Timeline for Can pullbacks resolve non-flatness?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 14, 2022 at 1:36 | comment | added | Andrew Stout | Nice answer! I'm sure you meant EGA IV_2 (same theorem number and page number). | |
| Oct 5, 2017 at 16:12 | vote | accept | Lawrence Jack Barrott | ||
| Oct 4, 2017 at 19:55 | answer | added | Jason Starr | timeline score: 2 | |
| Oct 4, 2017 at 19:22 | comment | added | Jason Starr | It is Th'eor`eme 11.8.1, p. 159 of EGA IV_3. For $Y$ reduced and Noetherian, if $X\to Y$ is not flat, then there exists a closed point $y\in Y$ and a DVR $R$ in $\text{Frac}(Y)$ dominating the local ring $\mathcal{O}_{Y,y}$ such that the base change of $X$ by $i:\text{Spec}\ R \to Y$ is not flat over $R$. For every proper, birational morphism $f:Z\to Y$, by the valuative criterion of properness, the morphism $i$ factors through $f$, i.e., $i$ equals $f\circ g$. Since the further pullback by $g$ is not flat, the pullback of $X$ by $f$ is not flat. | |
| Oct 4, 2017 at 17:32 | comment | added | Lawrence Jack Barrott | Is the general result in EGA or somewhere else? I'm looking for the general case, eg $X$ may very well not be proper over $Y$. | |
| Oct 4, 2017 at 13:57 | comment | added | Jason Starr | If $Y$ is reduced, and if $X\to Y$ is not flat, then also $Z\times_Y X\to Z$ is not flat. This is local, but for simplicity, consider the case when $X\to Y$ is projective and $Y$ is connected. Then $X\to Y$ is flat if and only if the Hilbert polynomials of geometric fibers are constant, cf. Theorem 9.9, p. 261 of "Algebraic geometry" by Robin Hartshorne. If two geometric fibers of $X\to Y$ have distinct Hilbert polynomials, then the same holds for $Z\times_Y X \to Z$ for points of $Z$ mapping to the two points of $Y$. | |
| Oct 4, 2017 at 13:18 | history | edited | Lawrence Jack Barrott | CC BY-SA 3.0 |
Added clarification and integrality hypotheses
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| Oct 4, 2017 at 13:15 | comment | added | Lawrence Jack Barrott | Oh yes, I should include some sort of integrality constraints for the picture I had in mind to work. I'll edit the question appropriately. | |
| Oct 4, 2017 at 12:03 | comment | added | Jason Starr | What precisely is your question? Do you just want an example? Consider the case where $Y$ is $\text{Spec}\ k[u,v]/\langle u^2,uv \rangle$, where $X$ is the closed subscheme $\text{Spec}\ k[u,v]/\langle u \rangle$, and where you blow up the ideal $\mathfrak{m} =\langle \overline{u},\overline{v} \rangle$ in $Y$. | |
| Oct 4, 2017 at 10:46 | history | asked | Lawrence Jack Barrott | CC BY-SA 3.0 |