Let the field of rational numbers be our base field $k$. I hope to find all rational curves on the following surface $S$ defined by $f$. You can find the motivation in the end. $f= (x^2y^2)z^3 + (5x^3y^3 + x^3y^2 + x^2y^3 + 2x^2y + 2xy^2 + xy)z^2+(- x^5y^5 + 10x^4y^4 + 5x^4y^3 + 5x^3y^4 + 3x^3y^2 + 2x^3y + 3x^2y^3 + 6x^2y^2 + 3x^2y + x^2 + 2xy^3 + 3xy^2 + 3xy + x + y^2 + y)z - x^6y^6 + 2x^5y^5 + 5x^5y^4 + 5x^4y^5 + 15x^4y^4 + 5x^4y^2 - 10x^3y^3 + 6x^3y^2 - 2x^3y + x^3 + 5x^2y^4 + 6x^2y^3 + 13x^2y^2 + 3x^2y + 3x^2 - 2xy^3 + 3xy^2 + x + y^3 + 3y^2 + y + 1.$ Note that $f$ is symmetric about $x$ and $y$. The Kodaira dimension of this surface seems to be positive so one expects only finitely many rational curves. Is there an algorithm to find all rational curves on surfaces of general type? A minor question is whether there are finitely many elliptic curves on $S$?
Here is my naïve attempt. We first exclude 2 trivial ones given by $x=0, y=0$. Let $f$ be an irreducible polynomial in $k[x,y,z]=k[x,y][z]$. By a codimension-1 factorization condition for $f$, I mean an element $r\in k[x,y]$ such that $f$ factorizes nontrivially in $\big(k[x,y]/(r)\big)[z]$. I found 5 codim-1 factorization conditions of genus 0: $$x-1,\ y-1,\ x-y,\ x^2y^2 - 2x^2y + xy + x – y,\ x^2y^2 - 2xy^2 + xy - x + y.$$ Here, by genus I mean the genus of the (possibly singular) curve they define (over $\mathbb{C}$). Such a factorization condition clearly yields a rational curve on $S$. Can you find more genus-0 conditions? I found many genus-1 conditions as well.
Motivation: any other rational curve should give rise to an elliptic curve over the function field $\mathbb{Q}(t)$ of positive rank with torsion $=\mathbb{Z}/10\mathbb{Z}$. It is explicitly given by $$xu^2v^2 + (x+1)uv(u + v) + u^2 + v^2 + (y + 1) (u + v) + y – zuv=0$$ I found the 5 conditions by “prescribing” torsion points, so in certain sense they are trivial ones.
