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By a variety over a field $k$, I mean a scheme that is separated and of finite type over $k$. I indicate changes of the base ring by subscripts.

Does there exist a smooth and projective variety $V$ over some field $k$, with $V(k)\neq\emptyset$, that is geometrically unirational, i.e., there exists a dominant rational map $$\mathbf{P}^n_{\overline{k}} \dashrightarrow V_{\overline{k}}$$ for some integer $n$, but not unirational, i.e., there does not exist a dominant rational map $$\mathbf{P}^n_k \dashrightarrow V$$ for any integer $n$.

I don't recall ever coming across an example of a $V$ as in the question. It is a classical fact that any such $V$ has to have dimension at least $2$. Furthermore, I think I would have known if any examples existed where the dimension of $V$ is $2$. On the other hand, while the condition $V(k)\neq\emptyset$ is certainly needed to keep the question from having a trivial answer in the positive (e.g., let $V$ be a smooth cubic surface in $\mathbf{P}^3_k$ that does not have a $k$-point), it seems too unnatural to me to be sufficient for a negative answer to the question in general.

NB. I realize that some people (including myself at times) use the term "(uni)rational" for the concept I'm referring to as "geometrically (uni)rational". However, when comparing the two properties, I somehow prefer talking about "geometrically unirational vs. unirational" to "unirational vs. unirational over the ground field".

Some edits made to reflect comments by Jason Starr and ayanta; see below.

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To prevent trivial examples, you should at least add "normal" to your list of hypotheses on $V$, but it is probably best to add the hypothesis "smooth". –  Jason Starr Jun 17 '13 at 13:41
    
(deleted previous comment expressing my lack of understanding) I see what you mean now. Thanks! –  René Jun 17 '13 at 14:58
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For an easier example than smooth cubic surfaces, consider a conic without a rational point. –  ACL Jun 17 '13 at 17:20
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Your "classical fact" is false! For $k_0$ of characteristic $p > 0$, let $k = k_0(t)$ and $V:=\{y^q=x-tx^p\}$ for a $p$-power $q > 2$. This is a smooth irreducible $k$-subgroup of $\mathbf{G}_a^2$ with closure in $\mathbf{P}^2_k$ that is regular (but not $k$-smooth). Thus, $V_{\overline{k}} \simeq \mathbf{G}_a$. But $V$ is not unirational over $k$ since $V(k)$ is finite and hence not Zariski-dense in $k$: if $p > 2$ then $V(k) = \{(0,0)\}$, whereas if $p=2$ then $V(k)=\{(0,0),(1/t,0)\}$ since $q > 2$ (for $q=2$ it is a smooth affine conic with a $k$-point, so rational!). –  user30180 Jun 17 '13 at 17:32
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@Jeremy: I am pretty sure that there are counterexamples coming from equivariant compactifications of $\textbf{PGL}_{r^2}$-torsors whose generalized Severi-Brauer scheme modeled on $\textbf{Grass}(r,r^2)$ has a rational point (these torsors must have index $r$). I will try to write up an example. These examples would have pretty large dimension (I guess $4^2-1 = 15$ is the minimum possible). –  Jason Starr Jun 17 '13 at 19:25

2 Answers 2

Here is a sketch of a counterexample, but I will need to add more details. Let $k$ be an infinite field. Let $G$ be a (smooth, connected) semisimple algebraic group scheme of adjoint type over $k$. Assume that $G$ is "quasi-split", i.e., there exists a closed smooth subgroup scheme $B$ in $G$ that is smooth over $k$ and whose base change to an algebraic closure of $k$ is a maximal, connected, solvable subgroup scheme. Let $\mathcal{T}_B$ be a $B$-torsor over $k$ whose associated $G$-torsor, $\mathcal{T}_G$, is nontrivial. Obviously I need to prove that there exists such a $(G,B,\mathcal{T}_B)$, but let's assume for the moment that it exists.

There is a "wonderful compactification" $\widehat{G}$ defined over $k$ that contains a copy of $G$ as a dense open subscheme and such that the natural action of $G\times G$ on $G$ (by left and right multiplication) extends to all of $\widehat{G}$. The scheme $\widehat{G}$ is smooth and projective over $k$. The minimal $G\times G$-orbit in $\widehat{G}$ is isomorphic to $(G/B)\times (G/B)$. Using the left action on $\mathcal{T}_G$ and the right action on $\widehat{G}$, form an action of $G$ on $\widehat{G}\times \mathcal{T}_G$. Since the action of $G$ on $\mathcal{T}_G$ is free, so is the action on $\widehat{G}\times \mathcal{T}_G$.

The geometric quotient $\widehat{\mathcal{T}_G} = (\widehat{G}\times \mathcal{T}_G)/G$ is a smooth, projective $k$-scheme that contains $\mathcal{T}_G$ as an open subscheme and is geometrically isomorphic to $\widehat{G}$. The stratum $(G/B)\times (G/B)$ becomes $(G/B)\times (\mathcal{T}_G/B)$. The point is, since $\mathcal{T}_G$ is induced from the $B$-torsor $\mathcal{T}_B$, the quotient $\mathcal{T}_G/B$ has a $k$-point, namely the image of $\mathcal{T}_B$. Thus, the "deepest stratum" in $\widehat{\mathcal{T}_G}$ has a $k$-rational point. Yet, since $\mathcal{T}_G$ is nontrivial, there are no $k$-rational points in this dense open subset. Over an infinite field $k$, this is enough to insure that $\widehat{\mathcal{T}_G}$ is not unirational. However, it is geometrically isomorphic to $\widehat{G}$, which is rational.

$\textbf{Edit}$. The "details" above turn out to be impossible in the adjoint case, as Xuhan points out: there is no such triple $(G,B,\mathcal{T}_B)$. In the intermediate case, it may be possible. However, even for perfect fields (where there is no issue with constructing wonderful compactifications via descent), the wonderful compactification of a non-adjoint group tends to be singular (although geometrically normal).

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@Jason: is there a literature reference for the wonderful compactification over a general (infinite) field? All references I have seen assume the field to be algebraically closed. Also, to have a nontrivial $B$-torsor over $k$ we cannot take $G$ to be either simply connected or adjoint, since in such cases ${\rm{H}}^1(k,B)=1$ (as the maximal $k$-tori of $B$ are "induced", and hence have vanishing ${\rm{H}}^1$, in these two extreme cases). So what sort of $G$ do you plan to try? –  user29283 Jun 18 '13 at 15:13
    
@Jason: In char. 0 singularities are fine since a resolution cannot have $k$-points if the original variety doesn't. Also, $G(k)\rightarrow(G/P)(k)$ is surjective for any connected reductive $k$-group $G$ and parabolic $k$-subgroup $P$ over any $k$, so ${\rm{H}}^1(k,P)\rightarrow{\rm{H}}^1(k,G)$ has trivial kernel. Hence, any nontrivial $B$-torsor over $k$ does the job in char. 0. If $\tilde{G}$ is the quasi-split simply connected central cover and $\tilde{T}$ a maximal $k$-torus in a Borel, it suffices to find a $k$-subgroup $\mu\subset Z_{\tilde{G}}$ so ${\rm{H}}^1(k,\tilde{T}/\mu)\ne 1$. –  user29283 Jun 19 '13 at 7:14
    
@xuhan: Regarding singularities, unfortunately the argument above only produces a $k$-point on the singular variety, not on the resolution. It might be possible to use the "stacky" description of the wonderful compactification ala Martens-Thaddeus (who realize the singular variety as the coarse moduli space of a smooth stack) to produce $k$-points that lift to a resolution. –  Jason Starr Jun 19 '13 at 11:57
    
@Jason: Oops, I got myself confused for what is needed with $k$-points, sorry. –  user29283 Jun 19 '13 at 15:12

Rosenlicht found forms of $\mathbb{G}_a$ over a nonperfect field $k$ that have only finitely many points, hence are not unirational over $k$ (but, of course, become rational over the algebraic closure of $k$).

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The question was revised to require $V$ to be projective and smooth, precisely to avoid such examples (noted in the comments to the question by ayanta). –  user29283 Jun 18 '13 at 15:03

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