Rationally connected spaces over non-algebraically-closed fields

The definition I most often see for what it means for a projective variety $X$ over a field $k$ to be rationally connected is that there exists a variety $M$ and a dominant morphism $f:\mathbb{P}^1\times\mathbb{P}^1\times M \to X\times X$. For instance, this is the definition given in both Kollar's "Rational curves on algebraic varieties" and Debarre's "Higher-dimensional algebraic geometry". Both books claim that, for $k$ algebraically closed, this implies the existence of a rational curve connecting any two general geometric points. However, I don't understand why the field needs to be algebraically closed for this to be true, it seems that a pretty straightforward argument could be made for arbitrary base field.

With this in mind, could anyone provide any examples of rationally connected varieties which don't necessarily possess rational curves between a general pair of points?

• You probably mean that $k$ is algebraically closed (rather than algebraically complete). – Mikhail Borovoi Apr 25 '16 at 22:03
• @Borovoi Yeah, I do. That was a pretty silly mistake for me to have made. Anyway, I fixed it. – user90845 Apr 25 '16 at 22:05
• There are counterexamples over $k=\mathbb{R}$ coming from the fact that $X(\mathbb{R})$ can be disconnected, yet $\mathbb{P}^1(\mathbb{R})$ is connected (for the analytic topology). For instance, inside $\mathbb{P}^1\times \mathbb{P}^2$ with homogeneous coordinates $([s,t],[u,v,w])$, consider the zero scheme of $s^3(u^2+v^2) = t(t^2-s^2)w^2$. The image in $\mathbb{P}^1(\mathbb{R})$ of the real points are the open intervals $-1\leq t/s\leq 0$ and $0\leq s/t \leq 1$. On the other hand, it is not difficult to construct a morphism $u:\mathbb{P}^1\times M \to X$ with $u^{(2)}$ dominant. – Jason Starr Apr 25 '16 at 22:14
• Your definition is copied wrong from both sources as Jason politely tried to point out. What is fun about the "correct" definition is that it does not a priori guarantee the existence of a single nonconstant morphism $\mathbf{P}^1_k \to X$ over $k$; for example $M$ need not have any $k$-rational point. So I do not see why you say "pretty straightforward". (Also, please be careful, if you are working over a general field $k$, then there are at least two intelligent ways to define varieties over $k$ and you need to make sure your references use the same definition...) – darx Apr 25 '16 at 23:21
• I have a counter-example even simpler than Jason's. Let $X$ be the real projective conic without real point ($u^2+v^2+w^2=0$ in homogeneous coordinates). Over every field $K$ over which $X$ gets a rational point, this point provides an isomorphism between $X_K$ and $\mathbb P^1_K$ (by considering the intersection of a line going through this point with $X$, this is the old-fashioned rational parametrization of a conic). This construction then gives rise to a dominant map $\mathbb P^1_{\mathbb R}\times_{\mathbb R}X\to X$. – Antoine Ducros Apr 26 '16 at 7:14