Let \(f: X \to Y\) be a proper, flat morphism of varieties over \(k\). Suppose that for some point \(y \in Y\) the fibre \(X_y\) is smooth over \(k(y)\). Show that there is an open neighborhood \(U\) of \(y\) in \(Y\) such that \(f: f^{-1}(U) \to U\) is smooth.
Solution
Let \[ V=\{x\in X:f\text{ is smooth at }x\}. \]
1. Every point of the fibre \(X_y\) lies in \(V\).
Because \(X\) and \(Y\) are varieties, \(f\) is locally of finite presentation. It is flat everywhere by hypothesis.
Fix \(x\in X_y\). The local fibre of \(f\) at \(x\) is the local ring of the \(k(y)\)-scheme \(X_y\) at \(x\). Since \(X_y\) is smooth over \(k(y)\), it is geometrically regular at \(x\). Thus at \(x\) the morphism \(f\) is flat, locally of finite presentation, and has geometrically regular fibre. By the local criterion for smoothness for smooth morphisms, \(f\) is smooth at \(x\).
Therefore \[ X_y\subseteq V. \]
2. The subset \(V\) is open in \(X\).
For a morphism locally of finite presentation, the locus where the morphism is smooth is open [@Har10a, Chapter III, §10]. Hence \(V\subseteq X\) is open. Equivalently, \[ Z=X\setminus V \] is closed.
3. The image \(f(Z)\) is closed in \(Y\) and does not contain \(y\).
The morphism \(f\) is proper, so it is closed. Since \(Z\) is closed in \(X\), \[ f(Z)\subseteq Y \] is closed.
1. Therefore\[ y\notin f(Z). \]
4. The required neighborhood is\[ U=Y\setminus f(Z). \]
By step 3, \(U\) is an open neighborhood of \(y\). Moreover, by definition of \(U\), \[ f^{-1}(U)\cap Z=\varnothing, \] so \[ f^{-1}(U)\subseteq V. \] Thus \(f\) is smooth at every point of \(f^{-1}(U)\), which says precisely that the restricted morphism \[ f:f^{-1}(U)\longrightarrow U \] is smooth.
5. Q.E.D.
Steps <1>1–<1>2 show that the smooth locus contains the whole fibre over \(y\), and steps <1>3–<1>4 use properness to remove the image of its complement from the base.