1
$\begingroup$

Let $X, Y$ be irreducible projective schemes over $\mathbb{C}$, $W \subset X \times Y$ a closed irreducible subscheme. Suppose that the natural projection map $pr_2:W \to Y$ is surjective on the underlying topological spaces. Note that for all $x \in W$, there is a natural map of tangent spaces $\phi_x:T_{W,x} \to T_{Y,pr_2(x)}$. If there is a point $y \in Y$ such that the $\mathbb{C}$-vector space generated by the union $\cup_{x \in pr_2^{-1}(y)} \phi_x(T_{W,x})$ does not coincide with $T_{Y,y}$, does this mean that $Y$ is nonreduced at the point $y$?

$\endgroup$
3
$\begingroup$

No. Take for $X$ and $Y$ smooth projective curves (say), $W\subset X\times Y$ a smooth curve such that $pr_2$ has degree 2. Any branch point $y$ of $pr_2$ satisfies your requirement.

$\endgroup$
5
  • $\begingroup$ I am a bit confused. I thought that a (schematic) branch point i.e., a point counted with multiplicity more than one is non-reduced. $\endgroup$ – Kali Aug 12 '14 at 14:01
  • $\begingroup$ No! The fiber is non-reduced. But as a point in $W$ (or $Y$), it is smooth. Think of $\mathbb{A}^1\rightarrow \mathbb{A}^1$, $t\mapsto t^2$. $\endgroup$ – abx Aug 12 '14 at 14:03
  • $\begingroup$ Am I correct to think that this means the tangent space to the scheme theoretic image is NOT the one generated by the union of the images of the tangent spaces, as above? $\endgroup$ – Kali Aug 12 '14 at 14:14
  • $\begingroup$ Yes, this doesn't hold in general. $\endgroup$ – abx Aug 12 '14 at 14:34
  • $\begingroup$ Thanks once again. Could you please suggest some reference that could help me understand when this does hold true (i.e., the tangent space to the scheme theoretic image is the same as the one generated by the union of the images of the tangent spaces as above). $\endgroup$ – Kali Aug 12 '14 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.