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We know that the intersection number of this curve $f=y-x^3$ and its tangent at the origin is $3$.

I'm trying to use this method described in the Fulton's book:

Following this definition we have the intersection number equals to $1$, because $m=1$ and $f_1=y$.

Am I missing something?

thanks

user42912
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  • The point $(0,0)$ in the curve is not a singular point (double, triple, …), because the form at the origin is just $Y$, which has degree $1$. The tangent at this point has an intersection number of $3$, because it's a flex. – egreg Jan 31 '14 at 22:37
  • @egreg is it possible a point has an intersection number 3 with the tangent line and having only one tangent with multiplicity 1 at this point in the same time? – user42912 Jan 31 '14 at 22:48
  • Yes, at flexes. – egreg Jan 31 '14 at 22:50

1 Answers1

3

You're confusing the concepts. Fulton is talking about singular points (double, triple, and so on). If you write the polynomial you have in the stated form, you get $$ F=Y-X^3=F_3(X,Y)+F_2(X,Y)+F_1(X,Y) $$ where $$ F_1(X,Y)=Y,\quad F_2(X,Y)=0,\quad F_3(X,Y)=-X^3 $$ which means that $(0,0)$ is a simple point. The tangent is indeed the line $Y=0$. When you compute the intersections, you solve $$ \begin{cases} Y-X^3=0\\ Y=0 \end{cases} $$ which gives $X^3=0$, so you conclude that the tangent and the curve have intersection number $3$ at the origin.

A different case is that of the Folium Cartesii, $$ F(X,Y)=X^3+X^2-Y^2 $$ where, with notation as before, $$ F_1(X,Y)=0,\quad F_2(X,Y)=X^2-Y^2,\quad F_3(X,Y)=X^3 $$ Then the origin is a double point and the tangents are given by $X^2-Y^2=0$, that is, they are $X=Y$ and $X=-Y$. A double point with simple tangents.

egreg
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