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Let $I=\langle a_1,\dots, a_s\rangle, J=\langle b_1,\dots, b_t\rangle$ be ideals of arbitrary commutative ring.

Then we know that $I+J=\langle a_1,\dots, a_s, b_1,\dots, b_t\rangle, IJ=\langle\{a_ib_j \mid 1 \leq i \leq s, 1\leq j \leq t\}\rangle$.

Also $IJ\subseteq I\cap J \subseteq I+J$.

I wonder about the generators of $I\cap J$. Is it possible that know the generators? Or is it finitely generated?

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    $\begingroup$ In a GCD domain you can define a $\mathrm{lcm}$ in which case $I\cap J = \langle\{\mathrm{lcm}(a_i,b_j)\}\rangle$. $\endgroup$ – JSchlather Sep 13 '12 at 0:59
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    $\begingroup$ @JSchlather Your claim is definitely wrong! $\endgroup$ – user26857 Mar 30 '14 at 19:08
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    $\begingroup$ Yes, JSchlather's formula is not correct. $(x,y) \cap (x+y) \neq (x(x+y),y(x+y))$ in $k[x,y]$. $\endgroup$ – Martin Brandenburg Nov 4 '14 at 21:55
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The intersection of any two finitely generated ideals in an integral domain $R$ is also finitely generated if and only if $R$ is coherent. An example of GCD domain which is not coherent can be found in Example 4.4 of this paper. So,

there are GCD domains which have finitely generated ideals whose intersection is not finitely generated.

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