Timeline for Geometrically unirational varieties that are not unirational
Current License: CC BY-SA 3.0
16 events
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| Jun 20, 2013 at 13:11 | comment | added | R.P. | I for one don't know. The next logical question would be the one where "$V(k) \neq \emptyset$" is replaced by "$V(k)$ is Zariski-dense". | |
| Jun 20, 2013 at 11:57 | comment | added | ACL | A question: If a geometrically unirational variety has a rational point, does one know whether its rational points are already Zariski dense? | |
| Jun 18, 2013 at 0:35 | answer | added | anon | timeline score: 0 | |
| Jun 17, 2013 at 23:49 | answer | added | Jason Starr | timeline score: 1 | |
| Jun 17, 2013 at 22:03 | history | edited | R.P. | CC BY-SA 3.0 |
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| Jun 17, 2013 at 22:01 | comment | added | R.P. | ayanta: you are absolutely right. I guess I should insist on smooth as well as projective. This is the case of which I was thinking anyway; however, while typing the question I didn't see a problem with omitting these hypotheses. Clearly I didn't think hard enough. The "classical fact" I was trying to get away with is that if $V$ is a geometrically unirational curve over any field $k$, it is actually rational; and if $V$ moreover has a $k$-point, then it is isomorphic to $\mathbf{P}^1_k$. Of course, this is only true when I require $V$ to be smooth and projective. | |
| Jun 17, 2013 at 21:51 | comment | added | user30180 | @Jason: My example is smooth. It is only the projective compactification that is merely regular: non-smooth at one point. But $V$ itself is smooth, and the question didn't say that $V$ is meant to be complete. So I guess that is a hypothesis which the OP meant to say but omitted. | |
| Jun 17, 2013 at 19:25 | comment | added | Jason Starr | @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). | |
| Jun 17, 2013 at 18:48 | comment | added | Jérémy Blanc | It seems to be open in dimension 2. In the article arxiv.org/pdf/1304.6798v1.pdf written two months ago we can read "With this assumption, it is not known if geometrically rational surfaces [...] are unirational over their field of definition." | |
| Jun 17, 2013 at 18:07 | comment | added | Jason Starr | @ayanta: Your type of example is why I suggested the OP add "smooth" to the hypotheses on $V$. | |
| Jun 17, 2013 at 17:32 | comment | added | user30180 |
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!).
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| Jun 17, 2013 at 17:20 | comment | added | ACL | For an easier example than smooth cubic surfaces, consider a conic without a rational point. | |
| Jun 17, 2013 at 14:58 | comment | added | R.P. | (deleted previous comment expressing my lack of understanding) I see what you mean now. Thanks! | |
| Jun 17, 2013 at 14:01 | history | edited | R.P. | CC BY-SA 3.0 |
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| Jun 17, 2013 at 13:41 | comment | added | Jason Starr | 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". | |
| Jun 17, 2013 at 12:46 | history | asked | R.P. | CC BY-SA 3.0 |