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→Computation modulo primes: there was a 2 missing in the formula for calculation qbar(x,y) with division polynomials. source: http://www-math.mit.edu/~musiker/schoof.pdf |
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<math>
\bar{q} (x,y) = (x_{\bar{q}},y_{\bar{q}}) = \left( x - \frac {\psi_{\bar{q}-1} \psi_{\bar{q}+1}}{\psi^{2}_{\bar{q}}}, \frac{\psi_{2\bar{q}}}{2\psi^{4}_{\bar{q}}} \right)
</math>
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===Additional Case: <math>l = 2</math>===
If you recall, our initial considerations omit the case of <math>l = 2</math>.
Since we assume <math>q</math> to be odd, <math>q + 1 - t \equiv t \pmod 2</math> and in particular, <math>t_{2} \equiv 0 \pmod 2</math> if and only if <math>E(\mathbb{F}_{q})</math> has an element of order 2. By definition of addition in the group, any element of order 2 must be of the form <math>(x_{0}, 0)</math>. Thus <math>t_{2} \equiv 0 \pmod 2</math> if and only if the polynomial <math>x^{3} + Ax + B</math> has a root in <math>\mathbb{F}_{q}</math>, if and only if <math>\gcd(x^{q}-x, x^{3} + Ax + B)\neq 1</math>.
==The Algorithm==
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