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[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

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    $\begingroup$ 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... $\endgroup$
    – Nefertiti
    Commented Nov 5, 2015 at 15:05
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    $\begingroup$ ...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. $\endgroup$
    – Nefertiti
    Commented Nov 5, 2015 at 15:05

1 Answer 1

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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.

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