Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.