4
$\begingroup$

Hartshorne in his chapter on surfaces defines a ruled surface(over an algebraically closed field) to be a smooth projective surface $X$ together with a surjective morphism $\pi:X\to C$, $C$ a smooth curve, such that the fiber over each point $y\in C$, call it $X_y$, is isomorphic to $\mathbb{P}^1$, and such that $\pi$ admits a section.

He then says that the existence of a section follows from Tsen's theorem under the earlier hypothesis.

When he says every point, does he include the generic point also? If yes, then the section will exist by defining it to be anything at the generic point and then using that a morphism from an open subset of a curve to a proper variety extends to the whole curve.

If by points he means only closed points, then how does the existence of a section follow from Tsen's theorem?

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ I don't have Hartshorne with me at the moment, so I can't say with any assurance what he intended. There are two possible definitions: (1) a ruled surface means that fibres over geometric points are $\mathbb{P}^1$'s, or (2) a ruled surface means that fibres over all points are $\mathbb{P}^1$'s. The good news is that these are in fact by Tsen's theorem. Assuming (2), you can realize the generic fibre as a plane conic in $\mathbb{P}^2$, and conclude that it has a rational point over $k(C)$ by the aforementioned theorem. Conics with points are isomorphic to $\mathbb{P}^1$. I think I'm outof spac $\endgroup$
    – Donu Arapura
    Commented Dec 13, 2010 at 0:54
  • 1
    $\begingroup$ Since Donu's comment ended so tragically, let me just add a tiny bit of clarification: you do require the generic fiber as well to be a projective line (at least geometrically, which, as Donu shows, is equivalent to its being $\mathbb{P}^1$ on the nose). The problem with Hartshorne of course is that he insists on working over algebraically closed fields, which makes this sort of thing seem a little suspect. $\endgroup$
    – Keerthi Madapusi
    Commented Dec 13, 2010 at 3:56
  • $\begingroup$ Keerthi, thanks for adding the clarification. $\endgroup$
    – Donu Arapura
    Commented Dec 13, 2010 at 12:47
  • 2
    $\begingroup$ The point is Tsen's theorem implies one needs to know only that the fibers over geometric (closed) points are projective lines. This is because, as Donu says, then the generic fiber is a conic (smooth and genus zero, so canonical bundle is of degree -2, its negative giving an embedding in the plane). Now Tsen's theorem implies such conics have a rational point since the base field is the function field in one variable over an algebraically closed field and hence $C^1$. $\endgroup$
    – Mohan
    Commented Dec 13, 2010 at 18:20

1 Answer 1

Reset to default
6
$\begingroup$

I thought I'd expand my earlier comment, which was not all that clear, and I'm not even sure where you'd look it up. Let's say that a ruled surface over a smooth curve $C$ is smooth projective morphism $f:X\to C$ all of whose fibres are isomorphic to $\mathbb{P}^1$. Then one checks that $\omega_{X/C}^{-1}$ is relatively very ample. So we get an embedding $X\to \mathbb{P}(f_*\omega_{X/C}^{-1})$ over $C$ as a family of conics. Assuming that you are over an algebraically closed field, you can apply Tsen to see that the generic fibre has a rational point. This implies that there is a rational section $C\dashrightarrow X$. Since $C$ is a curve, this extends to an honest section.

An alternative argument is to note that the obstruction to $X=\mathbb{P}(E)$, for some vector bundle $E$, lies in the Brauer group of $C$ via $$H^1(C_{et},GL_2)\to H^1(C_{et},PGL_2)\to H^2(C_{et},G_m)$$ and this vanishes by Tsen. One gets a section using this as well, and much more.

Improve this answer
$\endgroup$
1
  • $\begingroup$ I just realized that Mohan essentially said this in his comment. $\endgroup$
    – Donu Arapura
    Commented Dec 13, 2010 at 19:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .