Timeline for Fibre cardinality of an unramified morphism
Current License: CC BY-SA 3.0
11 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jan 22, 2012 at 8:45 | comment | added | Sándor Kovács | @Damian Rössler: Thank you. Just for the record: I wasn't fishing for an upvote, but I do appreciate your comment. Unfortunately written media lacks intonation and facial expressions so it is easy to misinterpret. I'm sorry to have misinterpreted your comment. Cheers! | |
| Jan 22, 2012 at 7:26 | comment | added | Damian Rössler | @Sandor Kovacs. Sorry, I didn't mean to compete with your answer. I understand now that you wanted to provide an elementary, scheme-free proof of the result; it wasn't clear to me that that was your intent. I was just puzzled by what seemed to be a convoluted way of proving a statement which follows easily from general, albeit less hands-on results. I voted your answer up. | |
| Jan 21, 2012 at 21:27 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
added 136 characters in body
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| Jan 21, 2012 at 21:24 | comment | added | Sándor Kovács | Finally, your comment has not much to do with my answer. It seems to suggest that you consider this a kind of competition. I don't. | |
| Jan 21, 2012 at 21:24 | comment | added | Sándor Kovács | An important point here is that the original question regards one of those statements that "is obvious" and one might use without proof in an argument. When one wants to prove such a statement it is preferable if one does not use machinery that is more complicated than the question at hand. Of course, you may argue (and perhaps this is what you are saying) that your proof is simpler than mine. Fine. However, I fail to see how that opinion of yours should imply that I should not post my own proof if I like. (cont'd) | |
| Jan 21, 2012 at 21:23 | comment | added | Sándor Kovács | Yes, $\phi_*\mathscr O_X$ is indeed locally free, but to make the further conclusions you would likely have to appeal to more advanced results. (I may be wrong, but I think this is actually irrelevant) In particular, I did not see a proof of the fact that the desired value is equal to the field extension. (I don't doubt that you can prove it. I am just saying I did not see it even mentioned.) (cont'd) | |
| Jan 21, 2012 at 21:23 | comment | added | Sándor Kovács | Dear Damian, first of all, thank you for your comment. It is always good to know that someone else actually reads what one writes. Second, I do not understand why you think that I think that it is necessary to reduce to the case of curves. I do think that such a reduction makes for a very simple proof. I would argue that the advantage of this proof is that it uses very little. The reduction step is nearly trivial, and so is the curve case. (cont'd) | |
| Jan 21, 2012 at 10:04 | comment | added | Damian Rössler | The quantity $\#\phi^{-1}(P)$ is independent of $P$ (for $P$ closed) because it is the rank of $\phi_*({\cal O}_X)$ (see my comments above). Note that $\phi_*({\cal O}_X)$ is flat and coherent and hence locally free, because $Y$ is a noetherian scheme and $\phi$ is flat and finite (and in particular affine). I dont understand why you think that it is necessary to reduce to the case of curves (?) | |
| Jan 21, 2012 at 1:19 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
edited body
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| Jan 21, 2012 at 0:46 | history | answered | Sándor Kovács | CC BY-SA 3.0 |