Let $X$ be a smooth projective plane curve$\{ [Z_0, Z_1, Z_2]\in \mathbb{CP^2}\ \vert \ p(Z_0, Z_1, Z_2) = 0 \}$ defined by non-singular homogeneous polynomial $p$ of degree $d$.

Then there exists a projective line $L\subset\mathbb{CP^2}$ intersecting $X$ at $d$ distinct points.

Is there a way to prove the above statement using only elementary knowledge of Riemann surfaces? I do not know much of algebraic geometry.

  • $\begingroup$ Do you have any ideas? $\endgroup$
    – user469065
    Commented Apr 28, 2018 at 11:35

2 Answers 2


This is straightforward and you do not require any theory of Riemann surfaces. Let $$ C := Z(p) = \{ x=[x_0:x_1:x_2]\in\Bbb C\Bbb P^2 ~\mid~ p(x) = 0 \}. $$ Observe that a line $L\subseteq\Bbb C\Bbb P^2$ can be described as $$ L = \{ [x_0+ty_0:x_1+ty_1:x_2+ty_2] ~\mid~ t\in\Bbb C\} $$ for certain points $x=[x_0:x_1:x_2]$ and $y=[y_0:y_1:y_2]$. The choice of the projective coordinates here does not matter since any potential scaling can be moved in to the parameter $t$. Let $$ p_L(T) := p( x_0+Ty_0, x_1+Ty_1, x_2+Ty_2) \in \Bbb C[T], $$ it is a polynomial of degree $d$ in one variable.

Now the points of $L\cap C$ are given by the $t\in\Bbb C$ such that $p_L(t)=0$, i.e. they are the zeros of $p_L$. You have to argue that $x$ and $y$ can be chosen such that $p_L$ has $d$ distinct roots.

The polynomial $p_L$ has a multiple root if and only if the discriminant of $p_L$ vanishes. The discriminant of $p_L$ is a polynomial in the coefficients of $p_L$ and the coefficients of $p_L$ are polynomials in the entries of $x$ and $y$ - therefore, the discriminant is a polynomial $$\Delta\in \Bbb C[X_0,X_1,X_2,Y_0,Y_1,Y_2].$$ Since there is at least one point $(x_0,x_1,x_2,y_0,y_1,y_2)$ where $\Delta$ is nonzero, you can pick these coordinates to define the line $L$ and it follows that this particular $L$ will intersect $C$ in $d$ distinct points.

This construction also shows that in fact, most of the lines you can choose will intersect $C$ in $d$ distinct points, because $\Delta$ vanishes only on a set of zero measure inside $\Bbb C^6$.


Here's an algebraic proof. Choose any point $P\in X$ that is simple (there are infinitely many such points), you can prove that all but a finite number of lines passing through $P$ intersect $X$ in $d$ points.

Here's a couple of hints:

You can suppose $P=[0:1:0]$ (applying a change of coordinates on $\mathbb{P}^2$ doesn't change the number of intersection points).

  • Consider the lines passing through $P$ of type $L_\lambda=\{[\lambda:t:1]\: |\: t\in \mathbb{C}\}\cup \{P\}$ for $\lambda\in \mathbb{C}$.
  • Write $X$ as a polynomial in $\mathbb{C}[Z_0,Z_2][Z_1]$.
  • Prove that $X(\lambda,t,1)$ for some $\lambda$ doesn't have double roots in $t$.
  • Deduce that $X$ intersects $L_\lambda$ in $n$ different points.

This is a more general result, and more algebraic, but I think it's an easy method to prove the property.

Note that this is not true in characteristics $p$, for example all lines passing through $P$ intersect $Z_0^{p+1}-Z_1^pZ_2$ in strictly less than $p-1$ points. Edit: Still, I believe that even in characteristic $p$ there is always a line intersecting a curve of degree $d$ in $d$ different points (over the algebraic closure)


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