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Apologies for what I expect to be an elementary question, but take into account that it's being asked by an old-school algebraic topologist who, probably to his detriment, has avoided algebraic geometry most of his career.

Here's the simplest case of what I'm looking for. Consider $n$ polynomials $f_1$, $\ldots$, $f_n$ on $\mathbb{C}^n$ and assume that they have an isolated common zero at the origin, but don't assume that the corresponding affine varieties intersect transversely there. There are (at least!) two definitions of the multiplicity of this zero:

Geometric definition: The multiplicity is the dimension over $\mathbb C$ of $\mathscr{O}_0/(f_1,\ldots,f_n)$, where $\mathscr{O}_0 = \mathscr{O}_{0,\mathbb{A}^n}$ is the local ring at the origin.

Topological definition: The functions $f_1$, $\ldots$, $f_n$ define a map $\mathbb{C}^n\to \mathbb{C}^n$. If we look at a small disc around the origin in the source, the boundary sphere does not hit the origin in the target, so its image lies outside a small disc around the origin in the target. Collapsing out the boundary in the source disc and everything outside of the boundary in the target defines a map $S^{2n}\to S^{2n}$ and we take the multiplicity to be the degree of this map.

I'm assuming that it's elementary and well-known that these two definitions agree. What I'm looking for is a reference I can point to so I can use this fact. (I'd also be interested in its proof, as it's currently not entirely obvious to me.)

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  • $\begingroup$ I think some kind of formal proof can be found in Griffits&Harris in Applications of Resudues $\endgroup$
    – Bad English
    Commented Apr 5, 2021 at 20:02
  • $\begingroup$ Non-formal idea is the following. If you fix a cycle of integration, then a small perturbation of the map gives a perturbation of the integral, which is an integral number. On the other hand you can always specify a perturbation $f_i^\varepsilon$ s.t. $(f^\varepsilon)^{-1}(0)$ is simple, hence it is enough to check that the dimension of the intersection algebra is the same after perturbation. It follows from the fact that $f_i$ is regular, hence a deformation of the Kozsul complex has the same dimension of the cohomology and is equal to the Euler characteristic, which is constant. $\endgroup$
    – Bad English
    Commented Apr 5, 2021 at 20:15
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    $\begingroup$ @BadEnglish, yes, Griffiths & Harris has exactly the statement I'm looking for, thanks! If you want to give that as an answer I'll mark it as the accepted answer. $\endgroup$
    – Steve Costenoble
    Commented Apr 6, 2021 at 0:43

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It is described in detail in Appendix E.3 of the Mond—Nuño-Ballesteros book https://link.springer.com/book/10.1007/978-3-030-34440-5

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    $\begingroup$ Thanks for the reference, that does also have the result I was looking for. $\endgroup$
    – Steve Costenoble
    Commented Aug 12, 2022 at 17:56

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