Family of hyperbolas isomorphic to A^2

[ay46]

This question is typically seen in the beginning of a commutative algebra course or algebraic geometry course.

Let $V = \mathcal{Z}(xy-z) \subset \mathbb{A}^3$. Here $\mathcal{Z}$ is the zero locus. Prove that $V$ is isomorphic to $\mathbb{A}^2$ as algebraic sets and provide an explicit isomorphism $\phi$ and associated $k$-algebra isomorphism $\tilde{\phi}$ from $k[V]$ to $k[\mathbb{A}^2]$ along with their inverses. Is $V = \mathcal{Z}(xy-z^2)$ isomorphic to $\mathbb{A}^2$?

Here is what I have so far: let $V = \mathcal{Z}(xy-z) \subset \mathbb{A}^3$. Consider the map: $\pi(x,y,z) = z$ where $\pi$ is a family of varieties i.e. a surjective morphism. This map give the hyperbola family: $\{\mathcal{Z}(xy-z) \subset \mathbb{A}^2\}_{z \in \mathbb{A}^1}$ and is injective. Does this provide an explicit isomorphism $\phi$? I am not sure how to proceed for the coordinate rings and how to define the inverses.

Thank you!

 

[jvicens]

Thank you for your question. You are correct in your observation that the map $\pi(x,y,z) = z$ provides an isomorphism between $V$ and the hyperbola family $\{\mathcal{Z}(xy-z) \subset \mathbb{A}^2\}_{z \in \mathbb{A}^1}$. This is because for any point $(x,y,z)$ in $V$, the value of $z$ uniquely determines the values of $x$ and $y$ (since $xy = z$), and vice versa.

To explicitly define the isomorphism $\phi$, we can use the map $\pi$ as follows: for any point $(x,y,z)$ in $V$, we can map it to the point $(x,y)$ in $\mathbb{A}^2$ by setting $\phi(x,y,z) = (x,y)$. Similarly, the inverse map $\phi^{-1}$ can be defined as $\phi^{-1}(x,y) = (x,y,z)$, where $z = xy$.

For the associated $k$-algebra isomorphism $\tilde{\phi}$, we can define it as follows: for any polynomial $f \in k[V]$, we can map it to the polynomial $\tilde{\phi}(f) \in k[\mathbb{A}^2]$ by replacing all occurrences of $z$ in $f$ with $xy$. Similarly, for any polynomial $g \in k[\mathbb{A}^2]$, we can map it to the polynomial $\tilde{\phi}^{-1}(g) \in k[V]$ by replacing all occurrences of $xy$ in $g$ with $z$.

Now, to answer your second question, $V = \mathcal{Z}(xy-z^2)$ is not isomorphic to $\mathbb{A}^2$. This is because the points in $V$ now satisfy the equation $xy = z^2$, which is a parabola rather than a hyperbola. Therefore, the map $\pi(x,y,z) = z$ is no longer an isomorphism, as it is no longer injective. This means that we cannot find an explicit isomorphism $\phi$ and associated $k$-algebra isomorphism $\tilde{\phi}$ as we did