# When is the product of two ideals equal to their intersection?

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.

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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.

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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

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 $\mathrm{Tor}^1(A/I, A/J)=0$.

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The condition with Tor is looking more complicated than the question. –  Mark Sapir Dec 13 '10 at 14:22
But it is more "geometric" since only $V(I)$ and $V(J)$ are involved. –  Martin Brandenburg Dec 13 '10 at 14:35
@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
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
@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. –  Ariyan Javanpeykar Dec 13 '10 at 17:49

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:

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

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