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Question: Find a rational parametrization of the conic whose equation in homogeneous coordinates is:

$x^2+y^2-xy-z^2=0$

Hence find all rational numbers $x, y$ such that:

$x^2+y^2-xy=1$

It seems I don't quite understand the rational parametrization lecture notes. Can someone show me how to deal with this example so that I see how to approach those type of problems? I think I am familiar with the basic concepts

Basically what I've done is to reduce it to $-z_1^2+\frac{1}{2}z_2^2+\frac{3}{2}z_3^2=0$

asdf
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1 Answers1

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For a parametrization of the affine curve you need a rational point, say $(-1,0)$. Then a line through $(-1,0)$ will intersect the conic in another point. For the line through $(-1,0)$ take $y=t(x+1)$ as the equation. Substitute the equation of the line into the equation of the affine conic. so $$x^2 + (t(x+1))^2-x(t(x+1))=1$$ After expanding and collecting terms we find $$x^2(1-t+t^2)+(2t^2-t)x+(t^2-1)=0$$ Divide through by $1-t+t^2$ to finally get $$x^2+\frac{(2t^2-t)}{(1-t+t^2)}x+\frac{(t^2-1)}{(1-t+t^2)}=0$$ We know one root of the quadratic is $-1$, and the product of the roots must be $\frac{(t^2-1)}{(1-t+t^2)}$, so we conclude $$x= \frac{(1-t^2)}{(1-t+t^2)}$$ Returning to the equation of the line $y=t(x+1)$ to find $y$ gives $$y = \frac{t(2-t)}{(1-t+t^2)}$$ So we have a rational parametrization of the conic in terms of the rational parameter $t$.

sharding4
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  • Thanks! This sure gives the solution, But is there a more "projective geometry"-ish approach? I am trying to understand the idea – asdf May 12 '17 at 20:42
  • Actually, the first one is the second equation in projective coordinates, is that so? – asdf May 12 '17 at 20:43
  • @asdf the first equation is the homogeneous equation (all terms have degree 2) for the projective curve. The second equation is the curve in one of the affine open subspaces (here $z\neq0$) – sharding4 May 12 '17 at 21:03
  • @asdf For the projective curve you should be able to use $x(t)=1-t^2$, $y(t)= t(2-t)$ and then $z(t)=1-t+t^2$ – sharding4 May 12 '17 at 21:08
  • @asdf: Building on the last comment, you can even homogenize to $x=s^2-t^2,y=t(2s-t),z=s^2-st+t^2$. In M2R=QQ[s,t,x,y,z] I=ideal(x-(s^2-t^2),y-t*(2*s-t),z-(s^2-s*t+t^2)) gens gb I gives your equation $x^2+y^2-xy-z^2$ back as an element in the Gröbner basis. – Jan-Magnus Økland May 13 '17 at 08:03