Timeline for Obstruction for points to be contained in smooth hypersurfaces in tterms of inseparability degree of residue field
Current License: CC BY-SA 4.0
7 events
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| Apr 11, 2023 at 15:36 | comment | added | JackYo | typo: smoothness criterion applied to left term... | |
| Apr 11, 2023 at 15:28 | comment | added | JackYo | And so in order to archieve such $x$ we are looking for residue field $\kappa(x)$ such that the purely inseparable part of $\kappa(x)/k$ has much more generators than $n-1$, since for purely inseparable extension $F/E$ the $F$-dimension of $\Omega^1_{F/E}$ equals the minimal number of generators of $F$ as $E$-algebra | |
| Apr 11, 2023 at 15:22 | comment | added | JackYo | @JasonStarr: and this condition follows immediately from the analysis of the exact sequence $\Omega^1_{\mathcal{O}_{X,x}/k} \otimes \kappa(x) \to \Omega^1_{\kappa(x)/k} \to \Omega^1_{\kappa(x)/ \mathcal{O}_{X,x}} \to 0$ induced by $k \to \mathcal{O}_{X,x} \to \kappa(x)$ + application of criterion of smoothness to right term, right? | |
| Apr 11, 2023 at 14:30 | comment | added | R. van Dobben de Bruyn | So for a concrete example, look at points like $P = [s^{1/p}:t^{1/p}:1]$ in $\mathbf P^2_{[x:y:z]}$ over $k = \mathbf F_p(s,t)$. It cannot be contained in a smooth curve since $\Omega_{\kappa(P)/k}$ is 2-dimensional: the point $z$ is cut out by the polynomials $f = x^p-s$ and $g = y^p-s$ in the affine chart $D(z) = \mathbf A^2_{(x,y)}$, so the map $\Omega_{\mathbf A^2/k}|_P \to \Omega_{\kappa(P)/k}$ is an isomorphism as $\mathrm df = \mathrm dg = 0$. | |
| Apr 11, 2023 at 14:23 | comment | added | Jason Starr | The statement needs to be clarified (as with many, many such statements in Qing Liu's book). For a finite field extension $L/k$, if the $L$-vector space dimension of $\Omega_{L/k}$ equals $r$, then $r$ is the smallest dimension of a smooth $k$-scheme $V$ that contains a closed point with residue field $L$ (as a $k$-extension). In particular, if a smooth hypersurface $V$ in projective space $\mathbb{P}^n$ contains a closed point with residue field $L$, then $n\geq 1+r$. | |
| Apr 11, 2023 at 14:11 | history | edited | JackYo | CC BY-SA 4.0 |
added 41 characters in body
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| Apr 11, 2023 at 14:05 | history | asked | JackYo | CC BY-SA 4.0 |