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Prove that, given an algebraic plane curve $X$, the map $f: \mathbb{A}^1 \to X$ defined as $f(t) = (t^2, t^3)$ and the map $f: \mathbb{A}^1 \to X$ defined as $g(t) = (t^2 - 1, t(t^2 -1))$ are birational.

My attempt For the first one I wrote $x= t^2$ and $y=t^3$ and then $\frac{y}{x}= t$. I think I can’t go on from here, I don’t even know if expressing $t$ in terms of only $x$ and $y$ is helpful.

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  • $\begingroup$ Please define the concept of "birationality". $\endgroup$ – Arman Malekzadeh Aug 6 at 11:11
  • $\begingroup$ A rational map is birational if it has an inverse rational map. $\endgroup$ – cip Aug 6 at 11:15
  • $\begingroup$ "I can't express $t$ in terms of only $x$ and $y$." But didn't you already do that when you wrote $t=\frac{y}{x}$? $\endgroup$ – Lazzaro Campeotti Aug 6 at 13:28
  • $\begingroup$ Yes, sorry, that’s definitely what I did. What I meant instead is that I don’t know if expressing $t$ like that is going to be helpful to find the inverse. $\endgroup$ – cip Aug 6 at 13:38
  • $\begingroup$ Hint: you already found the inverse. $\endgroup$ – Lazzaro Campeotti Aug 6 at 14:15

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