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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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Acutally, I voted to close this, but I then realized that it was not as trivial as I thought. Sorry! I'd retract my vote to close if I could... –  Andy Putman Jan 31 '13 at 16:01
    
I need this equation for my problem. However, I think it should have a counterexample. –  Pham Hung Quy Jan 31 '13 at 16:26
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It is true that $(\mathfrak{a}:x)\cap(\mathfrak{a}:y)=(\mathfrak{a}:(x,y))$ (see Atiyah-Macdonald p. 8, Ex. 1.12(v)). So the question is, what happens after both sides are multiplied by $\mathfrak{a}$? –  Mahdi Majidi-Zolbanin Jan 31 '13 at 17:03
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Let $R=\mathbb{C}[a,b,x,y]/\langle abx-aby\rangle$. And let $\mathfrak{a}=\langle ax,by\rangle$. Now $aby\in \mathfrak{a}(\mathfrak{a}:x)\cap \mathfrak{a}(\mathfrak{a}:y)$ but $aby\notin \mathfrak{a}(\mathfrak{a}:\langle x,y\rangle)$. This should be your counterexample. –  Patrik Feb 1 '13 at 14:37
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That's a nice example. Why don't you post that as an answer? –  Mahdi Majidi-Zolbanin Feb 1 '13 at 16:30

1 Answer 1

up vote 8 down vote accepted

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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