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A theorem in Manin's book on cubic forms leads to the conclusion that the least degree of a rational mapping from $\mathbb{P}^2$ to general cubic surfaces(e.g. Picard rank=1) over $\mathbb{Q}$ is 6.

There are two known ways to construct such a mapping: find out two plane rational curves over $\mathbb{Q}$, join each pair of rational points on the two curves by line; or find out two plane rational curves conjugate over some quadratic extension $K$ of $\mathbb{Q}$, join each pair of conjugate points over $K$ on the two curves by line.

Question: Is there any essentially different way to construct degree six mapping over $\mathbb{Q}$?

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I believe you get a six-to-one dominant, rational transformation from $\mathbb{P}^2$ to $X$ as soon as you have a single rational point $p$ on $X$ (except possibly if the point lies on a line or conic). Let $C'$ be the intersection of $X$ with the tangent $2$-plane $\Lambda_p$ to $X$ at $p$. The normalization $C$ of $C'$ is a genus $0$ curve that admits a degree $3$ zero-cycle, namely the inverse image of a hyperplane section of $C'$. Thus $C$ is a rational curve.

Consider the $\mathbb{P}^2$-subbundle $\Lambda \subset C \times \mathbb{P}^3$ of tangent $2$-planes $\Lambda_q$ to $X$ at points $q$ of $C$. The intersection of $\Lambda$ with $C\times X$ is a non-normal surface $S'$ fibered over $C$. The normalization $S$ is a conic bundle over $C$ that admits a degree $3$ zero-cycle, namely the inverse image of a hyperplane section. Thus $S$ is a rational conic bundle over the function field of $C$. In particular, $S$ is rational.

By my computations, the degree of the projection morphism from $S$ to $X$ equals $6$.

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    $\begingroup$ True even if $p$ lies on a line or conic; see Kollar, Unirationality of Cubic Hypersurfaces, 2002. The argument is, of course, harder. $\endgroup$ – David E Speyer Feb 1 at 16:09

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