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I was skimming through my solutions of the exercises in Chapter I of Hartshorne and I found two exercises I haven't been able to fully solve. Both exercises are about conics.

  • The first exercise (1.1 c) asks the following: Given an irreducible quadratic polynomial $f$ in $k[x,y]$, show that the affine coordinate ring of $k[x,y]/(f)$ is isomorphic to the coordinate ring of the parabola $y=x^2$ or the hyperbola $xy=1$.

  • The second exercise (3.1 c) asks to show that any conic in $\mathbf{P}^2$ is isomorphic to $\mathbf{P}^1$.

Both exercises I've been able to solve whenever the characteristic of the field is different from 2. For the first one I used a brute force calculation where I need to divide by two alot (as morphisms are not really allowed at this point). For the second one I used a symmetric matrix to reduce to the case where the defining polynomial is of the form $F(x,y,z)=ax^2+by^2+cz^2$. This approach also assumes a characteristic different from 2 to construct such a matrix.

My question now is how to do this in the case where the characteristic of the base field is 2. I can't seem to find a way to adapt my current methods to this case. Thanks in advance for any answers!

JKaerts
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    In characteristic $2$, reduce to $xy+z^2$. – Cantlog Oct 20 '13 at 09:10
  • Using this hint I've been able to solve the second problem. The first one I'm still struggling on. I'm trying to work in a completely affine setting which makes things more difficult. – JKaerts Oct 22 '13 at 07:47

1 Answers1

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I've written down details for the following if you need, but I think it's easier to see a sketch then flesh it out on your own. The basic idea is that the "changing coordinates" done below are just $k$-algebra automorphisms of $A(W)$, where $W$ is our conic.

Let $f = a_{11}x^2 + a_{12}xy + a_{22}y^2 + b_1x + b_2y + c$ and $W = Z(f) \subseteq \mathbb{A}^2$. The claim is that $A(W) \cong A(Z(y-x^2))$ if $a_{12} = 0$, and $A(W) \cong A(Z(xy-1))$ if $a_{12} \ne 0$.

If $a_{12} = 0$, then without loss of generality $a_{11} \ne 0$. Changing coordinates $x \mapsto x + \sqrt{a_{22}/a_{11}}y$, we can assume $a_{22} = 0$. $b_2 \ne 0$ since otherwise $f$ is reducible, so change coordinates $y \mapsto y + (b_1/b_2)x$ and $x \mapsto x + \sqrt{c}$ and we are in the case $f = y - x^2$.

If $a_{12} \ne 0$, letting $\alpha$ such that $\alpha^2 + a_{12}\alpha + a_{22} = 0$, change coordinates $x \mapsto x - \alpha y$ and $y \mapsto y + (a_{11}/a_{12})x$ to reduce to the case when $f = a_{12}xy + c$. $c \ne 0$ for otherwise $f$ is reducible, so we are in the case $f = xy - 1$ after rescaling.

Takumi Murayama
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  • I don't see how this can be right. E.g., if you take $f = xy-1$ and change to coordinates $u$ and $v$ such that $x = u + iv$ and $y = u - iv$, then $f$ becomes $(u +iv)(u - iv) - 1= u^2 + v^2 -1$ (which by your analysis should be transformable to the case $f = y - x^2$). – Rob Arthan Feb 17 '18 at 15:48
  • @RobArthan Maybe I am missing something, but the claim isn't that the hyperbola $xy-1$ transforms to the parabola $y-x^2$, but instead that hyperbolas and parabolas are two different types of conics. In particular, the example you wrote down should be considered to be in the hyperbola case $xy-1$, and I am not claiming that you can transform it to a parabola. – Takumi Murayama Feb 17 '18 at 16:02
  • I must be misunderstanding you then.You seem to be saying that you get a parabola if $a_{12} = 0$. But my example has $a_{12} = 1$ w.r.t. $x$ and $y$ and $a_{12} = 0$ w.r.t. $u$ and $v$. – Rob Arthan Feb 17 '18 at 16:15
  • @RobArthan I see; thank you for the clarification! I realize now that your example doesn't quite work since in characteristic 2, $u + iv = u - iv$ so your "change of coordinates" is not an automorphism of $\mathbf{A}^2$. Do you have some other examples? – Takumi Murayama Feb 17 '18 at 16:52
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    I see now: your answer is only for characteristic 2. (For other characteristics, you get a parabola precisely when the quadratic form $a_{11}x^2 + a_{12}xy + a_{22}y^2$ is singular - and you can't test that just by looking at $a_{12}$). Thanks for your time! – Rob Arthan Feb 17 '18 at 16:58
  • @RobArthan No problem; my apologies for the confusion! – Takumi Murayama Feb 17 '18 at 17:18
  • Very good the its answer in characteristic 2. I was looking for factorization of quadratic term in linear forms but its answer is better. Thanks – Bruno Rodrigues dos Santos Nov 30 '21 at 02:05
  • @TakumiMurayama Does it then follow from your argument that $Z(y-x^2)\ncong Z(xy-1)$ in a fields of characteristic two? – slowpoke May 22 '23 at 14:24