This example is taken from Silverman and Tate's Rational Points on Elliptic Curves.

Suppose we have the line

$C_1: x+y=2$

and the circle

$C_2: x^2 + y^2 = 2$.

We see that these intersect at the point $(1,1)$. We then put the curves in homogeneous coordinates and see that they are described by

$C_1: X+Y=2Z$ and $C_2: X^2 + Y^2 = 2Z^2$ and we get the single intersection point $(1:1:1) \in \mathbb{P}^2$.

Now, it is said in the text that since the line $C_1$ is tangent to the circle $C_2$ at $(1,1)$, this intersection point "should count double". Can anyone explain the intuition behind this statement?


Intuitively, if we have two irreducible plane curves $F$ and $G$ that intersect at a point $P,$ then the intersection multiplicity $I(P, F \cap G)$ is the maximum number of intersections of $F$ and $G$ near $P$ if we wiggle $F$ and $G$ a little.

For example, take $F$ to be a curve and $G$ to be a straight line that goes through $F$ at $P$ transversally. Then there is at most single point of intersection even if we wiggle $G$ a bit, so the multiplicity of this intersection is $1.$

Now take the cubic with a cusp $F= Y^2-X^3$ and the line $G= X.$ Then $G$ intersects $F$ at $P=(0,0)$ and if we shift $G$ a little to the right, it intersects $F$ in two places so $I(P,F\cap G)=2.$

  • $\begingroup$ Thanks, that's interesting. I had never heard this idea of ``wiggling'' our curves to explain the intersection multiplicity. Do you have a reference for that? $\endgroup$ – nigel May 9 '14 at 1:07
  • $\begingroup$ @nigelvr I actually have never read this anywhere. I first learned about Intersection Multiplicity when reading Section 3.3 of Fulton's Algebraic Curves. There he introduces it as a number that we want to satisfy 7 properties, and I figured out on my own what such a number does intuitively. When I later told my supervisor about it, he seemed to think it was obvious so I think it is well known but people don't like to write down imprecise things. $\endgroup$ – Ragib Zaman May 9 '14 at 2:13
  • $\begingroup$ Let me give you some warnings about where to be careful. I assume you are working with Real numbers. Suppose I took $F=Y^2-X^3$ again and now $G=Y.$ If I wiggle $G$ left and right, it stays the same. If I move it up and down, it still only intersects at $1$ point. If I rotate it a little around the origin, then I can get two intersections. So is the Intersection Multiplicity $2$ ? What other ways can someone "wiggle" the curve? It turns out (as it does for many things in algebraic geometry) to get a good consistent notion of Intersection Multiplicity, one needs to consider intersections.... $\endgroup$ – Ragib Zaman May 9 '14 at 2:18
  • $\begingroup$ ... not just in the original field (here the Real numbers) but in the algebraic closure of the field (here the complex numbers). If you do this, and wiggle $G$ to $G' = Y-\epsilon$ then you see that you get $3$ distinct intersection points, with $Y=\epsilon$ and $X$ being the $3$ cube roots of $\epsilon^2.$ So the intersection number here is actually $3.$ $\endgroup$ – Ragib Zaman May 9 '14 at 2:21

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