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.



necessary conditions for a quadric surface to be ruled (over a field of char 2)
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.


