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 $$Q$$?

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.

• Thank you for the answer. In the last part did you mean "this condition is necessary"? Sorry for my ignorance but what is an example of a non algebraically closed field without quadratic extensions? I am mainly interested in $k = \mathbb{C}(t)$ but I guess that $k(\sqrt{t})$ is a non trivial quadratic extension of $k$. Is this correct?
– Arty
Dec 22, 2021 at 11:40
• @Arty Yes, I meant necessary. An example is given by taking $\bicup_{n=1}^{\infty} \mathbb F_{p^{2^n}}$, or the field of numbers constructible with straightedge and compass - more generally, given any field, we can consider the field generated by all quadratic extensions, then the field generated by all quadratic extensions of that, and so on, and take the limit. Yes, over $\mathbb C(t)$ the counterexample I described should work taking $a = t$. Dec 22, 2021 at 13:10
• Thank you. From your answer it seems to me that the situation in higher dimension is similar. I mean if $Q^n\subset\mathbb{P}^{n+1}$ is an $n$-dimensional quadric with $n\geq 3$ and $k$ has a quadratic extension then $Q^n$ could contain no line. Is this correct?
– Arty
Dec 22, 2021 at 18:17
• @Art For higher $n$, having no quadratic extensions is still sufficient to ensure every quadric has a rational line, but may no longer be necessary. The same construction definitely doesn't work. Dec 22, 2021 at 18:32
• Note that for a specific field $k$ and quadric surface $Q$ you can using the description of Will to give way more concrete examples. If the characteristic of $k$ is not $2$ then the condition that the components Will talks about are defined over $k$ is equivalent to the discriminant of $Q$ being a square in $k$. And the condition that both components are isomorphic to $P^1$ is then equivalent to $Q(k) \neq \emptyset$.
– M.D.
Aug 20 at 15:03