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:
 * $\map \Rad {\mathfrak a + \mathfrak b} = \map \Rad {\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} }$

Proof
From Ideal of Ring is Contained in Radical:
 * $\mathfrak a \subseteq \map \Rad {\mathfrak a}$

From Sum of Larger Ideals is Larger:
 * $\mathfrak a + \mathfrak b \subseteq \map \Rad {\mathfrak a} + \map \Rad {\mathfrak b}$

From Radical of Ideal Preserves Inclusion:
 * $\map \Rad {\mathfrak a + \mathfrak b} \subseteq \map \Rad {\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} }$

Let $x \in \map \Rad {\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} }$.

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

By definition of sum of ideals, there exist $a \in \map \Rad {\mathfrak a}$ and $b \in \map \Rad {\mathfrak b}$ 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:
 * $\paren {a + b}^{p + q - 1} \in \mathfrak a + \mathfrak b$

Thus:
 * $x^{n \cdot \paren {p + q - 1} } \in \mathfrak a + \mathfrak b$

That is:
 * $x \in \map \Rad {\mathfrak a + \mathfrak b}$

Hence by definition of subset:
 * $\map \Rad {\mathfrak a + \mathfrak b} \supseteq \map \Rad {\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} }$

The result follows by definition of set equality.