Prove or disprove that $Z(z^2-xy)$ is isomorphic to $\mathbb{A^2}$.
This is exercise 24 from Dummit and Foote's section on Algebraic Geometry.
Work so far
I can prove that if the field is $\mathbb{Z}/2\mathbb{Z}$ then they isomorphic as $Z(xy-z^2)=Z(xy-z)$ and a previous exercise states that $Z(xy-z)$ is isomorphic to $\mathbb{A^2}$.
I can also prove that $K[x,y,z]/(xy-z^2)$ is not isomorphic to $K[x,y]$, so if I manage to show that $I(Z(xy-z^2))=(xy-z^2)$ whenever the field is infinite the result will follow.
Question
I have no clue about whether they are isomorphic or not for any finite field besides $\mathbb{Z}/2\mathbb{Z}$ and I could really use some help.
Thanks.
Let's see how much points it has $Z(xy-z^2)$ over the field $\mathbb{F}_q$ with $q$ elements. We will see it has the same number of points that $\mathbb A^2(\mathbb{F}_q)$.
For $z=0$, there are $2q-1$ pairs of $(x,y)$: the ones with $(0,y)$ and $(x,0)$.
For $z\ne 0$, there are $q-1$ pairs $(x,z^2/x)$, for $x\ne 0$. So we get $(q-1)^2$ such points.
So we get $2q-1+(q-1)^2=q^2$ points in $Z(xy-z^2)$, which is equal to $q^2$, which are the number of points in $\mathbb{A}^2(\mathbb{F}_q)$.
EDIT: In fact, one can find a bijection between both sets, even over an infinite field $K$, that explains this equality. You can send any point $(x,y,z)\in Z(xy-z^2)(K)$ with $z\ne 0$ to $(x,z)\in K^2$. This gives a bijection between all the points of $Z(xy-z^2)(K)$ with last coordinate $\ne 0$ with $(K^*)^2$, since $y=z^2/x$ is determined from $x$ and $z$. Then you send the points $(x,y,0)$ to $(x,y)$; note that for these points verify that either $x$ or $y$ are $0$, so $(x,y)$ lands in $K^2\setminus (K^*)^2$.
Now, the question one needs to answer is: this bijection, is in fact an isomorphism (as algebraic varieties)? And the answer is no, since it is not given globally by an algebraic function. In fact, there is no isomorphism between them since the verify distinct algebraic properties: the affine plane is a non-singular variety, but this variety is singular (in fact, it is what is called a cone, as a picture of it over $\mathbb R$ shows).