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Feb 4, 2012 at 15:48 vote accept gio
Feb 4, 2012 at 6:09 answer added Sándor Kovács timeline score: 2
Feb 4, 2012 at 0:48 comment added Charles Staats (Incidentally, under your other hypotheses, the dimension hypothesis is equivalent to the statement that $\mathscr{O}_{X,x}$ is flat over the local ring at $\pi(x)$.)
Feb 4, 2012 at 0:46 comment added Charles Staats There may be an obvious argument that the dimension hypothesis always holds or an obvious counterexample, but I'm not seeing either at the moment.
Feb 4, 2012 at 0:44 comment added Charles Staats I'm fairly sure the answer is yes IF $\operatorname{dim} X = \operatorname{dim} \overline{\pi_L(X)} + \operatorname{dim} \overline{\pi_L^{-1}(\pi_L(x))}$. Idea: take a regular sequence that locally generates the maximal ideal $\mathfrak{m}$ of $\pi(x)$, and pull it back it to $\mathscr{O}_{X,x}$. By standard properties of Cohen-Macaulay rings, the pullback sequence must still be exact; it also generates $\mathfrak{m}\mathscr{O}_{X,x}$. Smoothness of the fiber then allows us to extend the regular sequence to a regular sequence generating the maximal ideal of $\mathscr{O}_{X,x}$.
Feb 3, 2012 at 19:22 history edited gio CC BY-SA 3.0
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Feb 3, 2012 at 18:55 comment added Charles Staats Will: smooth schemes over a perfect field (and probably much more generally, but I want to speak carefully) are always reduced, since a regular local ring is always a UFD and, in particular, a domain.
Feb 3, 2012 at 18:47 comment added Will Sawin Is $\pi_L^{-1}$ taking the fiber? Because in that case, the projection of the nodal cubic $y^2=x^2(x+1)$ or the cuspoidal curve $y^2=x^3$ onto the $x$ axis is just the whole line, smooth, and the fiber at the origin is just the double point, presumably smooth.
Feb 3, 2012 at 17:27 history edited agt
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Feb 3, 2012 at 17:22 history asked gio CC BY-SA 3.0