Timeline for Generic behavior of the degree of a projective variety
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 5 at 19:06 | comment | added | user527391 | (Also, is the leading term of the Hilbert polynomial equal to the degree as defined in the book (maximal finite intersection cardinality)? I know it's equal to the degree, but with which definition?) | |
| May 5 at 6:41 | comment | added | user527391 | This would be way beyond the book, but OK. Could you write down a complete argument? | |
| May 4 at 15:14 | comment | added | R. van Dobben de Bruyn | @Aphelli ah, I didn't see that. I think you're right that this should be a better strategy. This will prove that whenever the intersection is zero-dimensional, its length will be the leading term of the Hilbert polynomial of $V$. Then use the generic smoothness to conclude that on a nonempty open, the zero-dimensional scheme you get is reduced, so its length equals its number of points. | |
| May 4 at 13:39 | comment | added | Aphelli | @R.vanDobbendeBruyn: how about the Hilbert polynomial argument that I sketched in a MSE comment (but never got around to fleshing out)? I’m a bit torn about it, since it seems to work – but it also seems too easy somehow? | |
| May 3 at 21:41 | comment | added | R. van Dobben de Bruyn | Ah, good point. I don't know actually, and it does make the statement sound a little fishy. I wonder if there's some other argument we're missing... | |
| May 3 at 12:55 | history | edited | user527391 | CC BY-SA 4.0 |
deleted 4 characters in body
|
| May 3 at 12:46 | comment | added | user527391 | How do you know that there is no point outside of U with finite fibers and possibly larger fiber cardinality? The claim in the book is really that the generic fiber cardinality is the maximal finite fiber cardinality. | |
| May 3 at 8:27 | comment | added | R. van Dobben de Bruyn | One way to argue is as follows: show that $X$ is a variety (i.e. integral) since the other projection $X \to V$ is a 'Grassmannian bundle' and $V$ is integral. If $\operatorname{char} k = 0$, this implies that $\pi$ is generically smooth, so there exists a dense open $U$ such that $\pi|_{\pi^{-1}(U)}$ is finite étale, so in particular flat with reduced geometric fibres. Then the number of points in the fibre is constant. | |
| S May 3 at 7:07 | review | First questions | |||
| May 3 at 7:10 | |||||
| S May 3 at 7:07 | history | asked | user527391 | CC BY-SA 4.0 |