# differing notions of the degree of a smooth projective plane curve

[Everything here will be over $\mathbb{C}$]

Hello, one definition of a smooth projective plane curve $X$ is, for $X \subset \mathbb{P}^{2}$.. $deg(X) =$ maximum number of intersection points (without multiplicity) of any hyperplane in $\mathbb{P}^{2}$.

When my smooth plane curve is defined by the zeros of a irreducible degree $d$ homogeneous polynomial, what is a good way to see that these two notions of degree agree?

Thanks!

Elliot

• Let $L$ be a line in $\mathbf P^2$. Restricting the homogeneous degree $d$ polynomial to $L$ gives a homogeneous degree $d$ polynomial in two variables. If this is not identically zero (which it won't be as long as $d \neq 1)$, it has exactly $d$ zeroes counted with multiplicity. You want to see that for some choice of line, you get $d$ zeroes without multiplicity: for this you need to write the condition that the restricted form has a multiple zero in terms of the coefficients of your linear form. This means calculating some resultant. You will see that the vanishing of the resultant... Commented Nov 5, 2015 at 15:05
• ...is a nontrivial algebraic relation between the coefficients of your linear form. Choosing any line whose defining form does not satisfy that relation, you get a line that intersects your curve in precisely $d$ points. Commented Nov 5, 2015 at 15:05

These two notions are equivalent due to Bézout's theorem. A general hyperplane has degree 1 in $\mathbb{P}^2$. Then the intersection number of your curve and a general hyperplane is $deg(curve)\times deg(hyperplane)=d\times 1=d$ by Bézout's theorem.