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:
 * $\operatorname{Rad} \left({\mathfrak a + \mathfrak b}\right) = \operatorname{Rad} \left({\operatorname{Rad} \left({\mathfrak a}\right) + \operatorname{Rad} \left({\mathfrak b}\right)}\right)$

Proof
The inclusion $\operatorname{Rad} \left({\mathfrak a + \mathfrak b}\right) \subseteq \operatorname{Rad} \left({\operatorname{Rad} \left({\mathfrak a}\right) + \operatorname{Rad} \left({\mathfrak b}\right)}\right)$ follows from:


 * Ideal of Ring is Contained in Radical:
 * $\mathfrak a \subseteq \operatorname{Rad} \left({\mathfrak a}\right)$


 * Sum of Larger Ideals is Larger:
 * $\mathfrak a + \mathfrak b \subseteq \operatorname{Rad} \left({\mathfrak a}\right) + \operatorname{Rad} \left({\mathfrak b}\right)$


 * Radical of Ideal Preserves Inclusion:
 * $\operatorname{Rad} \left({\mathfrak a + \mathfrak b}\right) \subseteq \operatorname{Rad} \left({\operatorname{Rad} \left({\mathfrak a}\right) + \operatorname{Rad} \left({\mathfrak b}\right)}\right)$

For the other inclusion, let $x \in \operatorname{Rad} \left({\operatorname{Rad} \left({\mathfrak a}\right) + \operatorname{Rad} \left({\mathfrak b}\right)}\right)$.

Then there exists $n \in \N$ such that the power $x^n \in \operatorname{Rad} \left({\mathfrak a}\right) + \operatorname{Rad} \left({\mathfrak b}\right)$.

By definition of sum of ideals, there exist $a \in \operatorname{Rad} \left({\mathfrak a}\right)$ and $b \in \operatorname{Rad} \left({\mathfrak b}\right)$ 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:
 * $\left({a + b}\right)^{p + q - 1} \in \mathfrak a + \mathfrak b$

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

That is:
 * $x \in \operatorname{Rad} \left({\mathfrak a + \mathfrak b}\right)$.