Consider a smooth quadric surface $Q\subset\mathbb{P}^3$ over a field $k$. Are there natural hypotheses one can put on $k$ in order to ensure the existence of a line defined over $k$ on $Q$?



Lines on quadric surfaces
Yes, it suffices that $k$ have no nontrivial quadratic extensions.


Since the surface is geometrically $\mathbb P^1 \times \mathbb P^1$, the space of lines is geometrically a union of two copies of $\mathbb P^1$. Arithmetically, the components are defined over a quadratic extension of $k$. If there are no nontrivial quadratic extensions then they are defined over $k$.


Each component is then a form of $\mathbb P^1$. Since the anticanonical bundle has degree $2$, a section of the anticanonical bundle has zero locus a scheme of length two. Because $k$ has no nontrivial extensions of degree $2$, the zero locus consists of either one or two $k$-points, so the space parameterizing lines has a $k$-point and thus there is a rational line.


At least in characteristic not $2$, this condition is necessary, as if $k$ has a quadratic extension then it contains a nonsquare $a$ and the form $x^2 -y^2 + z^2 - a w^2$. Any line on this surface must, over $k(\sqrt{a})$, be a line on $(x+y)(x-y) + (z- \sqrt{a}w)( z+\sqrt{a}w)$, thus of the form $\alpha (x+y) + \beta (z-\sqrt{a}w) = \alpha (x-y) - \beta (z+\sqrt{a}w)=0$ or of the form $\alpha (x+y) + \beta (z+\sqrt{a}w) = \alpha (x-y) - \beta (z-\sqrt{a} w) =0$, but the Galois group exchanges these two types of lines so this is impossible.


