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

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

2012-01-15T19:32:22Z

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'?

Answer by MP for How does one know the following surface contains no other lines?

2012-01-15T19:45:07Z

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.

Answer by Samir Siksek for How does one know the following surface contains no other lines?

2012-01-15T20:54:01Z

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 $x_4=0$.

Answer by Fedor Petrov for How does one know the following surface contains no other lines?

2012-01-15T23:43:23Z

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.