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 $Y$?

## 1 Answer

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.

connected, non-singular variety? $\endgroup$1more comment