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Let $I$, $J$ be finitely generated ideals in a ring $A$ (commutative with identity). I know that the intersection need not be finitely generated: can somebody give me an example?


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up vote 5 down vote accepted

This is a concrete answer.

Set $R=\mathbb Z+X^2\mathbb Q[X]$, $I=X^2R$, and $J=X^3R$. It's not hard to show that $I\cap J=X^5\mathbb Q[X]$. Assume that $I\cap J=f_1R+\cdots+f_nR$ and write $f_i=X^5g_i$ with $g_i\in\mathbb Q[X]$. Then write $g_i(0)=\frac{a_i}{b_i}$, where $a_i,b_i\in\mathbb Z$, $b_i\ge 1$, and $(a_i,b_i)=1$ for all $i=1,\dots,n$. Now define $g=\frac{1}{b+1}X^5$, where $b=b_1\cdots b_n$. Since $g\in I\cap J$ there exist $h_1,\dots, h_n\in R$ such that $g=f_1h_1+\cdots+f_nh_n$. By cancelling $X^5$ and sending $X$ to $0$ we get $\frac{1}{b+1}=\frac{a_1}{b_1}h_1(0)+\cdots+\frac{a_n}{b_n}h_n(0)$, where $h_i(0)\in\mathbb Z$, a contradiction.

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Thank you very much, it's a very clever example! – john Feb 7 '13 at 14:15

This is not a concrete answer.

Let $R$ be an integral domain. Then the intersection of two finitely generated ideals is finitely generated iff $R$ is coherent. If $V$ is a valuation ring of rank $>1$, then $V[[X]]$ is not coherent, so it contains two finitely generated ideals whose intersection is not finitely generated.

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I thought that $R[[X]]$ is noetherian when $R$ is noetherian? – Martin Brandenburg Feb 6 '13 at 22:27
@MartinBrandenburg That's right, but obviously here $V$ is considered non-discrete. – user26857 Feb 6 '13 at 22:31

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