Timeline for How can I show flatness for projective morphisms?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 22, 2016 at 14:33 | vote | accept | 54321user | ||
| Aug 18, 2016 at 16:54 | comment | added | abx | Yes, but what counts is the irreducible components of the total space. If it is reduced and irreducible, it is automatically flat. See Hartshorne, Proposition 9.7. | |
| Aug 18, 2016 at 16:32 | comment | added | 54321user | @abx but the closed fiber over $0$ is reducible. | |
| Aug 18, 2016 at 6:05 | comment | added | abx | That doesn't contradict Sasha's answer, the source of your map is irreducible (and smooth). | |
| Aug 17, 2016 at 22:20 | comment | added | 54321user | Isn't $\textbf{Spec}(\mathbb{C}[t][x,y]/(xy - t) \to \textbf{Spec}(\mathbb{C}[t])$ a flat morphism? If you try and resolve $\mathbb{C}$ as $\mathbb{C}[t] \xrightarrow{\cdot t} \mathbb{C}[t]$, you get vanishing $Tor_1$ | |
| Aug 17, 2016 at 21:59 | history | answered | Sasha | CC BY-SA 3.0 |