Calculating the Intersection Number of 2 Specific Sections on Elliptic Surface

Question

Take the elliptic surface defined by the equation $E_6: y^2= x^3 + t^6 + 1$ over the field $\mathbb{F}_5$ (or the algebraic closure thereof, it does not really matter for this question). I have the 2 sections/rational points given as $P=(-1,t^3)$ and $Q=(-t^2,1)$, which define curves on the surface $E_6$. I want to calculate their intersection number. I am having quite a hard time with this as most references leave out such specific examples. What I tried so far is the following: the only possible way they intersect over $\mathbb{A}^1$ is whenever $t = 1$ as the x and y coordinates need to line up. After this I am not so sure what to do. Do I need to calculate the self-intersection of the curve $y^2 =x^3+2$ (filling in $t=1$) or is it something else? Any help or pointers to references would be appreciated very much!

Answer 1

Below is a formal computation. We work in the ring $$ R = \Bbb F_5[x,y,t]/(-y^2 +x^3 + t^6 +1)\ . $$ (We write still $x,y,t$ for the genuine elements of $\Bbb F_5[x,y,t]$ also after working modulo the ideal generated by $-y^2 +x^3 + t^6 +1$.)

The point $A$ given by $(x,y,t)=(-1,1,1)$ "is" the maximal ideal $J_A=(x+1,\ y-1, \ t-1)$ generated by $(x+1)$, $(y-1)$, and $(t-1)$ in $R$. It "is" also the morphism $R\to \Bbb F_5$ induced by mapping $x,y,t$ to respectively $-1,1,1$. At Spec-level, this is a map in the other direction, $\operatorname{Spec}R\leftarrow \operatorname{Spec}\Bbb F_5$, and $\operatorname{Spec}\Bbb F_5=*$ "is" the (initial) point when working with (affine) schemes over $\Bbb F_5$.

Now we consider the two curves, $P$ and $Q$, from the question. Most references want one (polynomial) generator for each curve. But in our case, the curves are given by parametrizations. We have to associate corresponding ideals $J_P$ and $J_Q$ for them: $$ \begin{aligned} J_P&=(x+1, \ y-t^3)\ ,\\ J_Q&=(y-1, \ x+t^2)\ . \end{aligned} $$ The inclusions of ideals $J_P\subset J_A$, and $J_Q\subset J_A$ have the geometrical translations "the point $A$ is on the curve $P$", respectively "the point $A$ is on the curve $Q$". Now we build the intersection $$ \begin{aligned} J_P\cap J_Q &= (x+1,\ y-t^3,\ y-1,\ x+t^2) \\ &= (x+1,\ y-1,\ t^3-1,\ t^2-1) \\ &= (x+1,\ y-1,\ t-1) \\ &= J_A \ , \end{aligned} $$ The geometric intersection number is one.

Answer 2

First we check $P, Q$ are points on the curve. Substituting them in, we see they are zeroes of the curve.

Now define $m = \frac{1 - t^3}{-t^2 + 1}$ which is the gradient between the points. Then $y = m(x - x_0) + y_0 = \frac{t^3 - 1}{t^2 - 1}(x + 1) + t^3$. We can now parameterize this line so that it disappears when $s = 0$. Let $$x = s - 1$$ $$y = \frac{t^3 - 1}{t^2 - 1}s + t^3$$

Now substitute these new expressions into our original curve $$\left(\frac{t^3 - 1}{t^2 - 1}s + t^3\right)^2 = (s - 1)^3 + t^6 + 1$$ Performing the calculation in sage

sage: e = ((t^3 - 1)*s + t^3*(t^2 - 1))^2 == ((s - 1)^3 + t^6 + 1)*(t^2 - 1)^2
sage: e
((t^2 - 1)*t^3 + (t^3 - 1)*s)^2 == (t^6 + (s - 1)^3 + 1)*(t^2 - 1)^2
sage: e.expand()
t^10 + 2*s*t^8 + s^2*t^6 - 2*t^8 - 2*s*t^6 - 2*s*t^5 + t^6 - 2*s^2*t^3 + 2*s*t^3 + s^2 == t^10 - 2*t^8 + s^3*t^4 - 3*s^2*t^4 + t^6 - 2*s^3*t^2 + 3*s*t^4 + 6*s^2*t^2 + s^3 - 6*s*t^2 - 3*s^2 + 3*s
sage: e = e.expand()
sage: e -= e.rhs()
sage: e
2*s*t^8 + s^2*t^6 - s^3*t^4 - 2*s*t^6 + 3*s^2*t^4 - 2*s*t^5 + 2*s^3*t^2 - 2*s^2*t^3 - 3*s*t^4 - 6*s^2*t^2 + 2*s*t^3 - s^3 + 6*s*t^2 + 4*s^2 - 3*s == 0
sage: e.lhs().coefficient(s, 0)
0
sage: e.lhs().coefficient(s, 1)
2*t^8 - 2*t^6 - 2*t^5 - 3*t^4 + 2*t^3 + 6*t^2 - 3

So the first nonzero coefficient, occurs for $n = 1$. This is your valuation for $P$ on the curve. Now repeat from the beginning for $Q$. For more info see valuation.ipynb

Note we can also see this quickly by noting the coefficient of the curve at $P, Q$ is not $m$, so they are non-tangent and intersect at multiplicity 1.