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The problem asks to show that if $Y$ is a projective curve in $\mathbb{P}^2$ of degree $d$ and $L$ is a line such that $Y \neq L$, then $\sum_{P \in L \cap Y} (L \cdot Y)_P = d$.

The solution given here http://math.berkeley.edu/~reb/courses/256A/1.5.pdf, assumes that $L$ is given by $y=0$. This is without loss of generality, since we can always arrive there by a linear change of coordinates. Then the solution looks at $Y\ \cap U_z$, i.e. in the affine open chart where $z \neq 0$, and writes the equations that givees $Y \cap U_z$ as $f(x) + y(*)=0$. Then it claims that if $a$ is a root of $f$ of multiplicity $m$, we have that $(L \cdot Y)_{(a,0)}=m$.

Question 1: This claim would be true by $I.5.4(b)$, if the line $y=0$ were not in the tangent cone of $f(x) + y(*)=0$. But why would this be true in general? What is the ``without loss of generality" argument, if any? (Example that i am thinking of is $Y=Z(x^3z+yz^3+yx^3)$)

Question 2: I totally don't understand the last argument of this solution. Why look at the point $(0:1:0)$ and how do we get this local form of $f$?

Manos
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1 Answers1

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  1. The claim does not depend on $y=0$ being in the tangent cone or not. I am going to use $\alpha$ for what they call $x$ (the root of $f$). You are computing the length of the quotient of $k[x,y]_{(x-\alpha, y)}$ by the ideal $$(f(x) + y(*), y).$$ This is the length of the quotient of $k[x]_{(x-\alpha)}$ by $(f(x))$ which does not depend on anything except the multiplicity of the root.

  2. In the affine patch (i.e. #1) you don't see infinity! The line and the curve might intersect at the point of infinity ($z=0$), and so we must check there too. They should be checking the point $(1 : 0 : 0)$ because we are on the line $L$ given by $y=0$ after all. Now you need to homogenize and then set $x = 1$. Set $f(x) = \sum a_ix^i$. The equation for the curve $Y$ is

$$\sum_{i=0}^n a_ix^iz^{d-i} + y(*) = 0 \overset{x=1}{\Longrightarrow} \sum_{i=0}^n a_i z^{d-i} + y(*) = 0.$$

Note that in each place I use $(*)$ it is different. Now using the same computation as in 1 above, we see that the multiplicity at $[1 : 0 : 0]$ is just the multiplicity (in the elementary sense - count the roots) of $\sum_{i=0}^n a_iz^{d-i}$ at $z=0$. This is $d-n$.

RghtHndSd
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