I am reading some Commutative Algebra notes, and have come across the following result:
Let $R$ be a commutative unital ring, and let $a$ and $b$ be ideals of R. Then we have $(a \cap b) \cdot (a \cap b) \subseteq a \cdot b \subseteq a \cap b$ and thus $r(a \cdot b) = r(a \cap b)$. Here, $r( \bullet )$ is the radical. My question is where we get this result from, or how it follows from the inclusions
For two ideals $\mathfrak{a},\mathfrak{b}$ we always have
$$\mathfrak{a} \subset \mathfrak{b} \; \Rightarrow r(\mathfrak{a})\subset r(\mathfrak{b}).$$
But we also have $r(\mathfrak{a})\subset r(\mathfrak{a}^2)$, since $x\in r(\mathfrak{a})$ just means $x^n\in \mathfrak{a}$ for some $n$, and since $\mathfrak{a}$ is an ideal, then also $x^{2n}\in \mathfrak{a}^2$, thus giving $x\in r(\mathfrak{a}^2)$. Since $\mathfrak{a}^2 \subset \mathfrak{a}$,
$$ r(\mathfrak{a}) = r(\mathfrak{a}^2).$$
Then you can argue
$$r((\mathfrak{a} \cap \mathfrak{b})\cdot (\mathfrak{a} \cap \mathfrak{b})) \subset r(\mathfrak{a}\cdot \mathfrak{b}) \subset r(\mathfrak{a} \cap \mathfrak{b}) = r((\mathfrak{a} \cap \mathfrak{b})\cdot (\mathfrak{a} \cap \mathfrak{b})),$$
which then must entail equality.
