Somehow, this problem has been coming up a lot lately in different guises, which I'm taking as a sign that I ought to stop avoiding computational algebraic geometry. I could probably dig this up in one of my textbooks, but I thought I might get a more diverse perspective here, and somehow I couldn't find it in the archive.
Given an irreducible curve $C\subset\mathbb{A}^2$ and a rational map $\varphi:\mathbb{A}^2\to \mathbb{A}^2$ with nontrivial image, is there a standard way to give an equation for $\overline{\varphi(C)}$?
Here is a more concrete version of the question:
Given an irreducible, non-constant polynomial $P\in \mathbb{C}[X,Y]$, and rational functions $f,g\in \mathbb{C}(X,Y)$, how to compute an irreducible polynomial $Q$ such that, for all $a,b\in \mathbb{C}$, $P(a,b)=0 \implies Q(f(a,b),g(a,b)) = 0$?
I am assuming here that the image of an irreducible curve is either a single point, or a closed, irreducible curve with finitely many points missing. Since I'm specifying rational maps, technically the image could be empty—but I have a feeling I can do that case by hand. Anyway, the point is that I've tried to avoid pathologies in the question setup, but let me know if I failed to do so, and please tell me what I should have asked instead.
I would think there is some formulaic way to do this particular case (if we fix the degree, there should exist a polynomial expression in the coefficients of $P,f,g$ for the answer... right?), but I'm also interested in the more general question: What's a good way to compute images of affine or projective varieties? I have no doubt that the answer involves a software package, but I'm especially interested in ones that I can pick up easily, since my interest in this sort of heavy computation is still very casual.
Thanks in advance!
I think a good way to compute images of varieties is elimination (as an reference i would recommend appendix A.7 of the book "A Singular Introduction to Commutative Algebra" by G.-M. Greuel and G. Pfister).
For example, if you have an morphism $f: \mathbb{A}^n\rightarrow\mathbb{A}^m$, given by $m$ polynomials $f_1,\ldots ,f_m\in\mathbb{C}[x_1,\ldots ,x_n]$, and a variety $X\subset\mathbb{A}^n$ defined by the ideal $I$, then eliminating $x_1,\ldots ,x_n$ from the ideal $\langle I,y_1-f_1,\ldots ,y_m-f_m\rangle \subset\mathbb{C}[x_1,\ldots ,x_n,y_1,\ldots ,y_m]$ gives an ideal $J\subset\mathbb{C}[y_1,\ldots ,y_m]$, which defines the closure $\overline{f(X)}$ of the (set-theoretic) image $f(X)$ (this is quite literally taken from section A.7 of "A Singular Introduction to Commutative Algebra"). Basically the same works for projective varieties, see the same reference again... Finally, if your are looking for an appropriate computer algebra system, Singular, for example, provides an implemented function for elimination (I would really like to list some alternatives, but i only worked with Singular up to now). How elimination works is explained in "A Singular Introduction to Commutative Algebra" in many detail.
EDIT: I forgot to mention one point i want to catch up on now: The case described above of course describes a regular morphism of affine varieties (not a rational one -- which would have better corresponded to the explicit example given in your question). However, for projective varieties, the very same procedure is suitable to compute images of actually rational maps, as a rational map $\mathbb{P}^n\rightarrow \mathbb{P}^m$ for some $n,m$ can be descibed by $m+1$ homogenous polynomials $f_0,\ldots ,f_m\in\mathbb{C}[x_0,\ldots ,x_n]$ which may have common zeros, but "not too many", in the sense that $\mathbb{P}^n-V(f_0,\ldots ,f_m)$ is dense (in $\mathbb{P}^n$). The analogous procedure to that sketched above (for a projective variety $X\subset\mathbb{P}^n$ defined by a homogenous ideal $I$) then computes the closure $\overline{f\left( X-X\cap V(f_0,\ldots ,f_m)\right) }$ (As a possible reference, see "A Singular Introduction to Commutative Algebra" again).
Kind regards!