Timeline for Example of a nonsmoothable scheme
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 7, 2019 at 21:48 | vote | accept | flitwick | ||
| Jun 7, 2019 at 21:37 | comment | added | Mohan | @flitwick Finite and flat implies all fibers have the same dimension over $k$, so they all have the same dimension as that of $R$. | |
| Jun 7, 2019 at 21:00 | comment | added | flitwick | I think my problem was that I didnt realize that $X\rightarrow S$ is finite - why does this hold? But ok, if this map is finite, then the fibers constist of finitely many points. I have one fiber which is isomorphic to $R$ and if $R$ is smoothable, there is at least one other fiber which is smooth over $k$. You said that this other fiber is finite the (spectrum of a) ring of functions on finitely many points, because it is smooth and finite dimensional, but why is it finite dimensional? | |
| Jun 7, 2019 at 18:45 | comment | added | Mohan | @flitwick If $S$ is the parameter space, then total space $X$ maps (flat and in this case finite) to $S$ and a special fiber is your finite dimensional $R$ you started with and general fiber is the ring of functions on finitely many points. If no such $X,S$ exist, then $R$ can not be smoothened as in Iarrobino's example. | |
| Jun 7, 2019 at 17:58 | comment | added | flitwick | Thank you for your answer! The first three questions are now clear to me, but I am still confused about 4. You are talking about smooth algebras, but in which way does this algebra occur in the deformation diagram, is it the ring of global sections of the total space? | |
| Jun 7, 2019 at 17:34 | history | answered | Mohan | CC BY-SA 4.0 |