Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 factor-ring 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.

share|improve this question
    
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 $Tor$. –  Karl Schwede Dec 13 '10 at 15:53
    
@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? –  Mark Sapir Dec 13 '10 at 16:23
    
Mark, yes it can be. It's also certainly implemented in various computer algebra programs such as Macaulay. –  Karl Schwede Dec 13 '10 at 16:55
    
@Karl: It is good to know. From time to time such questions arise in group theory too (say, when one studies solvable groups). –  Mark Sapir Dec 13 '10 at 17:40

1 Answer 1

up vote 7 down vote accepted

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 $( t\cdot I+(1-t)\cdot I ) \cap A$) 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 $k[X]$.

share|improve this answer
    
@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. –  Mark Sapir Dec 13 '10 at 16:26
    
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. –  J.C. Ottem Dec 13 '10 at 16:53
    
@J.C.: Thank you! –  Mark Sapir Dec 13 '10 at 17:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.