This question was inspired by this one. Given two ideals $A,B$ in a finitely generated commutative ring $R$. Is it possible to decide whether $A\cap B=AB$? Here $R$ is given by generators and relations, i.e. as a factorring of the free commutative ring over a finitely generated ideal, $A$, $B$ are given by their (finite) generating sets. I do not remember seeing this question in standard books on Groebner bases, but perhaps it is hidden there.

$\begingroup$ Well, if you can compute $A \cap B$, then computing $AB$ is quite fast, and then use standard equality testing methods. As David Speyer pointed out in the previous question, you could also compute $Tor$. $\endgroup$ – Karl Schwede Dec 13 '10 at 15:53

$\begingroup$ @Karl: That is all in the ring of polynomials (say, over $\mathbb Z$). In general, you have three ideals $R, A, B$ in the polynomial ring, and you need to check $(A+R)\cap (B+R)=(AB)+R$. But I guess that can also be done using Groebner bases. Right? $\endgroup$ – user6976 Dec 13 '10 at 16:23

$\begingroup$ Mark, yes it can be. It's also certainly implemented in various computer algebra programs such as Macaulay. $\endgroup$ – Karl Schwede Dec 13 '10 at 16:55

$\begingroup$ @Karl: It is good to know. From time to time such questions arise in group theory too (say, when one studies solvable groups). $\endgroup$ – user6976 Dec 13 '10 at 17:40
I think this problem can in fact be handled by Gröbner basis theory in the case $A$ is a polynomial ring. Since $I\cdot J \subseteq I\cap J$ for any two ideals, one can simply compute a Gröbner basis of $I\cap J$ (which is computed as the elimination ideal $( t\cdot I+(1t)\cdot I ) \cap A$) and then checking whether each generator belongs to $I\cdot J$ (again using Gröbner basis algorithm).
EDIT: As Mark points out in his comment, this argument can be used to solve the problem for the general case $A=k[X]/R$ by considering the ideals $I+R$ and $J+R$ in $k[X]$.

$\begingroup$ @J.C.: It is nice. If you have three ideals $R, I, J$ in the polynomial ring, can you also compute $(I+R)\cap (J+R)$? It would be enough. $\endgroup$ – user6976 Dec 13 '10 at 16:26

$\begingroup$ Yes, it is possible. The union of the gröbner bases of I and R, is the grobner basis of the union $I+R$. This shows that one can compute $I+R$ and $J+R$ and finally their intersection by the above argument. $\endgroup$ – J.C. Ottem Dec 13 '10 at 16:53

$\begingroup$ @J.C.: Thank you! $\endgroup$ – user6976 Dec 13 '10 at 17:40