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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