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Let $X$ be a smooth variety and $V, W$ two closed irreducible and reduced subvarieties represented by ideal sheaves $I$ and $J$. Serre defines an intersection multiplicity for an irreducible component $Z$ of $V\cap W$ as $$ \mu(Z;V,W)=\sum_{i=0}^\infty (-1)^i \operatorname{length}_{\mathcal{O}_{X,z}} (\operatorname{Tor}^i_{\mathcal{O}_{Z,z}}(\mathcal{O}_{X,z}/I,\mathcal{O}_{X,z}/J)) $$ where $z$ is the generic point of $Z$.

The first summand of this sum is $$ \operatorname{length}_{\mathcal{O}_{X,z}} (\mathcal{O}_{X,z}/I \otimes_{\mathcal{O}_{X,z}}\mathcal{O}_{X,z}/J) = \operatorname{length}_{\mathcal{O}_{X,z}}(\mathcal{O}_{Z,z}) $$ and this is what has a geometric interpretation as the intersection multiplicity for me. Can someone explain to me at a concrete geometric example, why the "naive definition" isn't sufficient? In which often appearing cases is it sufficient?

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up vote 13 down vote accepted

See this MO questions and its answer for a discussion of this.

The basic point is that you have go to non-Cohen--Macaulay contexts to find examples where the higher Tors contribute.

E.g. consider the union of two planes in $\mathbb A^4$ meeting a point, e.g. $x = y = 0$ and $z = w = 0$. Now intersect them with a third plane which meets them just as this point, e.g. $x = z, y = w$. Then the intersection multiplicity should be $2$; for each of the first two planes separately, the intersection with the third plane is a transverse intersection in a single point, so the multiplicity is one. And the multiplicity should be additive when we take the union of the two planes.

But if you compute the tensor product of your question, you will get a length of $3$, not of $2$. It is corrected back to $2$ by the Tor terms in the formula.

(Note that the union of the two planes meeting just at a point is the most basic example of a non-Cohen--Macaulay algebraic set.)

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