Radical of Sum of Ideals

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

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

Then for the radical of their sum we have:
 * $\sqrt{\mathfrak a + \mathfrak b} = \sqrt{\sqrt{\mathfrak a} + \sqrt{\mathfrak b}}$

Proof
The inclusion $\sqrt{\mathfrak a + \mathfrak b} \subseteq \sqrt{\sqrt{\mathfrak a} + \sqrt{\mathfrak b}}$ follows from:
 * Ideal of Ring is Contained in Radical: $\mathfrak a \subseteq \sqrt{\mathfrak a}$
 * Sum of Larger Ideals is Larger: $\mathfrak a + \mathfrak b \subseteq \sqrt{\mathfrak a} + \sqrt{\mathfrak b}$
 * Radical of Ideal Preserves Inclusion: $\sqrt{\mathfrak a + \mathfrak b} \subseteq \sqrt{\sqrt{\mathfrak a} + \sqrt{\mathfrak b}}$

For the other inclusion, let $x \in \sqrt{\sqrt{\mathfrak a} + \sqrt{\mathfrak b}}$.

Then there exists $n \in \N$ such that the power $x^n \in \sqrt{\mathfrak a} + \sqrt{\mathfrak b}$.

By definition of sum of ideals, there exist $a \in \sqrt{\mathfrak a}$ and $b \in \sqrt{\mathfrak b}$ with $x^n = a + b$.

Let $p, q \in \N$ with $a^p \in \mathfrak a$ and $b^q \in \mathfrak b$.

By the Binomial Theorem, $(a+b)^{p+q-1} \in \mathfrak a + \mathfrak b$.

Thus $x^{n \cdot (p+q-1)} \in \mathfrak a + \mathfrak b$.

That is, $x \in \sqrt{\mathfrak a + \mathfrak b}$.