Non-isomorphic two-variable varieties in characteristic 2 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.
 A: 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 $W\simeq Y$.
A: 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]
