Let $k=\overline{k}$ be a field and let $X=V(x^2+by^2+c)$ and $Y=V(x^2+by+c)$ where $b,c \in k^{\times}$. Are the varieties $X$ and $Y$ isomorphic, i.e. are their coordinate rings isomorphic?
Ideas:
I have showed that $X$ and $Y$ correspond to prime ideals, and have been looking for geometric intuitions. I imagine $X$ as some kind of generalized 4D complex ellipse and $Y$ as a ditto parabola. Is this intuition right?
I want show whether there is a bijection of $k[x,y]/I(X)$ and $k[x,y]/I(Y)$ by showing whether there is a bijection that maps an ellipse to a parabola. If I try to solve this in the 2D case I would try this direction:
Find a parametrisation of an ellipse and an a parabola. Since the ellipse loops, there must be a point on it whose image is at infinity, and therefore no bijection exists.
Am I on the right track? Is there a geometric interpretation of this problem for $k=\mathbb{C}$?
EDIT: I'm making progress. What I would like comments on mostly is whether my intuition regarding the geometry is correct. E.g, can we think of $x^2+y^2+1$ as some figure in 4D?
a) If $char.k\neq 2$ the curves $X$ and $Y$ are not isomorphic.
The trick is to look at their projective closures $\bar X, \bar Y\subset \mathbb P^2$ which are both isomorphic to $ \mathbb P^1$, like all smooth projective plane conics.
Since $X$ is obtained from $\bar X$ by removing two points, $X$ is isomorphic to $\mathbb A^1$minus one point.
Since $Y$ is obtained from $\bar Y$ by removing one point, $Y$ is isomorphic to $\mathbb A^1$.
Hence $X$ is not isomorphic to $Y$.
b) If $char.k= 2$ the curves $X$ and $Y$ are both isomorphic to $\mathbb A^1$.
Indeed $X$ is the line in $\mathbb A^2$ with equation $x+\sqrt b\cdot y+\sqrt c=0$.
While $Y$ is the graph of the function $\mathbb A^1\to \mathbb A^1:x\mapsto y=-\frac 1b (x^2+c)$, and thus (like all graphs) isomorphic to its source $\mathbb A^1$.
NB
In b) I have considered $X$ as a variety in the naïve sense of being a subset of $\mathbb A^2$.
In the scheme theoretic sense $X$ is a non reduced line of multiplicity $2$, and thus is not isomorphic to $Y$.