Let $R$ be a commutative Noetherian ring, $\mathfrak{a}$ an ideal and $x, y$ elements. Is it true that
$$\mathfrak{a}(\mathfrak{a}:x) \cap \mathfrak{a}(\mathfrak{a}:y) = \mathfrak{a}(\mathfrak{a}:(x,y))?$$



ideal operations
No it is not true.


Let $R=\mathbb{K}[a,b,x,y]/\langle abx-aby\rangle$ and $\mathfrak{a}=\langle ax,by\rangle$.


Now $abx\in\mathfrak{a}(\mathfrak{a}:x)\cap\mathfrak{a}(\mathfrak{a}:y)$ but $abx\notin\mathfrak{a}(\mathfrak{a}:\langle x,y\rangle)$.


