Timeline for If $\Omega_{X/Y}$ is locally free of rank $\mathrm{dim}\left(X\right)-\mathrm{dim}\left(Y\right)$, is $X\rightarrow Y$ smooth?
Current License: CC BY-SA 2.5
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 3, 2009 at 1:53 | comment | added | Anton Geraschenko | Ok, it turns out there's a problem with the calculation d(a/b)=(bda - adb)/(b^2)=0. That works for showing that the relative differentials of the fraction field over the ring are zero, but the coefficients of the right hand side of that calculation may not be in the normalization. As Charley pointed out to me, the normalization of the cusp, Spec(k[x,y]/(y^2-x^3)) is Spec(k[y/x]), but you can't write d(y/x)=(xdy-ydx)/x^2 because x/x^2 and y/x^2 are not in the normalization. | |
| Oct 1, 2009 at 3:27 | vote | accept | Anton Geraschenko | ||
| Oct 3, 2009 at 1:53 | |||||
| Oct 1, 2009 at 1:16 | history | answered | Ishaidc | CC BY-SA 2.5 |