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Let $n\ge 1$ be an integer and let us work over the field of complex numbers. Let $\mathcal{R}_n$ denote the set of rational conic bundles $\pi\colon X\to \mathbb{P}^n$ (morphisms such that the generic fibre is a geometry irreducible conic and $X$ is rational), up to square equivalence: two conic bundles $\pi_1\colon X_1\to \mathbb{P}^n$ and $\pi_2\colon X_2\to \mathbb{P}^n$ are equivalent if and only if there are birational maps $\psi\colon X_1 \dashrightarrow X_2$ and $\varphi\colon \mathbb{P}^n\dashrightarrow \mathbb{P}^n$ such that $\varphi\circ \pi_1=\pi_2\circ \psi$. This corresponds to have a commutative diagram (where horizontal arrows are birational) $\require{AMScd}$ \begin{CD} X_1 @>\psi>> X_2\\ @V \pi_1 V V @VV \pi_2 V\\ \mathbb{P}^n @>>\varphi> \mathbb{P}^n. \end{CD}

If $n=1$, then $\mathcal{R}_n$ consists of a single point (result of Enriques).

Question: Is the set $\mathcal{R}_n$ uncountable for $n\ge 2$?

For $n=2$, the answer is yes, as we can take the discriminant locus to be any smooth irreducible cubic, and these are pairwise not equivalent under birational maps of $\mathbb{P}^2$ if they are not isomorphic.

For $n>2$, I guess that the answer is again yes, but I do not know any reference for this.

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    $\begingroup$ Naïvely I would think that the birational type of the discriminant is the same for equivalent conic bundles, but your question seems to imply that this is not the case. Could you explain? $\endgroup$ – abx Jul 20 '18 at 7:27
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    $\begingroup$ Do you require the birational maps to be morphisms $X_1 \to X_2$, or just rational maps $X_1 \dashrightarrow X_2$? (In the former case, you should probably take the equivalence relation generated by these, because the relation you define is not symmetric.) $\endgroup$ – R. van Dobben de Bruyn Jul 20 '18 at 11:08
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    $\begingroup$ @abx yes, the birational type of the discriminant is the same for equivalent conic bundles (except a priori if you have divisors of $\mathbb{P}^n$ that could be contracted by a birational map, but if you take some which are not uniruled you do not have this problem). The difficulty here is then to find divisors complicated enough to have moduli but simple enough so that the total space is rational. For conic bundles over surfaces, the variety is rational if the discriminant is of degree $\le 3$, so one can take smooth cubics. Already for conic bundles over threefolds I do not know. $\endgroup$ – Jérémy Blanc Jul 21 '18 at 7:04
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    $\begingroup$ @R.vanDobbendeBruyn The birational maps are just rational maps. I tried with "dasharrow" but it was not accepted by MO, so I put plain maps, which could have been confusing. I added then yes "rightdasharrow", which now works (and added a commutative diagram but here again I do not know how to put dashed arrows). The relation defined is symmetric. $\endgroup$ – Jérémy Blanc Jul 21 '18 at 7:10
  • $\begingroup$ Then what about taking these inequivalent 3-dimensional conic bundles over $\mathbb{P^2}$, say $X_{C}$ with discriminant a plane cubic $C$, and considering $X_C\times \mathbb{P}^n$? It is a rational conic bundle over $\mathbb{P}^2\times \mathbb{P}^n$ with discriminant $C\times \mathbb{P}^n$, and you can apply the same argument. Then use a birational map $\mathbb{P}^2\times \mathbb{P}^n--\rightarrow \mathbb{P}^{n+2}$ if you insist on the base being some $\mathbb{P}^m$. $\endgroup$ – abx Jul 21 '18 at 13:45

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