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Given a quadric surface $Q$ over a field $F$ of characteristic $2$, assume it is irreducible and reduced, we say it is ruled, if $Q$ is birational to $C \times \mathbb{P}^1$ for some $C$.

A sufficient condition for such a quadric to be ruled is that there is a rational point (use the projection from this point).

My question is, is this also necessary?

People have studied ruledness of a quadric hypersurface over fields of characteristic other than $2$. Maybe there are also results in characteristic $2$. In that case, I would also be very interested in knowing a reference.

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    $\begingroup$ That is not a necessary condition. Let $C\subset \mathbb{P}^2_F$ be a smooth, plane conice that has no $F$-point, so that also $C\times C$ has no $F$-point. On $C\times C$, the complete linear system of the Cartier divisor class of the diagonal $\Delta(C)$ embeds $C\times C$ into $\mathbb{P}^3$ as a smooth quadric. However, $C\times C$ is birational to $C\times \mathbb{P}^1$. Fix a line $L\subset \mathbb{P}^2$. Send general $(p,q)\in C\times C$ to $(p,r)\in C\times L$, where $r$ is the unique point such that $\text{span}(p,q)$ equals $\text{span}(p,r)$. This is birational. $\endgroup$
    – Jason Starr
    Commented Dec 29, 2015 at 16:21
  • $\begingroup$ @JasonStarr Dear Jason, thanks for your answer. Do you know some results giving necessary or sufficient conditions for a quadric to be ruled? $\endgroup$
    – Yuti Zahin
    Commented Dec 29, 2015 at 19:22
  • $\begingroup$ @JasonStarr, Dear Jason, I still do not understand what you mean by $\text{span}(p, q)$ and $\text{span}(p, r)$. Do you mean the line spanned by the two points? But the points $p$ and $q$ are not points in $\mathbb{P}^3$. $\endgroup$
    – Yuti Zahin
    Commented Dec 29, 2015 at 20:41
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    $\begingroup$ The curve $C$ is in $\mathbb{P}^2$. The line $L$ is in $\mathbb{P}^2$. The span is the line spanned by the two points in $\mathbb{P}^2$. $\endgroup$
    – Jason Starr
    Commented Dec 29, 2015 at 20:56
  • $\begingroup$ @SándorKovács. Is this about misspelling "conic"? $\endgroup$
    – Jason Starr
    Commented Dec 30, 2015 at 8:01

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The condition you state is not a necessary condition. You may find much more about these types of questions in Manin's "Cubic Forms".

First of all, over a finite field every quadric hypersurface has a rational point by Chevalley's theorem (or you can probably reduce this case to Wedderburn's earlier theorem). Thus, assume that the field is infinite. Then, by Bertini's Theorem, for a sufficiently general hyperplane section $C$ of your surface, $C$ is a smooth plane conic.

For the numerical polynomial $P(n) = n+1$, computing Hilbert polynomials with respect to the restriction to $Q$ of $\mathcal{O}_{\mathbb{P}^3}(1)$, the Hilbert scheme $\text{Hilb}^P_{Q/F}$ is a finite, flat scheme over $C$ of degree $2$. Of course $C$ is geometrically simply connected. Thus, either the Hilbert scheme is isomorphic to two disjoint copies of $C$ (the "split" case), or the Hilbert scheme is isomorphic to $C\times_{\text{Spec}\ F}\text{Spec}\ E$ for a degree $2$ separable extension of $F$ (the "non-split case").

In the split case, $Q$ is isomorphic to the product surface $C\times C$ embedded into $\mathbb{P}^3$ by the complete linear system of the Cartier divisor $\Delta(C)\subset C\times C$. If the surface $Q$ has an $F$-point, then that $F$-point projects to an $F$-point of $C$. Thus, for every smooth, geometrically integral curve $C$ over $F$ of arithmetic genus $0$ that has no $F$-point, $Q=C\times C$ is an example of a smooth quadric surface that has no $F$-point. On the other hand, identifying $C$ with either one of the connected components of $\text{Hilb}^P_{Q/F}$, also $Q$ is isomorphic to $C\times \mathbb{P}^1$.

Edit. The OP contacted me. There is a mistake in what I wrote above. It is still correct that for every conic $C$ with no $F$-rational point, the surface $Q=C\times C$ is birational to $C\times \mathbb{P}^1$ yet admits no $F$-rational point. However, for a singular quadric surface $Q$ and a smooth hyperplane section $C$, the Hilbert scheme $\text{Hilb}^P_{Q/F}$ may fail to be an étale double cover of $C$. The Hilbert scheme may be geometrically nonreduced (everywhere). So the answer above is quite incomplete: it does not address the non-split case, and it does not address the nonreduced case.

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