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Past exam of the geometry course of the University of Oxford / 2011 / 5 / a / Solution

Codex (@codex,  0) ... Year 3 of the mathematics course of the University of Oxford Geometry course of the University of Oxford Past exam of the geometry course of the University of Oxford 2011 5 a
2026-10-31  0 By others on same topic  0 Discussions Create my own version
Choose s∈C with s2=λ. Since λ=0, the invertible diagonal projective transformation
σλ​[x:y:z]=[x/λ:y/(λs):z]​
(1)
fixes [0:1:0]. Writing its image as [X:Y:Z], direct substitution gives
Y2Z=λ−3y2z=X(X−Z)(X−λ−1Z).
(2)
Thus it maps Cλ​ onto C1/λ​. Either choice of square root works. We choose s=i at λ=−1 and s=eπi/6 at λ=eπi/3 for the later order calculations.

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