find skew lines on a cubic surface for a parametrization

I consider the hypersurface $Y = V(y(x^2+z^2)-x) \subset \mathbb{A}^3_k$ ($k = \mathbb{C}$)

I've read that if you have two skew lines on a non-singular cubic surface $Y$, given by a polynomial of degree 3, then you can find a rational map to a plane.

To find a parametrization of $Y$, first i need to find two skew lines lying on $Y$. How can i find them?

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Dear user, just start with making the bet that the lines may be written as $$x = Ay + B, \quad z = Cy + D,$$ so that $y$ is a good parameter along the line. It's infinitely unlikely that this is not possible for a line. Of course, I will have to study potential lines inside $y={\rm const}$ planes, too. This will turn out to contain the essence of the problem. Let's start with the "non-singular ones", however.

When $y$ is a good parameter

Substitute these values to your $Y$. You will get $$Y / V = (A^2+C^2) y^3 + 2(AB+CD) y^2 + (-A+B^2+D^2) y - B$$ It's a cubic polynomial with four coefficients that must vanish if $Y=0$ holds automatically, and we have four parameters $A,B,C,D$ to adjust the line(s). So it looks good, let's solve it.

The vanishing of the cubic coefficient tells us that $A=\pm iC$. Assuming the former, the vanishing of the quadratic one tells us that either $B=D=0$ - which can't work because we would also get $A=C=0$ and it wouldn't be a line but a point - or $B=\pm iD$ with the same $\pm$ sign as before. However, in the latter case, the linear coefficient would give us $A=0$ anyway, so we would have $A=B=C=D=0$.

Alternatively, if $A=C=0$, the vanishing of the quadratic coefficient is automatic but again, we will derive that $A=B=C=D=0$ - no line, too bad. To find any lines, we will have to look at "singular" lines in the sense that they can't be parameterized by $y$.

When $y$ is a bad parameter

The variable $y$ is a bad parameter along the line if the lines belong to the $y=U$ plane where $U$ is a constant. Substitute it to $Y$: $$Y/V = U(x^2+z^2) - x$$ This is a quadratic curve inside the $y=U$ plane. Can it be straight? Well, the quadratic polynomial would have to factorize to $(1+v_1 x + v_2 z)(1+v_3 x + v_4 z)$. We would have $$v_1v_3=v_2v_4=U, \quad v_3+v_1=-1, \quad v_2+v_4=0, \quad v_2v_3+v_1v_4=0$$ There's also the absolute term $1$ that doesn't cancel - I expect that you allow $Y$ to be shifted by this - otherwise I won't find any lines.

With this disclaimer, you see that $v_2=-v_4=\sqrt{U}$ from the first and third equation, $v_1=v_3=-1/2$ from the fourth and then second equation, $U=1/4$ from the first equation again. Then $v_2=-v_4=1/2$ - without loss of generality, I can order the two factors in this way (permutation of $v_2,v_4$ just flips the two lines).

So I have found two lines given by $y=1/4$, $2-x\pm z=0$. They're two because the $\pm$ sign can be chosen in any way. The derivation of $v_1,v_2,v_3,v_4$ was sloppy when $U=0$. In this case, $y/V=-x$, and we find another line, $x=y=0$.

So you may take the two skew lines. One of them is given by $x=0$, $y=0$, and the other is given by $y=1/4$, $2-x\pm z =0$ where you can choose either $\pm$ of them. They clearly don't intersect because they lie at different constant values of $y$ and they're nor parallel, so they're skew lines.

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