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Let $L=\mathbb{P}^l\subset\mathbb{P} ^ N _ {\mathbb{C}}$ be a linear space and let $M=\mathbb{P}^{N-l-1}$ be a linear space skew to $L$, i.e. $L\cap M=\emptyset$. Let $X\subseteq\mathbb{P}^N_{\mathbb{C}}$ be a closed irreducible variety not contained in $L$ and let $$ \pi_L:X\dashrightarrow\mathbb{P}^{N-l-1}=M $$ be the linear projection, i.e. the rational map defined on $X\setminus L$ by $$ \pi_L(x)=\langle L,x\rangle\cap M.$$ Let $x\in X\setminus L$ a point.

Is true that if $\overline{\pi_L(X)}$ and $\overline{\pi_L^{-1}(\pi_L(x))}$ are smooth varieties, then $X$ is smooth at $x$?

Thanks.

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  • $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Commented Feb 3, 2012 at 18:47
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    $\begingroup$ 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. $\endgroup$
    – Charles Staats
    Commented Feb 3, 2012 at 18:55
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    $\begingroup$ 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 $\mathscr{O}_{X,x}$. $\endgroup$
    – Charles Staats
    Commented Feb 4, 2012 at 0:44
  • $\begingroup$ 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. $\endgroup$
    – Charles Staats
    Commented Feb 4, 2012 at 0:46
  • $\begingroup$ (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 $\pi(x)$.) $\endgroup$
    – Charles Staats
    Commented Feb 4, 2012 at 0:48

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The necessary condition is indeed the one given by Charles.

This is true if $$\dim X=\dim\overline{\pi_L(X)}+\dim\overline{\pi_L^{-1}(\pi_L(x))}\tag{$\star$}$$

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Assume that $(\star)$ holds. The target is smooth, so in particular $\pi_L(x)$ is a complete intersection. This implies that then (by the condition $(\star$)), $\overline{\pi_L^{-1}(\pi_L(x))}$ is also a complete intersection (at least near $x$), but if it is smooth, then it follows that then $X$ has to be smooth near $x$. (This is an iteration of the idea, that if a Cartier divisor is smooth, then so is the ambient space and essentially the same as what Charles suggested.)

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Here is an example that without assuming $(\star)$ the statement is not true. Let $X\subseteq \mathbb P^3_{u:v:w:t}$ defined by $uv=w^2$ and let $x=[0:0:0:1]$. Let $L=\{[1:0:0:0]\}$ and $M=(t=0)\subset \mathbb P^3$. Now $(\pi_L)|_X$ is dominant onto $M$, so $\overline{\pi_L(X)}=M\simeq \mathbb P^2_{v:w:t}$, in particular smooth. Furthermore, $\pi_L(x)=[0:0:1]$ and hence $\overline{\pi_L^{-1}(\pi_L(x))}=\{[\alpha:0:0:\beta]|[\alpha,\beta]\in \mathbb P^1\}\simeq \mathbb P^1$, also smooth. However, $X$ is obviously not smooth at $x$.

Remark: One can also see where the above argument breaks down for this example. Since $(\star)$ is not assumed, it does not follows that $\overline{\pi_L^{-1}(\pi_L(x))}$ is a complete intersection (and it isn't).

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  • $\begingroup$ Sándor, do you mean $M = (u = 0)$? $\endgroup$
    – Charles Staats
    Commented Feb 4, 2012 at 23:40

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