Lines on a singular cubic surface How many lines the cubic surfaces $xyz=w^3 \in \mathbb P^3$ has? I found only three: $x=w=0$, $y=w=0$ and $z=w=0$. How to prove that there are no other lines? Also, this surface is singular, is it possible to calculate these lines with some natural multiplicities, summing up to 27?
 A: One (cheap) way is to do this in Macaulay2. In this case there are actually 27 lines with some multiplicities, effectively making the lines you found all the lines.
Here's the code:
R = QQ[x,y,z,w]
f = ideal(x*y*z-w^3)
iF = Fano(1,f)

The output is the ideal of the Fano variety of lines on your surface. Its ideal is
o12 = ideal (p p  - p p  + p p , p , p p p , 2p p p  - p p , p p p , p p , p ,
              2 3    1 4    0 5   5   3 4 5    1 4 5    0 5   1 2 5   0 5   4
      --------------------------------------------------------------------------
         2                                                       2     3
      p p  - 2p p p , p p p  + p p p , p p p  - p p p , p p p , p p , p ,
       1 4     0 4 5   1 3 4    0 3 5   1 2 4    0 2 5   0 2 4   1 4   3
      --------------------------------------------------------------------------
                 2       2     2     2     2         2   2         2     2
      p p p , p p  - 3p p , p p  - 3p p , p p  - 3p p , p p  + 3p p , p p  +
       0 1 3   1 2     4 5   0 2     4 5   1 2     3 5   0 2     3 4   0 1
      --------------------------------------------------------------------------
        2     2       2         2       2       2       2     2       2
      3p p , p p  + 3p p , p p p , p p p , p p p , p p p , p p p , p p p )
        3 5   0 1     3 4   0 4 5   0 3 5   0 2 5   0 1 5   0 4 5   0 3 5

o12 : Ideal of QQ[p , p , p , p , p , p ]
                   0   1   2   3   4   5

Which is kinda unreadable. This is the ideal of the Fano variety of lines as a subvariety of the Grassmannian in its Plücker embedding. But writing "degree iF" gives 27, and using the command "primaryDecomposition" we get three components. Their radicals are gotten by writing "decompose":
i15 : decompose iF

o15 = {ideal (p , p , p , p , p ), ideal (p , p , p , p , p ), ideal (p , p ,
               5   4   3   1   2           5   4   3   0   2           5   4
      --------------------------------------------------------------------------
      p , p , p )}

As said, this is probably a cheap way, and there may be more direct methods to see this. However, it is not automatic that there are even finitely many lines on a singular cubic surface, as for example the example $x^3+y^3+z^3=0$ shows. This is a cone in $\mathbb P^3$ (since $w$ is not included), and here there are infinitely many lines.
A: Here is a direct way of showing the cubic surface $S=\{xyz=w^3\}$ with three cusp singularities has only three lines: $x=w=0$, $y=w=0$ and $x=w=0$.
Suppose there is another line $L$ in $S$ different from the above, then $L$ must contain a point $p$ of the form $$p=(x_0,y_0,z_0,1),$$
and $L$ lies in the tangent hyperplane $T_pS$. The condition $p\in S$ requires that $x_0y_0z_0=1$, so $T_pS$ has equation
$$\frac{x}{x_0}+\frac{y}{y_0}+x_0y_0z-3w=0.$$
The intersection $C=T_pS\cap S$ is a cubic plane curve and $C$ contains $L$ as an irreducible component, therefore $C$ is reducible. However, by isolating $w$ in the linear equation and canceling $w$ in the cubic surface equation, one finds the equation of $C$ is
$$xyz=\frac{1}{27}(\frac{x}{x_0}+\frac{y}{y_0}+x_0y_0z)^3$$
in the projective plane with coordinate $x,y,z$. Of course, $C$ is singular at $p=(x_0,y_0,\frac{1}{x_0y_0})$ since it is tangency point. However, a direct computation of partial derivatives shows that $p$ is the only singularity of $C$ and the Hessian at $p$ is always nondegenerate, so it is an ordinary double point. Therefore $C$ is irreducible, which is a contradiction.
A: A line not in the plane at infinity has a parametric equation of the form
$$x= x_0 + a t\\
y = y_0+ b t\\
z = z_0 + c t$$
If it were contained in the (finite) part of this surface $x y z = 1$, then we would have
$$(x_0 + a t)(y_0 + b t)(z_0 + c t) = 1$$
for every $t$. Note that $a$, $b$,$c$ are not all $0$, so on the LHS we have a polynomial in $t$ of degree $\ge 1$. We get a contradiction.
