Let $k$ be an algebraically closed field of characteristic $p>0$.
We consider the curve $$C = V(X^pZ^{p-1}-Y^{2p-1})\subset\mathbb{P}^2(k)$$ Show that $C$ is a rational cuve.
We did this problem in class as follows:
$C$ is rational if there exists a rational parametrization. We want $\varphi_i(s,t)$ such that $\{(\varphi_0(s,t),\varphi_1(s,t),\varphi_2(s,t))|(s:t)\in\mathbb{P}^1_{k}\}=V(f)$.
Choose $\varphi_0(s,t)=s^{2p-1}, \varphi_1(s,t)=s^pt^{p-1}, \varphi_2(s,t)=t^{2p-1}$.
$\Rightarrow f(\varphi_0,\varphi_1\varphi_2)=(s^{2p-1})^p*(t^{2p-1})^{p-1}-(s^pt^{p-1})^{2p-1}=s^{2p^2-p}t^{2(p-1)^2}-s^{2p^2-p}t^{2(p-1)^2}=0.$
My question is, how did we find this parametrization? Only by 'looking closely' or is there any method or trick to find it quickly?
Thanks and best, Luca
Since rationality can be tested on any non-empty open subset let's just study the affine part $C_0\subset \mathbb A^2_k=\{Z\neq0\}$ of your curve $C$.
The equation of $C_0$ is then $x^p=y^{2p-1}$.
But in general the equation $x^a=y^b$ cries out to be parametrized by $x=s^b, y=s^a$ so that in your case the parametrization is $x=s^{2p-1}, y=s^{p}$.
If you add the homogeneizing parameter $t$ in order to have a morphism defined on the whole projective line $\mathbb P^1_k$(and not only on $\mathbb A^1_k$) you obtain your parametrization.
And you know what? The assumptions that the characteristic is $p$ and that $p$ is a prime number are completely irrelevant!