Let $R$ be a commutative Noetherian ring and $I$, $J$ be two ideal of $R$. If $x\in R$, then is $((I+J):x)=(I:x)+(J:x)$?
I would be very grateful if someone comment me.
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1$\begingroup$ No. In $R=k[x,y]$ with $I=(x-y)$, $J=(y)$. The right hand side is $I+J=(x,y)$. The left hand side is $(x,y):x=(1)$. Note, these are graded ideals. $\endgroup$– Zach TeitlerCommented Mar 8, 2022 at 14:45
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2 Answers
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Theorem: Let $I,J\subset R$ be any ideals in a commutative Noetherian ring $R$, and let $x\in R$ be any element. Then we have the following implications:
(i) $I\cap J +(x)=(I+(x))\cap (J+(x)) \ \Leftrightarrow \ (I+J:x)=(I:x)+(J:x)$
(ii) $I+(x)$, $J+(x)$, and $I\cap J+(x)$ are all radical $\Rightarrow$ $I\cap J+(x)=(I+(x))\cap (J+(x))$.
Proof:
For (i), form a grid of short exact sequences of quotients of $R$, where the vertical sequences are multiplication by $x$ and the horizontal sequences are Mayer-Vietoris:
$$\begin{array}{ccccc}
& 0 & 0 & 0 \\
& \uparrow & \uparrow & \uparrow \\
& R/(I\cap J)+(x)\rightarrow & R/I+(x)\oplus R/J+(x) \rightarrow & R/(I+J+(x))\rightarrow & 0\\
& \uparrow & \uparrow & \uparrow \\
0 \rightarrow & R/I\cap J\rightarrow & R/I\oplus R/J\rightarrow & R/I+J\rightarrow
& 0\\
& \uparrow & \uparrow & \uparrow \\
0\rightarrow & R/I\cap J:x\rightarrow & R/I:x\oplus R/J:x\rightarrow & R/I+J:x & \\
& \uparrow & \uparrow & \uparrow \\
& 0 & 0 & 0 \\
\end{array}$$
Note that the top row is exact on the left iff $I\cap J+(x)=(I+(x))\cap (J+(x))$ and the bottom row is exact on the right iff $I+J:x=(I:x)+(J:x)$. Then (i) follows by the $3\times 3$ lemma.
To see (ii), you can use the $\mathcal{V}$ functor from algebraic geometry, where $\mathcal{V}(I)=\left\{P\in\operatorname{Spec}(R) \ | \ P\supseteq I\right\}$, and properties $\mathcal{V}(I\cap J)=\mathcal{V}(I)\cup\mathcal{V}(J)$ and $\mathcal{V}(I+J)=\mathcal{V}(I)\cap \mathcal{V}(J)$, to see that the ideals $I\cap J+(x)$ and $(I+(x))\cap (J+(x))$ have the same radical.
It would be nice to say that Zach Teitler's example fails because $I\cap J+(x)=(x-y)\cap (y)+(x)=(x,y^2)$ is not radical, but unfortunately, it would seem the implication in (ii) cannot be reversed. For example, in $R=k[x,y]$ take $I=(x)$, $J=(y)$, and take $f=x^2$, then $I\cap J+(f)=(xy,x^2)$ is not radical, yet we still have the equality $I\cap J+(f)=(I+(f))\cap (J+(f))$.
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$\begingroup$ Dear Chris , thank you so much for your nice Theorem and Lemma $\endgroup$ Commented Mar 10, 2022 at 8:53
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If $I$ and $J$ are monomial ideals, then the question is true.
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$\begingroup$ Sorry, but no, it is not sufficient for $I$ and $J$ to be monomial ideals. In $R = k[x,y]$ with $I=(x)$, $J=(y)$, the left hand side is $(I+J):(x+y) = (x,y):(x+y) = (1)$. Since $I:(x+y) = I$ and $J:(x+y) = J$, the right hand side is $I+J=(x,y)$. It's true that if $R = S[x_1,\dotsc,x_n]$ is a polynomial ring, $I$ and $J$ are monomial ideals, and $a \in R$ is a monomial, then $(I+J):a = (I:a)+(J:a)$. $\endgroup$ Commented Mar 10, 2022 at 7:30
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$\begingroup$ Sorry, my last comment was unclear. I was taking the ideals $I=(x)$ and $J=(y)$, and the element $x+y$. The original question uses "$x$" for the element, and for the value of "$x$" I am taking $x+y$. Perhaps it would be better to restate the question as: is $(I+J):a = (I:a)+(J:a)$? And then I am taking $a=x+y$. Or perhaps I should have used a polynomial ring with different variable names... Anyway, I apologize for any confusion. $\endgroup$ Commented Mar 10, 2022 at 8:04
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$\begingroup$ Dear Zach, Thank you so much for your nice explanations. $\endgroup$ Commented Mar 10, 2022 at 8:55
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1$\begingroup$ I think the question is true when $I$ and $J$ are monomial ideals and $x$ is monomial element of a polynomial ring. $\endgroup$ Commented Mar 10, 2022 at 8:57