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Let $k$ be an imperfect field of char $p>0$ and $x \in \mathbb{P}^n_k$ be closed point of projective space.

In this discussion Qing Liu wrote that

Over an imperfect field, a reduced point can not be contained in a smooth hypersurface if its residue field has "too high" inseparability degree.

I'm not sure in which sense this statement it is true. The central point there which confuses me, is that it seems to suggest that the inseparability degree of the residue field of $x$ gives a kind of "measure" how difficult it is to find a smooth hypersurface $V \subset \mathbb{P}^n_k$ containing $x$.

Problems:

  1. Which relevance has here the posed assumption that $x$ should be reduced? All points of $\mathbb{P}^n_k$ are reduced, so I not understand why it should be extra added. Maybe there is some crucial relevance I not see up to now.

  2. How to interpret it? It seems that Qing Liu's statement on high inseparabality degree of the residue field is a necessary, not sufficient statement for the property to be not contained in a smooth hypersurface, since there are of course points $x \in \mathbb{P}^n_k$ with residue field $\kappa(x)$ of arbitrary big inseparability degree beeing contained in a hyperplane. Or maybe it's a statement for general hypersurfaces containing $x$, I don't know, that's just a guess; cp. also with point 3 below)

Here a counterexample: the point $x \in \mathrm{Spec}(k[w,v]) \subset \mathrm{Proj}(k[W,V, U])$ (with $w:=W/U, v:=V/U$ local coords) associated to maximal ideal $\mathfrak{m}_x := (w^{p^n} - t, v) \subset k[w,v]$ where $t \in k$, the residue field is $\kappa(x) = k(t^{\frac{1}{p^n}})$, but $x$ is obviously contained in smooth hyperplane $(V=0)$.

  1. For every point $x \in \mathbb{P}^n_k$ Bertini's theorem would give us at least a regular hypersurface containing $x$, but it is known that over imperfect field not every regular scheme is smooth. But that's a binary condition: imperfect base field implies that regular not equivalent to smooth.

What does it have to do with how big the inseparability degree of the residue field is, and not just that it is not separable? If I'm not missing the point it seems that Qing Liu want to say that the obstruction for a regular hypersurface containing the point $x$ to be even smooth sits in the inseparability degree of the residue field of $x$. But is it true (at least as a statement for general hypersurfaces, since point 2 gives counterexple) and how to see it?

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    $\begingroup$ 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 $n\geq 1+r$. $\endgroup$
    – Jason Starr
    Commented Apr 11, 2023 at 14:23
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    $\begingroup$ 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 $\mathrm df = \mathrm dg = 0$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 11, 2023 at 14:30
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    $\begingroup$ @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? $\endgroup$
    – JackYo
    Commented Apr 11, 2023 at 15:22
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    $\begingroup$ 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 $\endgroup$
    – JackYo
    Commented Apr 11, 2023 at 15:28
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    $\begingroup$ typo: smoothness criterion applied to left term... $\endgroup$
    – JackYo
    Commented Apr 11, 2023 at 15:36

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