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Is there a general method for trying to finding parametric equations for polynomial equations?

The classic example of $x^2 + y^2 = 1$ having multiple parameterizations:

\begin{array}{rcl} (x,y) &=& (\sin t, \cos t) \\ (x,y) &=& (\cos t, \sin t) \\ (x,y) &=& \left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right) \end{array}

which are easily seen to work by plugging in. Note that the first two involve elementary functions, but the last one only rational functions.

I'm curious about higher degree polynomials. Is there a general method to find a parameterized set of equations in rational functions? That is, is there an algorithm for extracting the parameter functions for $x$ and $y$ from the equation (or arbitrary sets of equations over $v\ge1$)? And if not, is there a characterization of which implicit sets can be parametrized?

Take the (random off the top of my head) example:

$$x y^3 + x^2 y + x y = 1$$

Is there a set $x(t), y(t)$ for this that can be found algorithmically? (I may well be naive here, not realizing something obvious, or maybe the entire field of algebraic geometry is geared towards possible solutions to this).

One can do the reverse process of creating an implicit set of equations via Gröbner basis calculation, trying to eliminate $t$. Or likewise in the forward direction if you can eliminate variables one-by-one with GB that would suffice to parameterize via one of the original variables. But in general if elimination isn't possible, is there still an algorithm to (even partially) parameterize?

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    $\begingroup$ A curve over an algebraically closed field can be parametrized by rational functions iff its genus is zero. Your example has genus $2$, so it has no rational parametrization. For more details, see this paper by Lemmermeyer or this book by Winkler, Sendra, and Perez-Diaz. $\endgroup$ – André 3000 Oct 3 '18 at 16:18
  • $\begingroup$ @Nosrati Thanks for finding that (I searched on Math.SE but didn't find that one). I've edited to emphasize what I think is not being asked there. $\endgroup$ – Mitch Oct 3 '18 at 16:50
  • $\begingroup$ Don't worry about duplication. The people's vote is dominant :) $\endgroup$ – Nosrati Oct 3 '18 at 16:53
  • $\begingroup$ @Nosrati That question doesn't have 'the' answer. And there is one, or at least a better attempt at one which I will try to supply there. $\endgroup$ – Mitch Oct 4 '18 at 15:27
  • $\begingroup$ Sorry. I didn't know people choose duplication :(, do you want to reopen it? Did you see two references had given by @Quasicoherent?. $\endgroup$ – Nosrati Oct 4 '18 at 15:30