1
$\begingroup$

Let $f:X \to Y$ be a proper surjective morphism of reduced connected noetherian schemes. Assume $Y$ is irreducible. Let $y \in Y$ be a closed point. Denote by $X_y$ the fiber over $y$ to the morphism $f$. Suppose that $X_y$ is a non-singular, irreducible variety and for every closed point $x \in X_y$, the tangent space $T_xX$ have the same dimension. Does this imply that any irreducible component of $X$ intersecting $X_y$ contains the whole of $X_y$? Can we drop the assumption of reducedness of $X$ or $Y$?

$\endgroup$
  • $\begingroup$ What if $X$ has 2 connected components, each of them mapping surjectively to $Y$? $\endgroup$ – abx Mar 19 '15 at 10:07
  • $\begingroup$ @abx: Sorry, forgot to mention $X$ is connected. If $X$ has two connected components with each mapping surjectively to $Y$ then the fiber over $y$ will not be irreducible, as mentioned in the question. $\endgroup$ – Ron Mar 19 '15 at 10:12
  • $\begingroup$ So, are you assuming that $X_y$ is a connected, non-singular variety? $\endgroup$ – Francesco Polizzi Mar 19 '15 at 10:16
  • 1
    $\begingroup$ @Polizzi: I thought by definition, variety is an integral scheme. $\endgroup$ – Ron Mar 19 '15 at 10:19
  • $\begingroup$ Some authors require irreducibility in the definition of variety (e.g, Hartshorne), but some others do not (e.g, Shafarevich), so it is better to specify. $\endgroup$ – Francesco Polizzi Mar 19 '15 at 10:27
2
$\begingroup$

That is certainly not true. Let $Y$ be $\mathbb{A}^2$ with coordinates $(s,t)$. Denote by $[u_0,u_1]$ homogeneous coordinates on $\mathbb{P}^1$. Let $X$ be the hypersurface of $\mathbb{A}^2\times \mathbb{P}^1$ with homogeneous defining equation $stu_1 = 0$. Let $f:X\to Y$ be the restriction to $X$ of the projection to $Y$. Let $y$ be the closed point with associated maximal ideal $\langle s,t \rangle$.

The fiber $X_y$ of $f$ over $y$ is $\mathbb{P}^1$. The Zariski tangent space to $X$ at every point of $X_y$ equals the entire Zariski tangent space to $\mathbb{A}^2\times \mathbb{P}^1$, and this is $3$-dimensional. The irreducible component where $u_1$ equals $0$ intersects $X_y$, but it does not contain $X_y$.

There is a similar example where $Y$ is an irreducible nodal curve and $y$ is the node.

$\endgroup$

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.