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I am looking for an "arithmetic" analogue of a well known result on threefolds with a conic bundle structure. The following result can be found in [Iskovskikh - On the rationality problem for conic bundles, Lemma 1]. Note that Iskovskikh has some extra condition of relative minimality which I am pretty sure I don't need for the result I want.


Let $X$ be a smooth irreducible threefold over $\mathbb{C}$ with a morphism $\pi:X \to B$ to a smooth rational surface $B$ such that every fibre is a (possibly degenerate) conic.

Then, then there exists a reduced normal crossings divisor (the "discriminant curve") $\Delta \subset B$ such that for any $b \in B$ we have:

(a) $\pi^{-1}(b) \cong \mathbb{P}^1$, if $b \not \in \Delta$
(b) $\pi^{-1}(b)$ is two intersecting lines if $b \in \Delta \backslash Sing (\Delta) $
(c) $\pi^{-1}(b)$ is a non-reduced line if $b \in Sing(\Delta) $
(d) In particular, there are only finitely many non-reduced fibres.


In my situation, I have a smooth conic bundle surface $p:S \to \mathbb{P}^1$ defined over $\mathbb{Q}$, and I have chosen a regular model $\pi: X \to \mathbb{P}^1_{\mathbb{Z}}$, i.e. the morphism $\pi$ restricted to the generic fibre is exactly the morphism $p$ and every fibre is a conic.

Does an analogue of the above result hold in my case? If so, does anyone have a reference to where it has been worked out in the literature?

I hope it is clear, but just to clarify that I want a reduced normal crossings divisor $\Delta \subset \mathbb{P}^1_{\mathbb{Z}}$ which satisfies the appropriate analogues of conditions (a), (b), (c) and (d).

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  • $\begingroup$ Just to notify, replied to yours comment at mathoverflow.net/questions/96987/… will delete this later $\endgroup$ Commented May 17, 2012 at 10:53
  • $\begingroup$ @DanielLoughran Did you ever find what you were looking for? I am interested in this as well. $\endgroup$ Commented Jun 15, 2022 at 17:43
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    $\begingroup$ @Somatic Custard: I found way around this so didn't need it in the end (spreading out sufficed for my application). I suspect one can adapt the proof of Proposition 1.2 from Beauville - Variétés de Prym et jacobiennes intermédiaires to achieve this, but I didn't attempt it. Probably one should invert $2$ to make life easier. Let me know how you get on! $\endgroup$ Commented Jul 1, 2022 at 9:57

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