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This is a follow-up question to my previous post: Sufficient conditions to tell whether a surface contains a line

Suppose that $f(x_1, x_2)$ is an irreducible binary form with integer coefficients of the variables $x_1, x_2$ over $\mathbb{Q}$ say, and suppose that $g(x_1, x_2)$ is another binary form with integer coefficients (not necessarily the same degree as $f$) that does not have a common component as $f$. In the case I am interested in this condition is satisfied vacuously since we assume $\deg(g) < \deg(f)$. Now, can the surface $$f(x_1, x_2) = g(x_3, x_4)$$ (in weighted projective space) contain any lines?

Note: the case $f = g = x^d + y^d$ has been treated by Heath-Brown, I believe.

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    $\begingroup$ Unless I am mistaken, one cannot view what you have written down ($f(x_1,x_2)=g(x_3,x_4)$) as a being a surface unless $f$ and $g$ are homogeneous of the same degree, and also in which case you need to consider the corresponding variety as being projective. $\endgroup$
    – Daniel Loughran
    Commented Nov 29, 2011 at 23:25
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    $\begingroup$ You can consider it as a surface in a weighted projective space $\mathbb{P}(a,a,b,b)$ such that $a\text{deg}(f)$ equals $b\text{deg}(g)$ (and probably best to assume that $a$ and $b$ are relatively prime). $\endgroup$
    – Jason Starr
    Commented Nov 30, 2011 at 1:10
  • $\begingroup$ Yes; I do mean a surface in weighted projective space. $\endgroup$
    – Stanley Yao Xiao
    Commented Nov 30, 2011 at 14:58
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    $\begingroup$ But in weighted projective space what do you mean by a line? Just a rational curve? $\endgroup$
    – Felipe Voloch
    Commented Nov 30, 2011 at 15:14
  • $\begingroup$ Would the existence of a rational curve in a surface imply the existence of many rational solutions? $\endgroup$
    – Stanley Yao Xiao
    Commented Dec 3, 2011 at 4:18

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