Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is exercise 3.7 from Silvermans AEC (2nd edition).

Let $E$ be a nonsingular elliptic curve over $\mathbb{C}$ given by

$$ y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6.$$

The $n^{th}$ division polynomls $\psi_{n}$ are defined using

$ \psi_{1} = 1, \\ \psi_{2} = 2y+a_{1}x+a_{3}x, \\ \psi_{3} = 3x^{4} +b_{2}x^{3} +3b_{4}x^{2} + 3b_{6}x + b_{8}. \\ \psi_{4} = \psi_{2}(2x^{6} +b_{2}x^{5} + 5b_{4}x^{4} +10b_{6}x^{3} + 10b_{8}x^{2} +(b_{2}b_{8} - b_{4}b_{6})x + (b_{4}b_{8}-b_{6}^2)),$

then recursively by the formulas

$\psi_{2n+1} = \psi_{n+2}\psi_{n}^{3} - \psi_{n-1}\psi_{n+1}^{3} \\ \psi_{2n}\psi_{2} = \psi_{n-1}^{2}\psi_{n}\psi_{n+2}- \psi_{n-2}\psi_{n}\psi_{n+1}^{2}. $

Show that

$$ \psi_{m+n}\psi_{m-n}\psi_{r}^{2} = \psi_{m+r}\psi_{m-r}\psi_{n}^{2} - \psi_{n+r}\psi_{n-r}\psi_{m}^{2}.$$

Now, it seems that this should be done by considering div($\psi_{n}$), and by doing so (and considering $\psi_{n}$ as a function on $\mathbb{C} / \Lambda$) I can show

$$ \frac{\psi_{m+n}(z)\psi_{m-n}(z)}{\psi_{m}^{2}(z)\psi_{n}^{2}(z)} = \wp(nz) - \wp(mz). $$

which gives the result. However elliptic functions aren't covered until chapter 6.

So, my question is: How can this be done without using elliptic functions?

share|improve this question
    
Also posted to, and commented on (but closed) at MO, mathoverflow.net/questions/137211/… –  Gerry Myerson Jul 20 '13 at 12:32
add comment

1 Answer

up vote 5 down vote accepted

You can use the addition/duplication formulas and a bunch of algebra to solve the problem without using any complex analysis. The advantage of the algebraic proof is that it's valid in any characteristic

share|improve this answer
    
Thank you Joe!! –  Jeff Bleaney Jul 20 '13 at 20:11
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.