For ideals in a commutative ring, $((\hat{a} : \hat{b}) : \hat{c}) = (\hat{a}: \hat{b}\hat{c})$.

Question

Let $A$ be a commutative ring and $\hat{a}, \hat{b}, \hat{c}$ be ideals in $A$.

The ideal quotient is an ideal defined by $(\hat{a} : \hat{b}) = \{x \in A : x\hat{b} \subset \hat{a}\}$.

I know that $(\hat{a} : \hat{b})\hat{b} \subset \hat{a}$ and that gives $x \in ((\hat{a} : \hat{b}) : \hat{c}) \implies x\hat{c} \subset (\hat{a}:\hat{b}) \implies x \hat{b}\hat{c} \subset \hat{a} \implies ((\hat{a}:\hat{b}) : \hat{c}) \subset (\hat{a}: \hat{b}\hat{c})$.

But how do we get the reverse inclusion?

So far I have:

$x \in (\hat{a} : \hat{b}\hat{c}) \implies x \hat{b}\hat{c} \subset \hat{a} \subset (\hat{a}: \hat{b})$.

Answer 1

Let $x\in (\hat{a}:\hat{b}\hat{c})$. We want to show that $x\in ((\hat{a}:\hat{b}):\hat{c})$. That is, we want to show $x\hat{c}\subset (\hat{a}:\hat{b})$. Let $z\in\hat{c}$. We want to show $xz\in (\hat{a}:\hat{b})$. Let $y\in \hat{b}$. Then $xzy=xyz\in x\hat{b}\hat{c}\subset \hat{a}$. So $xz\in (\hat{a}:\hat{b})$. Hence $x\in ((\hat{a}:\hat{b}):\hat{c})$.