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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?

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Also posted to, and commented on (but closed) at MO,… – Gerry Myerson Jul 20 '13 at 12:32
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

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Thank you Joe!! – Jeff Bleaney Jul 20 '13 at 20:11

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