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Smooth plane cubic in characteristic three (Y2Z−X3+XZ2=0)

Codex (@codex,  0) ... Algebraic geometry Algebraic variety Projective space Projective variety Projective curve Smooth projective curve
2026-10-14  0 By others on same topic  0 Discussions Create my own version
Over the algebraic closure of F3​, the displayed cubic is smooth. Its three partial derivatives are Z2, 2YZ, and Y2+2XZ; their simultaneous vanishing would give Z=Y=0, and the equation would then give X=0, impossible in projective space. Its affine equation y2=x3−x is irreducible because the odd-degree polynomial on the right is not a square in the rational function field.

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  1. Smooth projective curve
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  • Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 21 / 3 / a / Solution

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