I'm trying to understand better conic bundles on quartic del Pezzo surfaces over non-algebraically closed fields.

Let $k$ be a field. A conic bundle surface is a smooth projective surface $S$ over $k$ equipped with a dominant morphism $S \to C$ to some smooth curve $C$, whose fibres are isomorphic to plane conics.

Let $S$ be a quartic del Pezzo surface over $k$ (i.e. $S$ is a smooth complete intersection of two quadrics in $\mathbb{P}^4$). If $S$ contains a conic $Q$ over $k$, then it admits a conic bundle. Namely, the pencil of hyperplanes through $Q$ gives rise to a conic bundle morphism $S \to \mathbb{P}^1$, determined by the conics residual to $Q$. I want to know whether every conic bundle on $S$ has this form.

Let $S$ be a quartic del Pezzo surface over $k$ equipped with a conic bundle $S \to C$. Then is $C(k) \neq \emptyset$?

Note that $C$ is clearly a smooth curve of genus $0$ in this case.

The answer to the analogue of my question is no for del Pezzo surfaces of degree $8$. Take $S = C \times C'$ where $C$ and $C'$ are conics without rational points. Then $S \to C$ is a conic bundle, but there is no conic bundle $S \to \mathbb{P}^1$.

For cubic surfaces, however, the answer is yes. A cubic surface with a conic bundle contains a line and thus a rational point. Hence the base of the conic bundle is always $\mathbb{P}^1$.

I'm guessing that the answer to my question is no, but I don't see how to construct explicit counter-examples. Note that a similar argument to the cubic surface case shows that such a counter-example must have no $0$-cycle of odd degree.

  • $\begingroup$ I don't quite see why do you think a conic bundle constructed from a conic $Q$ satisfies $C(k) \ne 0$. In fact, $Q$ is not a fiber of this conic bundle. On a contrary, fibers are conics residual to Q. $\endgroup$
    – Sasha
    Aug 8, 2016 at 19:31
  • $\begingroup$ I have clarified my construction. If you have a quartic del Pezzo surface with a conic $Q$ then it in fact has 2 conics bundles. Namely you consider the pencil of hyperplanes which contain $Q$ and take the residual intersection. This gives one conic bundle. But then you can of course run this construction with the residual conics to get another conic bundle. $\endgroup$ Aug 8, 2016 at 20:10
  • $\begingroup$ But both these conic bundles arise from pencils! And by definition, a pencil gives rise to a morphism $S \to \mathbb{P}^1$. $\endgroup$ Aug 8, 2016 at 20:11

1 Answer 1


Let $S_0 \subset P^3$ be a quadric surface (defined over $k$) with no 0-cycles of odd degree. Let $S_1 \subset P^3$ be another quadric surface (also defined over $k$), such that the intersection $E := S_0 \cap S_1$ is smooth and in the pencil generated by $S_0$ and $S_1$ there are no degenerate quadrics defined over $k$.

Let $S$ be the double cover of $S_0$ branched over $E$. Then I claim that $S$ has just two conic bundles defined over $k$, and the base of both has no $k$-points, thus providing the required counterexample.

  • $\begingroup$ Where do the conic bundles on $S$ come from? Are you taking something like $S_0 = C \times C$? $\endgroup$ Aug 9, 2016 at 7:34
  • 1
    $\begingroup$ Conic bundles on $S$ come from degenerate quadrics (by projecting from their vertices) in the pencil of quadrics passing through $S$ in its anticanonical embedding. Under the above condition only one such quadric is defined over $k$, hence only two conic bundles are defined over $k$, and for these two the bases are just the bases of conic bundles on $S_0$. And yes, $S_0$ is $C \times C$. $\endgroup$
    – Sasha
    Aug 9, 2016 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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