0
$\begingroup$

I am trying to do an exercise from Rick's Miranda Book that goes like this

Show that if $X$ is a line in the projective plane, then the intersection divisor of any other line with $X$ has degree one. In general, show that the intersection divisor of a homogeneous polinomyal $G$ of degree $d$ with a line $X$ has degree $d$.

My attempt for the first sentence was that we know that there is only one point $p$ in the intersection of those two lines so we need to calculate $ord_p(G/H)$, but i cant seem to do this in a decent way, so any help is aprecciated. Also for the second case should i try to transform the polinomyal in lines or something ? Thanks in advance.

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Fix homogeneous coordinates $(x:y:z)$. A line $X$ is given by the zeros of a homogeneous linear form that, up to a linear change of coordinates, we may suppose to be $z$. Then we may also give a global parameterization $\mathbb{P}^1 \rightarrow \mathbb{P}^2$, $(x:y) \mapsto (x:y:0)$.

It follows that the restriction of a degree $d$ homogeneous polynomial $G$ in $x,y,z$ to $X$ is just making $G(x,y,0)$. Now this is a homogeneous polynomial in two variables and you can deduce the result using the Fundamental Theorem of Algebra.

Ex.: Let $H = 3x-y+7z$. Then $H|_X = 3x-y$ which vanishes only at $(1:3)$.

Let $G = y^2(x^2-y^2)+z^4$ then $g = G|_X = y^2(x^2-y^2)$ which has two simple zeros $(1: \pm 1)$ and a double zero at $(1:0)$. To see this we can go to the affine chart $U = \{x=1\}$ where $g|_U = y^2(1-y^2)$ and compute its order in each point.

$\endgroup$
5
  • $\begingroup$ Hi thanks for the answer but i have a few doubts , so we have a line and we can change its coordinates so we can suppose its $z$, then i guess we can consider the function $z/x$ and we need to calculate the order of this at $0$, but i dont see how that does it , because if $z=0$ we know that one of the other coordinates suppose $x\neq 0$, so we are in the affine plane curve that is of the form $F(1,y,z)$ but we dont know anything about the derivatives to know if z is a function of $y$ or vice-versa. $\endgroup$
    – Someone
    Commented Apr 15, 2020 at 9:02
  • $\begingroup$ @Something I added two examples, see if it helps. $\endgroup$
    – Alan Muniz
    Commented Apr 15, 2020 at 13:10
  • $\begingroup$ Yes i understand the examples, its just from an arbitary point of view i dont see a good way of calculating the order of the function that we need at the point of interest. $\endgroup$
    – Someone
    Commented Apr 15, 2020 at 21:25
  • $\begingroup$ After restricting the meromorphic function to the curve its order may be calculated by choosing a local parameter hence a local expression for this fucntion. $\endgroup$
    – Alan Muniz
    Commented Apr 15, 2020 at 22:27
  • $\begingroup$ In the second example we are taking $y \in \mathbb{C} \mapsto (1:y:0) \in \mathbb{P}^2$ and pulling $\dfrac{G}{x^4}$ back. This gives a function only in $y$ for which we know how to compute the order. $\endgroup$
    – Alan Muniz
    Commented Apr 15, 2020 at 22:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .