# Concrete example of Weil Pairing

I am trying to find a concrete example of the Weil Pairing. What I have done until now is that I took $E=(x-1)(x-2)(x-3)$ over $F_5$. I took $E[2]=\{\infty,(1,0),(2,0),(3,0)\}$. I know that there exist a rational function f such that \begin{equation*} div(f)=2[T]-2[\infty] \end{equation*} How can I find this function f? How can I find the points of order 4? How should I continue?

• Have a look here. And as for finding the points of order $4$, are you allowed to use a computer? I yes, Sage or Pari/GP will do. Otherwise, you can find all points with pen and paper, and identify those of order $4$. Commented Oct 14, 2015 at 12:56
• thank I you! I use matlab. Is there an esay way to find the points of order 4? Is there a known algorithm to identify points of a certain order? Commented Oct 14, 2015 at 14:18
• I do not know Matlab. In Pari, you can obtain generators of the cyclic components of the group, from which it is trivial to obtain the points of any order. Commented Oct 14, 2015 at 14:25
• I did not know Pari. It looks interesting for such kind of computations! Could you explain me a litte bit more how to do this with Pari? Commented Oct 14, 2015 at 14:30
• The comments are not meant to be used for discussions, so please send an e-mail, I will answer tomorrow as it is late here (but FYI, you use the Pari function ellgroup()). Commented Oct 14, 2015 at 14:32

The automatic way to find f is by using the Miller algorithm.

But you can also see this directly here: Since the line y=0 is not tangent to E , y is a uniformizing function. You can write $(x-1) = y^2*((x-2)(x-3))^{-1}$, where $(x-2)*(x-3)$ has neither a zero nor a pole at (1, 0). This says exactely that (1,0) is a double zero of the function x-1. Since it has a double zero it must also have a double pole at $\infty$. This gives you \begin{equation*} div((x-1))=2[(1,0)]-2[\infty] \end{equation*}

In a similar way you can show this for the points (2,0) and (3,0)

Now you have all 2-torsion points, lets find the other points:
x=0 gives $y^2 = -1$, which solves to y=2 and y=3
x=4 gives $y^2 = 1$, which solves to y=1 and y=4
Since the group has 8 elements, at least two of those points must have order 4. That all 4 points have order 4 can be shown by a direct calculation(2 squarings are sufficient). One can also see this by a direct argument:
If the group would be cyclic, it would have 4 elements of order 8. This would let no place for an element of order 4. Therefore the group cannot be cyclic and all 4 elements, which are not 2-torsion elements, must have order 4.

• wow! thank you! this really helped me! One additional question. You said:Since the group has 8 elements, at least two of those points must have order 4. How do you know this? Commented Oct 23, 2015 at 16:03
• The group has 8 elements by simply listing them. The 2-torsion consists of four points. Any other point generates a cyclic subgroup of E, and therefore its order must divide the group order, The order o of this cyclic group must therefore be 4 or 8. This group must have a subgroup of order 4. That you have two elements of order 4 stems from the fact that all points, which are not on the x-line occur in conjugated pairs.
– user27950
Commented Oct 23, 2015 at 16:21