a \in \mathfrak a$.

By definition of power, $a^1 = a$.

Thus $a \in \sqrt {\mathfrak a}$.