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.

generatedby these, because the relation you define is not symmetric.) $\endgroup$ – R. van Dobben de Bruyn Jul 20 '18 at 11:08