# Intersection divisor of a homogeneous polynomial with a line in the projective plane

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.

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.

• 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. Commented Apr 15, 2020 at 9:02
• @Something I added two examples, see if it helps. Commented Apr 15, 2020 at 13:10
• 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. Commented Apr 15, 2020 at 21:25
• 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. Commented Apr 15, 2020 at 22:27
• 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. Commented Apr 15, 2020 at 22:31