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Let $X$ be a non-singular projective variety over a field $k$ (perhaps not of characteristic $2$), and let $\pi:Y\to X$ be a conic bundle over $X$ i.e. a proper morphism all of whose fibres are isomorphic to plane conics. Let $Z \subset X$ denote the discriminant locus, i.e. the closed subset of $X$ consisting of those point whose fibres are singular (this is a divisor on $X$). Finally let $U = X \setminus Z$.

Then, $\pi$ restricts to give a smooth morphism over $U$, which locally for the ├ętale topology is a trivial bundle of $\mathbb{P}^1$'s, hence we obtain an element $A \in H^1(U,PGL_2)$ associated to $X$. A standard argument using the exact sequence defining $PGL_2$ gives rise to an injective map $H^1(U,PGL_2) \to H^2(U,\mathbb{G}_m)=\mathrm{Br}~U$ by which we obtain an element of $\mathrm{Br}~U[2]$ (which by abuse of notation we also denote by $A$).

For any discrete valutation $v$ of the function field $k(U)$, we have a residue map $$\mathrm{res}_v:\mathrm{Br}~U \to H^1(k(v),\mathbb{Q}/\mathbb{Z}).$$ My first question is about how the geometric properties of $\pi$ are related to the algebraic properties of $A$.

Question 1. Let $v$ be a valuation of $k(U)$ corresponding to an irreducible divisor $D$ supported on $Z$. What is the a relationship between $\mathrm{res}_v(A)$ and the fibre of $\pi$ over $D$?

This is slightly vague so I will try to make more precise what I am after.

Naively, one might expect that $\mathrm{res}_v(A)$ is non-zero since the fibre over $D$ is singular. However, this is not quite true as birational conic bundles should give rise to the same Brauer group element on sufficiently small open subsets of $X$. So there may be some another conic bundle also giving rise to $A$ for which the fibre over $D$ is non-singular. The following is a more precise version of the previous question which hopes to get around this issue (we use the same notation as Question 1).

Question 2. Is $\mathbb{res}_v(A) =0$ if and only if the singular conic $\pi^{-1}(D)$ (considered over $k(v)=k(D)$) is split over $k(D)$?

Here by split I mean that the two singular lines are defined over $k(D)$, rather than conjugate over some quadratic field extension. Also I am slightly abusing notation, as really it is the generic fibre of $\pi^{-1}(D)$ over $D$ which is a conic over $k(D)$.

Finally, one might expect that conic bundles for which all singular conics $\pi^{-1}(D)$ are not split over $k(D)$ are in some sense "relatively minimal". Unfortunately I can't make this precise except in the case where $\mathrm{dim}~ Y = 2$, where relatively minimal means that one may not contract any of divisors lying in the fibres of $\pi$. I would like a higher dimensional analogue of this.

Question 3. Suppose that for some divisor $D \subset Z$ the corresponding singular conic $\pi^{-1}(D)$ over $k(D)$ is split. Then is it possible to contract one of the components of $\pi^{-1}(D)$? More precisely, does there exist a conic bundle $\pi':Y'\to X$ and a morphism $f:Y \to Y'$ (respecting $\pi$ and $\pi'$) such that $Z' = Z \setminus D$? Here $Z'$ denotes the discriminant locus of $Y'$.

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Perhaps you should search for "maximal order". –  Jason Starr Jan 9 '13 at 14:21
    
@Jason: Thanks for the hint. I have come across a paper by Artin-Mumford which seems to suggest that relatively minimal conic bundles should correspond to maximal orders in the associated quaternion algebra. Do you have any ideas for my other questions? –  Daniel Loughran Jan 10 '13 at 18:15
    
In some particular case (the base is a rational surface) something has been developed here arxiv.org/abs/1010.2417 but it is not much more than A-M. There's also another recent paper arxiv.org/abs/1101.1705 maybe closer to what you need. –  IMeasy Jan 13 '13 at 20:13
    
@MBeasy: Thanks for the references. I have had a look and I feel like I'm getting closer to being able to answer my questions. One thing which is still confusing me is how does one construct an order in the associated central simple algebra $A$ from the conic bundle? The key thing about this order is that it should be "defined everywhere" (i.e. not just on some open subset like the associated Brauer group element $A$ above). If anyone could help me with this I would be much obliged. –  Daniel Loughran Jan 14 '13 at 19:04
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