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Apr
26 |
comment |
Rationally connected spaces over non-algebraically-closed fields
In fact the $K_0$ group of $X$ is isomorphic to $K_0(\mathbb R)\bigoplus K_0(\mathbb H)$; and I am pretty sure that $X\times X\to X$ is the projective bundle associated to the $2$-dimensional vector bundle having for $K_0$-class the generator of $K_0(\mathbb H)$. In particular, it is not trivial. |
Apr
26 |
comment |
Rationally connected spaces over non-algebraically-closed fields
Wait... Is a trivial S-B scheme over $X$ isomorphic to $\mathbb P^n_X$ or to $\mathbb P(\mathscr E)$ for some vector bundle $\mathscr E$ ? I think that the second statement is true, but not necessarily the first one. |
Apr
26 |
comment |
Rationally connected spaces over non-algebraically-closed fields
They should, because what you wrote is nothing but "any conic with a point is isomorphic to the projective line" in the relative setting. |
Apr
26 |
comment |
Rationally connected spaces over non-algebraically-closed fields
Oh yes of course! Do you think that both maps more or less coincide ? |
Apr
26 |
awarded | Critic |
Apr
26 |
comment |
Rationally connected spaces over non-algebraically-closed fields
Maybe I am wrong; I think that my construction will give a dominant map $\mathbb P^1_{\mathbb R}\times_{\mathbb R}U\to X$ for some dense Zariski-open subset $U$ of $X$, but I am not completely sure that I can extend it to the whole of $\mathbb P^1_{\mathbb R}\times_{\mathbb R}X$ without blowing-up. Nevertheless, it still provides a dominant map, but its source is not proper. (The point is that the identification between $\mathbb P^1$ and the set of lines going through a point of the projective plane is not canonical; it can be described by a uniform formula only on a given affine chart). |
Apr
26 |
awarded | Commentator |
Apr
26 |
comment |
Rationally connected spaces over non-algebraically-closed fields
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 $\mathbb P^1_{\mathbb R}\times_{\mathbb R}X\to X$. |
Apr
23 |
comment |
Reduced scheme and closed points
Yes user89334, you are right. One should rewrite my proof, and replace "sober" with $T_0$, and "irreducible" with "non-empty". At the end, the conclusion should be that if $x$ and $y$ are two points of $G$, then since $\overline{\{x\}}=\overline{\{y\}}=G$, one has $x=y$ by the $T_0$ property. Hence $G$ consists of one point, which is obviously closed. Thank you! |
Apr
22 |
answered | Reduced scheme and closed points |
Apr
22 |
comment |
Non-Archimedean non-standard models for R
And I disagree with ACL: $F(t)$ cannot be real closed. For instance, $t$ will never be a square in $F(t)$. |
Apr
22 |
comment |
Non-Archimedean non-standard models for R
My construction is perhaps more explicit, and I can make it completely explicit (neither choice nor compactness involved): embed $F(t)$ into $F((s))$ with $s=1/t$ , and set $ K=\bigcup_n F((s^{1/n}))$. Equip $K$ with the ordering extending that of $F$ for which $s$ is positive and smaller than every positive rational number; this ordering extends that of $F(t)$, and makes $K$ a real closed field. Now you may take for $S$ the algebraic closure of $F(t)$ inside $K$. |
Apr
22 |
comment |
Reduced scheme and closed points
Thank you ACL! I have just slightly rewritten the proof. |
Apr
22 |
revised |
Reduced scheme and closed points
added 200 characters in body |
Apr
22 |
revised |
Reduced scheme and closed points
added 200 characters in body |
Apr
22 |
awarded | Necromancer |
Apr
21 |
answered | Non-Archimedean non-standard models for R |
Apr
21 |
awarded | Revival |
Apr
21 |
comment |
The target of a regular function in Non-archimedean analytic geometry
No. $P$ is meant to be a polynomial. And my formula defines a semi-norm, i.e., something that takes a polynomial and provides a non-negative real number. |
Apr
21 |
answered | Reduced scheme and closed points |