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

Consider a ring $A$ and an affine scheme $X=SpecA$ . Given two ideals $I$ and $J$ and their associated subschemes $V(I)$ and $V(J)$, we know that the intersection $I\cap J$ corresponds to the union $V(I\cap J)=V(I)\cup V(J)$. But a product $I.J$ gives a new subscheme $V(I.J)$ which has same support as the union but can be bigger in an infinitesimal sense. For example if $I=J$ you get a scheme $V(I^2)$ which is equal to "double" $V(I)$.

Vague Question : What is geometric interpretation of $V(I.J)$ in general?

Precise question : When is $I\cap J=I.J$? Everybody knows the case $I+J=A$ but this is absolutely not necessary. For example if $A$ is UFD and $f,g$ are relatively prime then $(f).(g)=(f)\cap(g) $ but in general $(f)+(g)\neq A$ (e.g. $f=X, g=Y \in k[X, Y]$)

Thank you very much.

share|improve this question
add comment

3 Answers

up vote 21 down vote accepted

To add to David Speyer's answer, since this story continues with a rather interesting and illustrious history:

When $A$ is regular, the Tor functor satisfies the following property:

(1) $\text{Tor}_1^A(M,N) = 0$ implies $\text{Tor}_i^A(M,N) = 0$ for $i>0$ for any two finitely generated modules.

(this is a theorem by Auslander in the geometric and unramified case and Lichtenbaum in the ramified case. (1) is called the rigidity of Tor).

It turns out that when $A$ is regular and local (so one can talk about depth), (1) implies

(2) $\text{depth} (M) + \text{depth}(N) = \dim A + \text{depth} {M\otimes_AN}$

This stunning formula looks exactly the same as the property of "proper intersection" in intersection theory, except that one uses depth instead of dimension. Note that if $M=A/I, N=A/J$ then $M\otimes N = A/(I+J)$, which represents the intersection of $V(I)$ and $V(J)$, so this is very geometric.

(3) Talking about intersection theory, by Serre formula for intersection multiplicity, as all the Tors vanish, one can compute the intersection multiplicity of $V(I), V(J)$ by counting the length at the minimal components (i.e. the naive way). So you will have a generalization of Bezout theorem.

Finally, if $V(I)$ and $V(J)$ only intersect at isolated closed points, (2) implies (1) locally on the support of the intersection, so

(4) If $V(I) \cap V(J)= \{m_1, \cdots, m_n \}$ then $I\cap J = IJ$ if and only if $A/I, A/J$ are locally Cohen-Macaulay at the points $m_i$s.

You can find the last statement in Serre's Local Algebra book, V.6, Theorem 4, p 110 of the English version.

PS: Also, David did not mention his own interesting contribution, here.

share|improve this answer
    
Nice exhaustive answer, so let me ask a stupid reference. I do not want to prove that if $f,g$ have no common factor in a UFD then $(f)\cap(g)=(f\cdot g)$ in a paper I am writing, but I found no explicit reference: do you know any? Adapting Serre's criterion seems a bit overkilling, for a UFD...Thanks. –  Filippo Alberto Edoardo Aug 27 '13 at 10:54
    
Filippo: Any thing in $(f)\cap (g)$ would be of the form $fx=gy$. By writing both sides a product of irreducibles one concludes that $g$ divides $x$... –  Hailong Dao Aug 28 '13 at 0:47
    
...hmm, I guess I agree and that NO reference is by far the best option. I was probably a bit puzzled when I asked, sorry. –  Filippo Alberto Edoardo Aug 28 '13 at 4:38
add comment

Answer to the precise question: When $\mathrm{Tor}^1(A/I, A/J)=0$.

Proof: We have the exact sequence $$0 \to I \to A \to A/I \to 0$$ Tensoring with $A/J$, we get $$0 \to \mathrm{Tor}^1(A/I, A/J) \to I/(I \cdot J) \to A/J \to A/(I+J) \to 0.$$ The left hand term is $0$ because $A$ is flat as an $A$-module.

Now, what is the kernel of $I \mapsto A/J$? Clearly, it is $I \cap J$. So the kernel of $I/(I \cdot J) \to A/J$ is $(I \cap J)/(I \cdot J)$. We see that $I \cap J = I \cdot J$ if and only if $\mathrm{Tor}^1(A/I, A/J)=0$.

share|improve this answer
3  
The condition with Tor is looking more complicated than the question. –  Mark Sapir Dec 13 '10 at 14:22
1  
But it is more "geometric" since only $V(I)$ and $V(J)$ are involved. –  Martin Brandenburg Dec 13 '10 at 14:35
1  
@Martin: So you think it is better than the question? If we have Groebner bases of $I$, and $J$, can we decide whether $IJ=I\cap J$? I think that can be an interesting question. In fact I am not sure that David's answer gives any algorithm to decide $IJ=I\cap J$. It must be decidable, though. –  Mark Sapir Dec 13 '10 at 14:47
4  
I agree that the condition with Tor is more geometric --- it can be viewed as a kind of `purity' of intersection (for instance, two smooth subvarieties of a smooth variety have this property if and only if their intersection has the expected dimension). Is there an accepted name for this condition? –  t3suji Dec 13 '10 at 15:38
2  
@t3suji. Let V and W be closed integral subschemes of a nonsingular quasi-projective irreducible variety. Then, for any irreducible component Z of VcapW, it holds that codim Z <= codim V + codim W. (See Serre's Local Algebra.) We say that V and W intersect properly in Z if equality holds. A stronger condition is being in general position. If V and W are in general position all the higher Tor's vanish. The cycle [VcapW] associated to VcapW is then equal to the product cycle [V][W]. As far as I know, this is standard language in intersection theory for algebraic varieties. –  Ari Dec 13 '10 at 17:49
show 3 more comments

A vague answer to the vague question:

When you want the union of $V(I)$ and $V(J)$ to behave well under deformations and to `count with multiplicity', then you may prefeer to use the ideal $IJ$ rather than $I\cap J$. Let me give an example:

Take $V=V(x)$, $W=V(x-t)$ and $T=V(t)$ denote $V_0:=V\cap T=V(x,t)=W\cap T=:W_0$. If you use intersection of ideals for the union of varieties you will get:

$(V\cup W)\cap T=V(x^2)$, and

$(V_0\cup W_0)\cap T=V(x)$.

While using product you will get:

$(V\cup W)\cap T=V(x^2)=(V_0\cup W_0)\cap T$.

share|improve this answer
add comment

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.