0

I was reading some lecture notes online and they mentioned that there are only two plane conics up to affine coordinate change (parabola and hyperbola). However, they cite a book that does not give an explanation of this fact, and I can't figure out how to prove it.

I know this means that every (irreducible) quadratic looks like the zero set of $y-x^2$ or $xy-1$. But why is this true?

Edit: I'm working in projective space with polynomials in $\mathbb{C}[x,y]$.

mkmath
  • 25
  • 1
    This makes some sense from a projective perspective. The thing that distinguishes these conics is how they interact with the line at infinity, and affine transformations leave it fixed. I’m a bit puzzled that there are only two, though. Which of them is isomorphic to an ellipse? – amd Apr 20 '18 at 23:50
  • Right, @amd: Over $\Bbb R$, I’d expect that you’d have three, corresponding to whether a certain discriminant is positive, zero, or negative. I’m further disquieted by the lack of mention of a base field. Over $\Bbb Q$, is $x^2+y^2=1$ really affinely equivalent to $x^2+y^2=3$? – Lubin Apr 21 '18 at 04:00
  • Three points: I would guess that the notes assume algebraically closed field of characteristic not 2, because the statement is not true otherwise. Secondly, over any algebraically closed field, every quadratic form is diagonalizable, so $xy$ is equivalent to $x^2+y^2$, and thus $xy-1$ represents an ellipse. Thirdly, over non-algebraically closed fields, the classification is sensitive to whether the conic has a rational point, as Lubin's example shows (but it's even worse- $x^2+y^2+c$ for some $c$ rational and not a square are all inequivalent as $c$ varies). – KReiser Apr 21 '18 at 05:11
  • To clarify, I'm working with polynomials in $\mathbb{C}[x,y]$ here. I'm still not understanding rigorously how this works. – mkmath Apr 21 '18 at 18:48
  • See this answer. – Jan-Magnus Økland Apr 22 '18 at 11:25

0 Answers0