Is every quadratic surface in $\mathbb{P}^3$ ruled?

Question

Let $\mathbb{P}$ be the projective line over an algebraically closed field $k$.

Is it true that every quadratic surface in $\mathbb{P}^3$ is ruled?

How can one see that this is the case?

This seems rather strange to me, but I was told this by a classmate.

Answer 1

Real case

The sphere is an unruled quadric, so the answer is no.

Wikipedia distinguishes several cases, in particular three non-degenerate ones:

The first case is the empty set.

The second case generates the ellipsoid, the elliptic paraboloid or the hyperboloid of two sheets, depending on whether the chosen plane at infinity cuts the quadric in the empty set, in a point, or in a nondegenerate conic respectively. These all have positive Gaussian curvature.

The third case generates the hyperbolic paraboloid or the hyperboloid of one sheet, depending on whether the plane at infinity cuts it in two lines, or in a nondegenerate conic respectively. These are doubly ruled surfaces of negative Gaussian curvature.

So it effectively depends on the sign pattern of the normal form, which is the sign pattern of the eigenvalues of the symmetric matrix representing the quadric.

To see that the statement is true for the last case, it is enough to show it for one representative of that class, since all others are related to that by some projective transformation, and a projective transformation will preserve the ruling.

Algebraically closed field

OK, as Marioano pointed out in a comment, there is a comment to the question indicating that the question is not about real projective space, but instead over an algebraically closed field. In that case, the concept of signs breaks down, and all non-degenerate quadrics are related by projective transformations, and hence equivalent. So if any of them is ruled, all are. Showing that at least one of them is, by giving an explicit ruling, should be fairly simple.

Degenerate conics will factor in a pair of planes, which are obviously ruled as well.