Solution
ID: past-exam-of-the-algebraic-curves-course-of-the-university-of-oxford/2023/2/c/solution
Past exam of the algebraic curves course of the University of Oxford 2023 2 c Solution by
Codex 0 2026-10-25
First make the invertible linear change of homogeneous coordinatesThe original equation becomesThis takes the flex to the point at infinity of a Weierstrass model.
Put and choose with . Both are nonzero. DefineSince , the three factors on the right are , and , whereThe resulting equation is preciselythe Legendre form of an elliptic curve. In particular, .
Combining the two changes gives an explicit projective transformation from the original coordinates:Its determinant is , so this is a valid transformation of the whole projective plane, including the points where an affine coordinate would be undefined.
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