An isomorphism of $\mathbb{K}$-algebras from $\mathbb {K}_D$ to $\mathbb{K}_C$ Let $C=\mathcal{Z}(Y^2-X^2-X^3)$ and  $D=\mathcal{Z}(X^2+Y^2-X)$. Show that $\phi:C\rightarrow D$, $\phi(a,b)=\left(\frac{a^2}{a^2+b^2}, \frac{ab}{a^2+b^2}\right)$  determines an birational isomorphism between $C$ and $D$.
I know that is enough to find an isomorphism of $\mathbb{K}$-algebras from $\mathbb {K}_D$ to $\mathbb{K}_C$, where $\mathbb {K}_C$ is the field of rational functions of $C$ and similar $\mathbb {K}_D$.
 A: Hint: Try to find a rational inverse for $\phi$.  Letting $C: y^2 = x^2(1+x)$ and $D: u = u^2 + v^2$, observe that if $(u,v) = \phi(x,y)$, then
$$
\frac{v}{u} = \frac{\frac{xy}{x^2+y^2}}{\frac{x^2}{x^2 + y^2}} = \frac{xy}{x^2} = \frac{y}{x} \, .
$$
Can you use this to solve for $x$ and $y$ in terms of $u$ and $v$?

 Note that the equation for $C$ implies $\frac{y^2}{x^2} - 1 = x$.

EDIT: I can't resist saying more about this example!  If one was not given the birational map between these two curves, here is how you might find it.  Both $C$ and $D$ are birational to the line $\mathbb{A}^1$.  This can be seen by parametrizing the curves using pencils of lines as follows.  Since $C$ has degree $3$ and the origin is a double point on $C$, then a line through the origin intersects $C$ in exactly one other point by Bézout's Theorem.  Writing this line as $y = tx$ and substituting, then
$$
t^2 x^2 = x^2 + x^3 \implies 0 = x^3 + (1-t^2)x^2 = x^2(x - (t^2-1))
$$
The factor of $x^2$ corresponds to the double point at the origin, while the other factor gives us the parametrization $x = t^2 - 1, y = tx = t(t^2-1)$.  Proceeding similarly with $D$, we find the birational maps
\begin{align*}
\varphi_1: C &\overset{\sim}{\dashrightarrow} \mathbb{A}^1 & \varphi_2: D &\overset{\sim}{\dashrightarrow} \mathbb{A}^1\\
(x,y) &\mapsto y/x & (u,v) &\mapsto v/u\\
(t^2-1, t(t^2-1)) &\leftarrow{\raise{.5pt}{\hspace{-4.5pt}\shortmid}} t & \left(\frac{1}{1+t^2}, \frac{t}{1+t^2}\right) &\leftarrow{\raise{.5pt}{\hspace{-4.5pt}\shortmid}} t \, .
\end{align*}
It turns out that your $\phi$ is the composition of these two maps:
\begin{align*}
(x,y) \overset{\varphi_1}{\longmapsto} y/x \overset{\varphi_2^{-1}}{\longmapsto} \left(\frac{1}{1+(y/x)^2}, \frac{y/x}{1+(y/x)^2}\right) = \left(\frac{x^2}{x^2+y^2}, \frac{xy}{x^2+y^2}\right) \, .
\end{align*}
So, in the end, there is a lovely geometric explanation of this birational map: take a point $P$ on $C$ (resp., $D$), and draw the line through the origin that passes through $P$.  This line will intersect $D$ (resp., $C$) in exactly one point $Q$ (besides the origin), and this point $Q$ is the image of $P$ under the birational map.  I've made an image using Sage depicting this below.
$\hspace 4.5 cm$
