$V=\mathcal{Z}(xy-z) \cong \mathbb{A}^2$. 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!

Hint: Use the projection
\begin{align*}
\pi : V &\to \mathbb{A}^2\\
(x,y,z) &\mapsto (x,y) \, .
\end{align*}
Can you find its inverse?
To find the induced maps on the coordinate rings, recall that these maps are just given by preccomposition, e.g., for $\pi : V \to \mathbb{A}^2$,
\begin{align*}
\widetilde{\pi} : k[\mathbb{A}^2] &\to k[V]\\
F &\mapsto F \circ \pi \, .
\end{align*}
To address the question in the comments, no, $V = \mathcal Z(xy-z^2)$ and $\mathbb A^2$ are not isomorphic.  This can be seen on the level of coordinate rings.  Note that the coordinate ring of $\mathbb{A}^2$ is $k[x,y]$ which is a UFD; we claim that the coordinate ring $k[V] = k[x,y,z]/(xy - z^2)$ is not a UFD.  The relation $xy = z^2$ gives two different factorizations, since $x,y,z$ are non-associate irreducibles.  (They all have degree $1$.)  This thread has further explanation.
