Timeline for Does the degree of a finite dominant morphism bound the induced degree on subschemes?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 11, 2017 at 3:38 | vote | accept | Eric Canton | ||
| Jun 11, 2017 at 3:38 | vote | accept | Eric Canton | ||
| Jun 11, 2017 at 3:38 | |||||
| Jun 10, 2017 at 0:34 | vote | accept | Eric Canton | ||
| Jun 11, 2017 at 3:38 | |||||
| Jun 9, 2017 at 21:45 | answer | added | Dragon | timeline score: 2 | |
| Jun 9, 2017 at 19:49 | comment | added | Eric Canton | I see. You raise a good point, though, about upper semicontinuity: compactness implies that fiber length admits a maximum, and so using the general bound on degree by fiber length, the degree on any subscheme is bounded by some (potentially huge) multiple of the degree of $f$. | |
| Jun 9, 2017 at 19:47 | comment | added | R. van Dobben de Bruyn | To make examples where $\tilde W$ is not reduced, you can consider maps like $k[x^n, x^{n-1}y, \ldots, y^n] \subseteq k[x,y]$. This is a finite morphism of degree $n$ of normal domains, but the scheme-theoretic fibre above $(0,\ldots,0)$ has length $\binom{n+1}{2}$. | |
| Jun 9, 2017 at 19:39 | answer | added | Mohan | timeline score: 5 | |
| Jun 9, 2017 at 19:29 | comment | added | R. van Dobben de Bruyn | It seems to me that this might be false in general, because (scheme-theoretic) fibre length is upper semi-continuous (so it can be really large on some closed subscheme). But it's pretty hard to make counterexamples, for example because of miracle flatness. | |
| Jun 9, 2017 at 15:11 | history | asked | Eric Canton | CC BY-SA 3.0 |