# How does one know the following surface contains no other lines?

In Heath-Brown's 2002 paper, "Rational points on curves and surfaces", he states

"We may observe that if $d \geq 3$, the surface $$x_1^d + x_2^d - x_2^{d-2} x_3 x_4 = 0$$ is absolutely irreducible, and contains no lines other than those in the planes $x_2 = 0, x_3 = 0$, and $x_4 = 0$."

I am wondering how he is able to be so conclusive as to state no other lines lie on the surface. Can anyone explain why this is 'obvious'?

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Which characteristic? Affine or projective? (I assume projective.) –  darij grinberg Jan 15 '12 at 19:44

A line is given by a pair of equations: \begin{equation*} a_1 x_1 +a_2 x_2+a_3 x_3 + a_4 x_4=0, \qquad b_1 x_1 + b_2 x_2 + b_3 x_3 + b_4 x_4=0. \end{equation*} Suppose this line is on $X$. If the minor $a_3 b_4-a_4 b_3$ is non-zero, then we may rewrite the equations of the line as \begin{equation*} x_3=a x_1+ b x_2, \qquad x_4=c x_1 + e x_2. \end{equation*} Substituting into the equation of the surface $X$ we see that the expression \begin{equation*} x_1^d+x_2^d-x_2^{d-2}(ax_1 +b x_2)(c x_1 + e x_2) \end{equation*} vanishes as polynomial in x_1 and x_2. This is clearly impossible by considering the coefficient of $x_1^d$. Hence the minor $a_3 b_4-a_4 b_3=0$. So we can suppose that one of the equations of the line is of the form $a x_1 + b x_2=0$. Assume that the line does not lie on either of the planes $x_1=0$ or $x_2=0$. Thus neither of $a$ or $b$ is zero and we may rewrite this equation as $x_1=c x_2$. Substituting in the equation for $X$ we see that the line lies on the conic \begin{equation*} (1+c^d) x_2^2 - x_3 x_4=0. \end{equation*} If $1+c^d \ne 0$, then the conic is irreducible and so does not contain a line. Hence $1+c^d=0$ and so the line is on one of the planes $x_3=0$ or $x_4=0$.

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Aah... algebraic geometry with bare hands calculations, like in the good old times! –  Georges Elencwajg Jan 15 '12 at 21:59

If line is parametrized by equations $x_i=a_it+b_is,1\leq i\leq 4$, then we see that $x_1^d$ is divisible by $x_2^{d-2}$ as polynomial in $t$ and $s$, so either $x_2=0$ and we are done, or $x_1=cx_2$ for some scalar $c$, we get $(c^d+1)x_2^2=x_3x_4$, which is possible only if $c^d+1=0$, in this case $x_3$ or $x_4$ vanishes, or if $x_3,x_4,x_2$ are proportional linear forms in $t$ and $s$, but in this case our line appears to be just a point.

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Any line on $X$ has at least one point in common with the plane $x_2=0$. Thus, any line on $X$ contains a point whose first two coordinates vanish. In particular, one of the equations of a line in $X$ can be chosen to be of the form $\lambda x_1+\mu x_2=0$ for a non-zero pair $(\lambda,\mu)$. From here, the result is immediate.

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It would be more clear to say "any line in $X$ is contained in a hyperplane of the form $\lambda x_1+\mu x_2=0$". –  GH from MO Jan 15 '12 at 20:41