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