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Mar 25, 2023 at 2:52 comment added JackYo @QingLiu: Could you explain why it is true that a reduced point $x$ of a scheme $X$ over imperfect field $k$ can not be contained in a smooth hypersurface if its residue field has too high inseparability degree? If we take $X := \mathrm{Spec}(k[x,y])$ and $x$ with $\mathfrak{m}_x = (x^{p^n} - t, y)$ for some $t \in k$ which is not a $p$-th power, then $\kappa(x) = k(t^{\frac{1}{p^n}})$ can has arbitrary large inseparability degree, but $x \in V(y)$ which is obviously a smooth hypersurface. Or did I misinterpreted your comment?
Feb 22, 2013 at 4:34 comment added roy smith yes we all assumed he meant complete intersection!
Nov 16, 2012 at 7:51 comment added Qing Liu @Jason Starr: Sorry I misunderstood your comment.
Nov 16, 2012 at 3:28 comment added Jason Starr @Qing Liu: The OP explicitly states that he is taking a generic element of the corresponding Hilbert scheme. Thus the zero dimensional scheme will be smooth over the ground field.
Nov 15, 2012 at 22:04 comment added Qing Liu @Jason Starr: probably everybody thinks the question is over an algebraically closed fields, but this is not stated explicitely in the OP. Over an imperfect field, a reduced point can not be contained in a smooth hypersurface if its residue field has too high inseparability degree.
Nov 15, 2012 at 16:41 comment added Naga Venkata @Starr: Is there some results when $Z$ is not zero dimensional?
Nov 15, 2012 at 15:15 comment added Jason Starr @Dustin: The OP does not ask whether $Z$ is a complete intersection. He asks whether $Z$ is an intersection of some (presumably huge) number $n$ of smooth hypersurfaces. The integer $n$ need not equal $m+1$, the codimension. The answer to the question is yes for $Z$ any finite collection of reduced points in $\mathbb{P}^3$.
Nov 15, 2012 at 14:23 history answered Dustin Cartwright CC BY-SA 3.0