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:

$\operatorname{Rad} \left({\mathfrak a}\right) \subseteq \operatorname{Rad} \left({\mathfrak b}\right)$

Proof

Let $x \in \operatorname{Rad} \left({\mathfrak a}\right)$.

We show that $x \in \operatorname{Rad} \left({\mathfrak b}\right)$.

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 $x \in \operatorname{Rad} \left({\mathfrak b}\right)$.

$\blacksquare$