Radical of Ideal Preserves Inclusion

Theorem
Let $A$ be a commutative ring with unity.

Let $\mathfrak a \subseteq \mathfrak b \subseteq A$ be an ideals.

Then we have an inclusion of their radicals:
 * $\sqrt {\mathfrak a} \subseteq \sqrt {\mathfrak b}$

Proof
Let $x \in \sqrt {\mathfrak a}$.

We show that $x \in \sqrt {\mathfrak b}$.

By definition of radical, there exists $n \in \N$ such that the power $x^n \in \mathfrak a$.

Because $\mathfrak a \subseteq \mathfrak b$, also $x^n \in \mathfrak b$.

Thus $x \in \sqrt {\mathfrak b}$.