# How surfaces intersect in projective spaces

Consider this parametrization $$\phi:\mathbb{P}^1\longrightarrow\mathbb{P}^3$$

$$(t_0:t_1)\longmapsto (t_0^3: t_0^2t_1:t_0t_1^2:t_1^3)$$

Let $\mathcal{C}$ be the image of $\phi$. I've proved that $\mathcal{C}=V(F,G,H)$, where $V$ means zeri locus and $F(X_0,X_1,X_2,X_3)=X_0X_3-X_1X_2, G=X_0X_2-X_1^2,H=X_1X_3-X_2^2$. If we call $\mathcal{Q}_F,\mathcal{Q}_G,\mathcal{Q}_H$ the quadrics with equations $F,G,H$ respectively, we have

$$\mathcal{C}=\mathcal{Q}_F\cap\mathcal{Q}_G\cap\mathcal{Q}_H$$ Now I want to prove that each pair of previous quadrics intersects in $\mathcal{C}$ union a projective line. For instance, let $L$ be the following line $$L:X_1=X_2=0$$ Then $L\subseteq \mathcal{Q}_G\cap\mathcal{Q}_H$. My professor said:

• $\operatorname{deg}(\mathcal{Q}_F\cap\mathcal{Q}_G)=4$

• $\deg{\mathcal{C}}=3$

• By Bezout Theorem, $\mathcal{Q}_G\cap\mathcal{Q}_H=\mathcal{C}\cup L$. My questions are:

1) Why $\operatorname{deg}(\mathcal{Q}_F\cap\mathcal{Q}_G)=4$?

2) Why $\deg{\mathcal{C}}=3$? To find the degree of a curve, I suppose I should know the equation of that curve, but I have only a parametric representation of $\mathcal{C}$, how can I deduce that degree is 3?

3) Bezout theorem which I listened about concerns the number of intersections of two plane curves of given degrees.....is this another different Bezout theorem? How can I apply it to quadrics??

• There is a much more general version of the Bezout theorem, which you can read about in the books of Harris or Hartshorne. For example, in your case it says that two surfaces in $\mathbf P^3$ of degrees $d_1$ and $d_2$ intersect in a curve of degree $d_1d_2$ (as long as we count with multiplicities). To see that your curve has degree 3, intersect it with a plane in $\mathbf P^3$, e.g. the plane $X_0=X_3$, and count the number of intersection points. (To make this a full proof, you have to check that plane intersects the curve transversely, but hopefully this gives the idea.) – user64687 Dec 4 '13 at 20:11