2
$\begingroup$

Let $\pi:\mathcal{X} \to S$ be a flat family of affine curves contained in $\mathbb{C}^n$ for $n \ge 3$ i.e., $\mathcal{X} \hookrightarrow \mathbb{C}^n_S$ and the inclusion commutes with the natural morphism to $S$. Let $p \in \mathbb{C}^n$ be a general point (not intersecting any $\mathcal{X}_s$) and $\phi:\mathbb{C}^n_S \backslash (p \times S) \to \mathbb{C}^{n-1}_S$ be the trivial deformation of the linear projection from the point $p$. Let $\mathcal{Y} \subset \mathbb{C}^{n-1}_S$ be the image of $\mathcal{X}$ under the morphism $\phi$. Is $\mathcal{Y}$ going to be $S$-flat (under the natural morphism to $S$)?

$\endgroup$
  • $\begingroup$ As I understand it, the linear projection from a point lands into $\mathbb{P}^{n-1}_\mathbb{C}$, not $\mathbb{C}^{n-1}$. $\endgroup$ – Laurent Moret-Bailly Feb 12 at 9:16
  • $\begingroup$ Linear projection does not always preserve flatness. Images of morphisms do not always preserve flatness. The reduced image of a linear projection can have strictly lower degree than the domain. This forces the linear projection to have non-reduced structure if you want to preserve flatness. The non-reduced structure depends on more than the fiber, and this quickly leads to examples where the linear projection is not flat. $\endgroup$ – Jason Starr Feb 12 at 11:33
  • $\begingroup$ @JasonStarr Can we comment about generic flatness i.e., will there exist a non-empty open subset $U$ of $S$ such that $\mathcal{Y}_U$ (the preimage of $U$ under the natural morphism from $\mathcal{Y}$ to $S$) is flat over $U$? $\endgroup$ – Chen Feb 12 at 13:08
  • $\begingroup$ "... will there exist a non-empty open subset $U$ of $S$ such that $\mathcal{Y}_U$ ... is flat over $U$?" I am not certain what you are asking. If $S$ is a reduced Noetherian scheme, that is true by Grothendieck's Generic Flatness Theorem. However, if $S$ is nonreduced, that can fail. $\endgroup$ – Jason Starr Feb 12 at 13:15
  • $\begingroup$ @JasonStarr Thanks. I was interested in the case $S$ is non-reduced. $\endgroup$ – Chen Feb 12 at 13:40

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.