Timeline for Fibre cardinality of an unramified morphism
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
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| Jan 25, 2012 at 21:10 | vote | accept | Jesko Hüttenhain | ||
| Jan 23, 2012 at 9:59 | comment | added | Damian Rössler | @Jesko : That's right. As Sandor writes (thank you for clarifying this), I mean $(\phi_*{\cal O}_X)_y:=(\phi_*{\cal O}_X)\otimes\kappa(y).$ | |
| Jan 22, 2012 at 18:39 | comment | added | Sándor Kovács | @Jesko: I think Damian meant $(\phi_*\mathscr O_X)_y:=(\phi_*\mathscr O_X)\otimes \kappa(y)$. That makes more sense for the rest of the comment as well. (That's a finite dimensional vector space over $\kappa(y)$, while the localization is not). | |
| Jan 22, 2012 at 12:51 | comment | added | Jesko Hüttenhain | @Damian: I do not see why $X_y=\mathrm{Spec}((\phi_\ast\mathcal{O}_X)_y)$. Assuming $X=\mathrm{Spec}(A)$ and $Y=\mathrm{Spec}(B)$, I would say that $X_y=\mathrm{Spec}(A\otimes_B\kappa(y))$ while $(\phi_\ast\mathcal{O}_X)_y=A\otimes_B B_y$. | |
| Jan 22, 2012 at 3:45 | comment | added | Keerthi Madapusi | Sandor--You're of course correct. The hypothesis is right there in the result I was citing. | |
| Jan 21, 2012 at 16:35 | answer | added | Damian Rössler | timeline score: 6 | |
| Jan 21, 2012 at 0:46 | answer | added | Sándor Kovács | timeline score: 7 | |
| Jan 21, 2012 at 0:22 | comment | added | Sándor Kovács | @Keerthi: Re: "In fact, more is true. If you have a map $f:X\to Y$ of irreducible varieties with the target smooth, and with all fibers of $f$ equi-dimensional of dimension $\dim X−\dim Y$, then $f$ is flat. See EGA IV.6.1.5.". This is actually not true as stated. You need $X$ to be Cohen-Macaulay. Check EGA IV.6.1.5. (It's not just that you need this because it is stated, but the statement is not true otherwise. If $f:X\to Y$ is finite and $Y$ is non-singular, then $f$ is flat iff $X$ is Cohen-Macaulay. | |
| Jan 20, 2012 at 18:04 | comment | added | Damian Rössler | Note that my comments above are strictly based on the hypotheses (no regularity hypothesis is made on $X$ or $Y$). | |
| Jan 20, 2012 at 17:45 | comment | added | Keerthi Madapusi | In fact, more is true. If you have a map $f:X\to Y$ of irreducible varieties with the target smooth, and with all fibers of $f$ equi-dimensional of dimension $\dim X-\dim Y$, then $f$ is flat. See EGA IV.6.1.5. | |
| Jan 20, 2012 at 17:38 | comment | added | Keerthi Madapusi | A finite dominant map between non-singular varieties is automatically flat. This follows from the fact that a finite injective map between regular local rings is flat. | |
| Jan 20, 2012 at 17:18 | comment | added | Damian Rössler | With you current assumptions (finite, unramified, dominant), the morphism $\phi$ is flat if and only if your equality holds. Apply Hartshorne's criterion III, Th. 9.9 to see this (in your case, the Hilbert polynomial of the fibre is $\#\phi^{-1}(P)$). | |
| Jan 20, 2012 at 17:04 | history | edited | Jesko Hüttenhain | CC BY-SA 3.0 |
added 17 characters in body
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| Jan 20, 2012 at 17:00 | comment | added | Damian Rössler | Flatness is automatic if $Y$ is a non-singular curve (Hartshorne, III, Prop. 9.7) and $\phi$ is dominant. This is not true in higher dimensions, though. Note also that a closed immersion is a finite unramified morphism but it is clearly not flat is general. | |
| Jan 20, 2012 at 16:55 | comment | added | Damian Rössler | You may compute $X_y={\rm Spec}\ (\phi_*({\cal O}_X))_y$ and $(\phi_*({\cal O}_X))_y$ is a vector space of dimension $r$ over $\kappa(y)$; it is also an (étale !) algebra over $\kappa(y)$ and hence it must be a direct sum of separable extensions of $\kappa(y)$. So if $\kappa(y)$ is alg. closed (so that it has only the trivial separable extension) then it must be a direct sum of $r$ copies of $\kappa(y)$, the direct sum being viewed as a $\kappa(y)$-algebra. | |
| Jan 20, 2012 at 16:45 | history | edited | Jesko Hüttenhain | CC BY-SA 3.0 |
added important assumption
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| Jan 20, 2012 at 16:45 | comment | added | Jesko Hüttenhain | I definitely want $\phi$ to be finite. I will correct that right away - doesn't it follow from my other assumptions that $\phi$ is flat? @Damian: Could you elaborate more on why $X_y$ is a disjoint union of $r$ copies of $\mathrm{Spec}(k)$? | |
| Jan 20, 2012 at 16:44 | comment | added | Damian Rössler | (my comments were written before I read Laurent Moret-Bailly's remark) | |
| Jan 20, 2012 at 16:42 | comment | added | Damian Rössler | (...continued). If $y$ is the generic point of $Y$, then $X_y$ is a disjoint union of spectra of separable extensions of $\kappa(y)=K(Y)$ and since $Y$ has only one generic point (because it is integral), you can see that $X_y$ consists of the spectrum of one separable extension of $K(Y)$, which again must be of degree $r$. | |
| Jan 20, 2012 at 16:39 | comment | added | Damian Rössler | I think you want to assume that $\phi$ is finite (otherwise an open immersion, which is not the identity, gives a counterexample). Also, I think you should assume that $\phi$ is flat (and thus étale). If you make that hypothesis, then $\phi_*({\cal O}_X)$ is locally free of rank $r$ (by assumption). If $y$ is any point on $Y$ (including the generic one !), then the scheme-theoretic fibre $X_y$ is then a finite $\kappa(y)$-algebra. If $y$ is a closed point then necessarily $X_y$ is a disjoint union of $r$ copies of $\kappa(y)=k=$your alg. closed ground field. | |
| Jan 20, 2012 at 16:27 | comment | added | Laurent Moret-Bailly | The field extension doesn't make sense unless $\varphi$ is dominant. Even so, it clearly fails if $\phi$ is an open immersion and $X\neq Y$. So, what do you asssume exactly? | |
| Jan 20, 2012 at 16:14 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Jan 20, 2012 at 16:06 | history | edited | Charles Staats |
added reference-request tag
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| Jan 20, 2012 at 15:54 | history | asked | Jesko Hüttenhain | CC BY-SA 3.0 |