Consider a ring $A$ and an affine scheme $X=\operatorname{Spec}A$ . 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 $IJ$ gives a new subscheme $V(IJ)$ 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 $V(IJ)$ in general?

Precise question: When is $I\cap J=IJ$?

Everybody knows the case $I+J=A$ but this is absolutely not necessary. For example if $A$ is a 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.


4 Answers 4


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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    $\begingroup$ 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. $\endgroup$ Aug 27, 2013 at 10:54
  • $\begingroup$ 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 $x$... $\endgroup$ Aug 28, 2013 at 0:47
  • $\begingroup$ ...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. $\endgroup$ Aug 28, 2013 at 4:38
  • $\begingroup$ In (4), I guess there is a missing "proper intersection" hypothesis, dim(A/I) + dim(A/J) = dim(A) (as in the Serre Local Algebra reference). $\endgroup$ Feb 18, 2022 at 4:45
  • $\begingroup$ @VictorWang: the depth formula forces this hypothesis, as dim is at least depth for each module and dim A=depth A. $\endgroup$ Feb 25, 2022 at 17:47

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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    $\begingroup$ The condition with Tor is looking more complicated than the question. $\endgroup$
    – user6976
    Dec 13, 2010 at 14:22
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    $\begingroup$ But it is more "geometric" since only $V(I)$ and $V(J)$ are involved. $\endgroup$ Dec 13, 2010 at 14:35
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    $\begingroup$ 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? $\endgroup$
    – t3suji
    Dec 13, 2010 at 15:38
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    $\begingroup$ @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. $\endgroup$ Dec 13, 2010 at 17:49
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    $\begingroup$ @Ariyan. Thanks! (I somehow always forget the "proper intersection" terminology, that's why my use of "purity" in place of... "propriety" or "properness".) $\endgroup$
    – t3suji
    Dec 13, 2010 at 18:00

I apologize in advance for resurrecting such an old question but I absolutely could not resist the urge of sharing a precise characterization for $I \cap J=IJ$, that I read recently in a beautiful short paper, when $I,J$ are monomial ideals in a polynomial ring.

First some terminology: Let $k$ be a field and $R=k[x_1,...,x_n]$ . Every monomial ideal $I$ of $R$ has a unique minimal monomial set of generators, usually denoted by $G(I)$ . For a set of monomials $T$ in $R$, let $\newcommand{\Supp}{\operatorname{Supp}}\Supp (T) :=\{i | x_i $ divides $m$ for some $m \in T \}$ .

With this, we can state the characterization: Let $I,J$ be monomial ideals in $k[x_1,...,x_n]$ , then $I \cap J=IJ$ if and only if $\Supp (G(I)) \cap \Supp (G(J))$ is empty.

This is Theorem 2.2 in https://link.springer.com/article/10.1007/s12044-019-0509-5

Here's the arxiv link https://arxiv.org/abs/1705.00488


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 prefer 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:

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

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

While using product you will get:

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

  • $\begingroup$ I am not totally convinced of this. If we consider the intersection of $X=V(y-x^2)$ with $Y_t=V(y-t)$ in the affine plane we have $X\cap Y_t$ consist of two points for every $t\neq 0$ so maybe we should expect a double point at the origin in the union $X\cup Y_0$ but the local ring there is reduced. However, I am changing from intersection to union thinking in this way so I am not sure if my intuition is correct. $\endgroup$ May 19, 2018 at 13:53

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