# Why to use such a complex definition of intersection multiplicity?

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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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.