Timeline for Alterations factor as modification + finite map
Current License: CC BY-SA 2.5
6 events
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| Mar 10, 2011 at 13:35 | comment | added | Karl Schwede | HNuer, sorry, I should have responded to this earlier. Suppose that $k$ is an algebraically closed field of characteristic $p$, then consider the extension $k[x] \subseteq k^{1/p}[x^{1/p}] \cong k[x^{1/p}]$. The induced map of specs gives us exactly one point in each fiber. In particular, the points of $\text{Spec} k[x]$ are the ideals $(x - a)$ for $a \in k$ and $(0)$. $(0)$ has exactly one point above it, $(0)$. $(x-a)$ has exactly one point above it, $(x^{1/p} - a^{1/p})$. The argument you are giving works for varieties in characteristic zero. | |
| Mar 1, 2011 at 20:33 | comment | added | HNuer | The proof in that exercise, following Hartshorne's hint, first has you show that the function field of one variety is a finite field extension of the other and the index of the extension is precisely the number of points in the generic fiber, in this case we've shown it's one. I believe that does do the trick, but please tell me where I went wrong. Thanks | |
| Mar 1, 2011 at 15:32 | comment | added | Karl Schwede | Be careful, the exercise you mentioned above does NOT imply that the function fields of $Y$ and $Z$ are equal. Consider for example the Frobenius map (it's the identity on points, but raises sections to their $p$th powers). Basti's proof is right, it uses the construction of the Stein factorization (sheafy-spec). | |
| Mar 1, 2011 at 14:10 | comment | added | HNuer | Yeah, the second after I posted I saw yours and realized that they were the same when you unwind the definition. Thanks for the help anyway, though. :) | |
| Mar 1, 2011 at 13:52 | comment | added | Sebastian Petersen | Ah - it seems we wrote almost in parallel :-) | |
| Mar 1, 2011 at 13:45 | history | answered | HNuer | CC BY-SA 2.5 |