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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.

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  • $\begingroup$ 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. $\endgroup$
    – Daniel Loughran
    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 $S \to \mathbb{P}^1$. $\endgroup$
    – Daniel Loughran
    Aug 8, 2016 at 20:11

1 Answer 1

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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.

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  • $\begingroup$ Where do the conic bundles on $S$ come from? Are you taking something like $S_0 = C \times C$? $\endgroup$
    – Daniel Loughran
    Aug 9, 2016 at 7:34
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    $\begingroup$ 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

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