Timeline for Is the degree of a finite morphism stable by base change
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 9, 2011 at 13:13 | vote | accept | Taicho | ||
| Oct 9, 2011 at 13:02 | answer | added | Georges Elencwajg | timeline score: 22 | |
| Oct 9, 2011 at 12:45 | comment | added | Taicho | i don't know...I define the degree to be the degree of the finite field extension $K(Y) \subset K(X)$ if $X$ and $Y$ are integral. I see why one requires the flatness condition now. | |
| Oct 9, 2011 at 12:31 | comment | added | user2035 | How do you define the degree if you assume neither integral schemes nor $f$ flat? | |
| Oct 9, 2011 at 11:37 | comment | added | Taicho | How do you map $X$ to $\mathbf{A}^1$ with degree 1? Won't one point in $\mathbf{A}^1$ have two points lying over it? For a finite morphism of integral schemes which isn't flat consider the normalization of k[x,y]/(xy), where k is a field. | |
| Oct 9, 2011 at 11:21 | comment | added | Allen Knutson | Indeed, you'd better assume integral. Otherwise consider $X = A^1 \coprod pt$ mapping to $A^1$ degree $1$, and $S$ a point mapping to the image of $pt$. Now I just need a finite morphism of integral schemes that isn't flat... | |
| Oct 9, 2011 at 11:03 | history | asked | Taicho | CC BY-SA 3.0 |