Exercise 10.13

Suppose that the characteristic of F is not 2 , and consider the curve defined by a x 2 + bxy + c y 2 = 1 , where a , b , c ∈ F ∗ . If b 2 − 4 ac ∈ F 2 , show that there are one or two points at infinity depending on whether b 2 − 4 ac is zero. If b 2 − 4 ac = 0 , show that the point at infinity is singular.

Answers

Proof. Let 𝒞 be the curve defined by f ( x , y ) = a x 2 + bxy + c y 2 − 1 . The homogeneous equation of the projective closure 𝒞 ¯ of 𝒞 is

f ¯ ( t , x , y ) = a x 2 + bxy + c y 2 − t 2 .

The points [ 0 , u , v ] at infinity are given by the equation

a u 2 + buv + c v 2 = 0 .

Assume that Δ = b 2 − 4 ac = δ 2 ∈ F 2 . Since a ≠ 0 ,

a u 2 + buv + c v 2 = a [ ( u + b 2 a v ) 2 − b 2 − 4 ac 4 a 2 v 2 ] = a [ ( u + b 2 a v ) 2 − δ 2 4 a 2 v 2 ] = a ( u − − b + δ 2 a v ) ( u − − b − δ 2 a v ) = a ( u − αv ) ( u − βv ) ,

where α = − b + δ 2 a , β = − b − δ 2 a are the two roots of a X 2 + bX + c .

Therefore the points at infinity are p = [ 0 , α , 1 ] and q = [ 0 , β , 1 ] .

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If b 2 − 4 ac ≠ 0 (hyperbolic case), then α ≠ β and p ≠ q , so that 𝒞 has two points at infinity.
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If b 2 − 4 ac = 0 (parabolic case), then α = β , and 𝒞 has one (double) point at infinity r = [ 0 , α , 1 , ] , where α = − b 2 a is the root of multiplicity 2 of a X 2 + bX + c . Thus r = [ 0 , − b , 2 a ] .

Since

∂ f ¯ ∂t ( t , x , y ) = − 2 t , ∂ f ¯ ∂x ( t , x , y ) = 2 ax + by , ∂ f ¯ ∂y ( t , x , y ) = bx + 2 cy ,

then

∂ f ¯ ∂t ( 0 , − b , 2 a ) = 0 , ∂ f ¯ ∂x ( 0 , − b , 2 a ) = − 2 ab + 2 ab = 0 , ∂ f ¯ ∂y ( 0 , − b , 2 a ) = − ( b 2 − 4 ac ) = 0 .

This shows that the point at infinity r = [ 0 , − b , 2 a ] is singular.

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2022-07-19 00:00
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