How do I invert the rational map $(x,y)\mapsto \frac xy$ over the variety $C=(y^3=x^4+x^3)$?

Question

The following question comes from Miles Reid's Algebraic Geometry, Chapter $4$ problem $8$.

I'm asked to come up with a polynomial map $\varphi$ which inverts $\phi:C \dashrightarrow\mathbb A^1$, $\phi(x,y)\mapsto\frac xy$, where $C$ is the affine variety given by $y^3=x^4+x^3\subset\mathbb A^2$.

So far, I know that the domain of $\phi$ is $C\setminus\{(0,0)\}$, and on this domain, $\phi(x,y)=(x+1)^{-1/3}$.

But, to parameterize the inverse as a polynomial map, there must be some polynomial in $t=(x+1)^{-1/3}$ that equals $x$, and another that equals $y$.

However, the only operation I can think of to apply to $t$ is cubing it to get rid of the root, but that leaves us with $t^3=\frac1{x+1}$, which can't be (at least to my knowledge) sent to $x$ without a power series, and definitely not in a polynomial.

Am I just looking at this problem wrong? The inverse map $\varphi$ is supposed to establish an isomorphism between $\mathbb A^1\setminus\{3\text{ pts}\}\to C\setminus\{(0,0)\}$, but this only confuses me further.

Answer 1

You're very close to the answer, but you should be solving for $x$ in terms of $t$, not vice versa. You're looking for functions $x(t), y(t): \mathbb{A}^1 \dashrightarrow C$ such that $\frac{x(t)}{y(t)} = t$.

Let $t = x/y$. Rearranging the equation for $C$, then $$ y^3 = x^4 + x^3 = x^3(x+1) \implies t^3 = \frac{x^3}{y^3} = \frac{1}{x+1} $$ just as you found. Then $x + 1 = \frac{1}{t^3}$, so $x = \frac{1}{t^3} - 1 = \frac{1 - t^3}{t^3}$. Using the fact that $t = x/y$, then $$ y = \frac{x}{t} = \frac{1}{t}\left(\frac{1}{t^3} - 1\right) = \frac{1-t^3}{t^4} \, . $$ Thus the inverse map is \begin{align*} \mathbb{A}^1 &\dashrightarrow C\\ t &\mapsto \left(\frac{1 - t^3}{t^3}, \frac{1-t^3}{t^4}\right) \, . \end{align*}

EDIT: As the OP pointed out, the map I gave is not a polynomial map. To fix this, we can conjugate by the inversion map $\mathbb{A}^1 \setminus \{0\} \to \mathbb{A}^1 \setminus \{0\}$, $t \mapsto 1/t$. Letting $u = 1/t$, then \begin{align*} \frac{1 - t^3}{t^3} &= u^3\left(1 - \frac{1}{u^3}\right) = u^3 - 1\\ \frac{1-t^3}{t^4} &= u^4 \left(1 - \frac{1}{u^4}\right) = u(u^3 - 1) \, . \end{align*} We obtain the polynomial map \begin{align*} \mathbb{A}^1 &\dashrightarrow C\\ u &\mapsto (u^3 - 1, u(u^3 - 1)) \, . \end{align*} However, this map is no longer inverse to $C \dashrightarrow \mathbb{A}^1, (x,y) \mapsto x/y$, but rather to $(x,y) \mapsto y/x$. So we compose with (the inverse) of the inversion map, which has the effect of swapping the $x$- and $y$-coordinates. Thus the desired map is \begin{align*} \mathbb{A}^1 &\dashrightarrow C\\ t &\mapsto (t(t^3 - 1), t^3 - 1) \, . \end{align*}