I am trying to solve the following problem:

Let $C$ be a non-degenerate line (resp. conic) in $\mathbb{C}\mathbb{P}^2$ and $D$ a projective curve in $\mathbb{C}\mathbb{P}^2$ of degree $d$ such that $C$ is not a subset of $D$. Without using the resultant prove that the intersection $C\cap D$ has at most $d$ (resp. $2d$) points. [Hint: You may find it useful to use suitable parametrizations.]

The previous part of the question (which I did manage to do) was to show that, for any set of $n+2$ points $p_1,\ldots,p_2$ in general position in $\mathbb{C}\mathbb{P}^n$, there exists a unique projective transformation that sends the first $(n+1)$ of them to $(0,\ldots,0,1,0,\ldots,0)$ (where the $1$ is in the $i$-th place for $p_i$) and the $(n+2)$-th to $(1,1,\ldots,1)$.

I'm assuming that these are the coordinates that make this question easy to do, but I can see which points to choose to be in general position. It makes sense to start with the points in $C\cap D$, but we can only pick four points in there in general position, so I'm not too sure what use this is for the $d\geqslant3$ case.

Any help would be appreciated, be it hints or solutions.

  • $\begingroup$ For starters, try to understand a particular case. For instance, suppose that the line is the x-axis or y-axis and see what happens. $\endgroup$ – user1971 Jun 10 '15 at 17:49

There is a projective transformation that takes any line $L$ to $\mathbb V(x)$.

And there is a projective transformation that takes any nondegenerate conic $C$ to $\mathbb V(x^2 + y^2 + z^2)$.

The first is just linear algebra, and the second can be proved using your result before and the fact that there is a unique conic passing through any $5$ general points. (Alternately, it's equivalent to the statement that any symmetric bilinear form can be diagonalized, which is easy enough directly).

Now write down an arbitrary equation for $D$, and do it directly.

  • $\begingroup$ Sorry, but what exactly is $\mathbb{V}(x)$? That's not something I've come across before. $\endgroup$ – Tim Jun 11 '15 at 18:17
  • $\begingroup$ It's just notation for the line in projective space defined by the vanishing of the $x$ coordinate. So the set of all $[0: y: z]$. $\endgroup$ – Phil Tosteson Jun 11 '15 at 18:27
  • $\begingroup$ Gotcha, thanks! $\endgroup$ – Tim Jun 11 '15 at 18:42

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