Timeline for A kind of Stein factorization for non-proper morphisms
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Apr 30, 2014 at 13:29 | comment | added | Sebastian Petersen | OK, now I get it. Maybe I should require $S\to T'$ dominant instead of surjective? I am thinking for a moment... | |
| Apr 30, 2014 at 12:01 | comment | added | Jason Starr | "Then I can take $T'=S$?" No, you cannot. The induced morphism from $S$ to $T$ is not a finite morphism, and it does not have geometrically connected generic fiber. So you cannot take $T'$ equals $T$, and you also cannot take $T'$ equals $S$. | |
| Apr 30, 2014 at 5:53 | comment | added | Sebastian Petersen | Then I can take $T'=S$? | |
| Apr 29, 2014 at 19:53 | comment | added | Jason Starr | I think you should try some examples on your own. For instance, what happens if $T$ is $\text{Spec}(\mathbb{Z})$ and $S$ is the maximal open subscheme of $\text{Spec}(\mathbb{Z}[x]/\langle x^3 + x + 1 \rangle)$ that is smooth over $T$? | |
| Apr 29, 2014 at 19:03 | history | asked | Sebastian Petersen | CC BY-SA 3.0 |