I was reading the wikipedia page on intersection number: http://en.wikipedia.org/wiki/Intersection_number#Intersection_multiplicities_for_plane_curves

and I was confused by property number 6 they want there. I cannot justify why this is a natural property to have.

I have seen the definition of intersection number via dimension of the complete local ring of the scheme theoretic intersection over the base field. From this definition property 6 is natural, but I would like to see a low-brow (i.e. in classical language), intuitive reason for expecting 6.


  • 1
    $\begingroup$ I think a possible explanation is that you parametrise the curve $Q=0$ by $(x(t),y(t))$ and define the intersection number of $P=0$ and $Q=0$ at $(x(t_0),y(t_0))$ to be the vanishing order of $P(x(t),y(t))$ at $t=t_0$, then 6 is trivial. However, 1 is not obvious. $\endgroup$ – Yuchen Liu May 12 '12 at 9:17
  • $\begingroup$ Thanks! That's almost satisfactory for me, except this one example that still bugs me: intersection of $y = 0$ and $x^2 = 0$ at the origin. Why is the "correct" parametrization of $x^2 = 0$ $(0,t^2)$? $\endgroup$ – user27126 May 12 '12 at 9:46
  • $\begingroup$ I think my explanation requires that both $P$ and $Q$ are irreducible at $(x(t_0),y(t_0))$ in sense of germs of holomorphic functions, and you can use property 5 to reduce any case to that special case. In your example, $x^2=x\cdot x$, so all you need is to parametrise $x=0$, calculate the intersection number of $y=0$ and $x=0$ then double it. $\endgroup$ – Yuchen Liu May 15 '12 at 6:56

The point is that the system of equations $P = Q = 0$ is equivalent to $P + RQ = Q = 0$, meaning you can solve one if and only if you can solve the other. Another point of view is that the intersection multiplicity of $P$ and $Q$ at $p$ should only depend on the ideal $I = (P,Q)$, suitably localized to only look at what is happening near $p$ (and not at other intersection points). But on the level of ideals, $(P,Q) = (P + RQ, Q)$. This last equality is really what my first sentence is expressing more loosely.

  • $\begingroup$ Thanks. What you said sort of makes sense but somehow it still doesn't fully click. I guess deep down I am not super convinced that the intersection of say, Spec R/I and Spec R/J should look like Spec R/I+J. (So that I don't feel strong about the ideal viewpoint). Do you mind saying a few words on why that should be true intuitively? (Say, why I+J rather than some other ideal with the same radical?) $\endgroup$ – user27126 May 29 '12 at 19:55
  • $\begingroup$ The basic idea is to have a "principle of continuity." For example, if you intersect $y = x^2 - c$ and $y = 0$, you get two points of intersection when $c \neq 0$ (and the intersection is transverse). So we'd like to say that $y = 0$ and $y = x^2$ intersect "twice" (or better: with multiplicity 2). That's clearly the case here if we use the ideal definition. With more work, you can show that in the case of plane curves over $\mathbb{C}$, the ideal theoretic viewpoint agrees with the "perturb intersection points so they become transverse and then count the number of intersections" approach. $\endgroup$ – Michael Joyce May 29 '12 at 21:55
  • $\begingroup$ Thanks again. I guess I can summarize my confusion here: $\endgroup$ – user27126 May 29 '12 at 23:30
  • $\begingroup$ 1. To your first sentence: Somehow I am not very convinced that P = Q = 0 and P+RQ = Q = 0 are the same thing in terms of multiplicity. I'll try to think more about it. 2. When you switch to idealic viewpoint, I feel troubled why (P,Q) really captures the intersection information, as opposed to some other ideal with the same radical. 3. Your last comment seems to be unrelated to the topic hand - you are claiming that intersection number should be continuous in a "good" family. That can convince me $y = 0$ and $y=x^2$ intersects twice (which I believe), but this continuity is not even clear $\endgroup$ – user27126 May 29 '12 at 23:34
  • $\begingroup$ .. from the definition with ideals. $\endgroup$ – user27126 May 29 '12 at 23:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy