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Let $K$ be an algebraically closed field of characteristic $2$, and we will consider affine varieties of $\mathbb{A}^2$. Let $X = Z(y-x^2)$ and $Y = Z(xy-1)$. I have shown through some exhausting case analysis that the use of completing the square can show that these are the only irreducible quadratic varieties in characteristic $\neq 2$, but the text I'm working from seems to indicate that this isn't the case in characteristic $2$, as it asks if a proof of the fact can be valid in characteristic $2$ (which to me seems to indicate, in the way math problems do, that the answer is no).

I would like a gentle hint towards finding such a variety if possible. A friend and I have considered $W = Z(x^2 + xy + y)$ but we don't have much technical machinery to show that $W \not \simeq X$ or $Y$. Does anyone have any ideas?

Edit: Running with this variety for a moment, note that $(1,y)$ is never a point in $W$ as then $1 + y + y = 0 \rightarrow 1 + 2y = 0 \rightarrow 1 = 0$ as $K$ has characteristic $2$, and certainly this is a contradiction. So now suppose $x \neq 1$, then we may solve for $y$: $y = \dfrac{x^2}{x+1}$ so $W = \left\lbrace\left(x,\dfrac{x^2}{x+1}\right)\right\rbrace$. I have some hope for this one as the second component is NOT a regular function of $x$, which might help, but I'm not sure.

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  • $\begingroup$ I am not sure this would help but I think considering the regular functions on a variety can help $\endgroup$ – happymath Sep 7 '15 at 5:46
  • $\begingroup$ @happymath I've done some consideration of the regular functions, but nothing pops out at me. I'll add some detail I've put into about the points on this variety. $\endgroup$ – walkar Sep 7 '15 at 5:50
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As is often the case, once an abstract proof is found it is psychologically easier to find a down-to-earth one. Here goes:

Change variables to $x=u+1,y=u+v$ and the equation $x^2+xy+y=0$ transforms to $uv+1=0$, or $uv-1=0$, since signs are irrelevant in characteristic $2$.
So you get the equation of $Y$ (with different letters).

[I have left the other answer which is quite general and might be of use for other questions]

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Your psychological interpretation of the question is false: the affine conic $W$ is isomorphic to $Y$. Here is why:

There is, up to isomorphism, only one projective smooth conic $C$ over an algebraically closed field $k$ of characteristic $2$, and it has equation $z^2=xy$: see here.
Consequently there are only two affine conics in $\mathbb A^2_k$: they are obtained by removing the intersection of $C$ with a line in $\mathbb P^2_k$.
According as that line is tangent to the conic or cuts it in two points you obtain an affine conic isomorphic to $X$ or to $Y$.
In your case the projectified conic has equation $x^2+xy+yz$ and cuts the line at infinity $z=0$ in the two points $(0:1:0)$ and $(1:1:0)$.
We are thus in the second case of the dichotomy above and thus $W\simeq Y$.

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  • $\begingroup$ Darn, we were convinced characteristic 2 made something go bad here. We haven't learned anything about projective space so I can't use any of that, but it doesn't matter that I can't use 1 for an x-coordinate as per my analysis above (since you gave the point (1,1,0))? $\endgroup$ – walkar Sep 7 '15 at 11:59
  • $\begingroup$ Dear @walkar, in what course (given by whom) were you assigned this exercise? $\endgroup$ – Georges Elencwajg Sep 7 '15 at 12:49
  • $\begingroup$ It is an Introduction to Algebraic Geometry course offered by Dale Cutkosky. $\endgroup$ – walkar Sep 7 '15 at 13:36
  • $\begingroup$ Thanks for the information, walkar. I was curious because it is quite unusual to discuss conics over fields of characteric two, even in advanced textbooks. Actually I don't know a single book which does that. $\endgroup$ – Georges Elencwajg Sep 7 '15 at 17:13
  • $\begingroup$ This is a fourth part to a single exercise, so it's not like it's the focus of the whole chapter, but he does mention in the introduction to his notes that he was going to point out differences between zero and positive characteristic often. $\endgroup$ – walkar Sep 7 '15 at 17:33

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