I'm a beginner in this subject and I think this "easy" exercise could help me to have more practice in basic algebraic curves.

Let $F=X^{p+1}+Y^{p+1}+Z^{p+1}$ be a Fermat curve in the field $k$, with $char(k)=p\gt 0$.

I've already showed every point in the Fermat curve is a flex using Hessian $H_F$ of $F$. Now, I would like to calculate the intersection number $(F,T_PF)_P$ of a generic point $p=(x_0:y_0:z_0)$ and I know that $T_PF=x_0^pX+y_0^pY+z_0^PZ$.

I've already tried to use these properties, but I didn't find any common factor to simplify the calculations. I've already tried also to parametrize $T_PF$ without any success.

I need help



I proved every point is a flex proving that the Hessian is zero everywhere, because this curve is not singular, then it has to be flex at every point. My doubt is to find the intersection point $(F,T_PF)_P$ for each point which should be more or equal than 3, because every point is a flex.


In the wikipedia link that you give, they describe some ways of computing these intersection numbers in practice. The one I'm going to use is described in this paragraph:

One realization of intersection multiplicity is through the dimension of a certain quotient space of the power series ring $K[[ x,y]]$. By making a change of variables if necessary, we may assume that the point $p$ is $(0,0)$. Let $P(x, y)$ and $Q(x, y)$ be the polynomials defining the algebraic curves we are interested in. If the original equations are given in homogeneous form, these can be obtained by setting $z = 1$. Let $I = (P, Q)$ denote the ideal of $K[[x,y]]$ generated by $P$ and $Q$. The intersection multiplicity is the dimension of $K[[x, y]]/I$ as a vector space over $K$.

Step 1: Since your equations are given in homogeneous form, we start by picking an affine coordinate patch to work in. I will assume your point $P$ lies in the affine coordinate patch where $Z\neq 0$. In this patch your curve and tangent plane are defined by the polynomials $$F(x,y) = x^{p+1} + y^{p+1} + 1,\,\,\,\,\,\,\,\,\,\,T(x,y) = x_0^px + y_0^py + 1,$$ respectively, where your point $P$ is $(x_0,y_0)\in \mathbb{A}^2$.

Step 2: Having fixed our point $P = (x_0,y_0)$, we now want to work (for simplicity) in coordinates where in fact $P$ is the origin. So let $z,w$ be the coordinates $z := x - x_0$ and $w := y - y_0$. We can now express the polynomials $F$ and $T$ in these coordinates (check these!): $$F(z,w) = z^{p+1} + x_0z^p + x_0^pz + w^{p+1} + y_0w^p + y_0^pw$$ $$ T(z,w) = x_0^pz + y_0^pw$$

Step 3: Now the intersection multiplicity of the curve with its tangent plane at $P$ is equal to the dimension of the $k$-vector space $$R:= k[[z,w]]/(F(z,w),T(z,w)).$$ To compute this dimension, first observe that one cannot have both $x_0 = 0$ and $y_0 = 0$, since then $P$ would not lie on the curve. Without loss of generality, assume that $x_0\neq 0$. From the equation of the tangent plane, we get that in the ring $R$, $$z = -\frac{y_0^p}{x_0^p}w.$$ In particular, $$R \cong k[[w]]/(F(-(y_0/x_0)^pw, w)).$$ If I haven't messed up my algebra (it should be checked!), $$F\left(-\frac{y_0^p}{x_0^p}w, w\right) = (-1)^{p+1}\frac{y_0^{p(p+1)}}{x_0^{p(p+1)}}w^{p+1} + (-1)^p\frac{y_0^{p^2}}{x_0^{p^2-1}}w^p + w^{p+1} + y_0w^p.$$ If the $w^p$ term is not $0$, then it follows that dimension of $R$ is $p$. Otherwise, the dimension will be $p+1$. The $w^p$ term vanishes precisely when $(-1)^py_0^{p^2} = -y_0x_0^{p^2-1}$. Since this equation does not vanish identically along your curve, we see that generically the intersection multiplicity should be $p$.

  • $\begingroup$ I haven't studied this method yet, this will be very helpful to me in some weeks, thank you very much. $\endgroup$ – user122342 Feb 12 '14 at 14:38

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