# Hartshorne exercise III.10.2

This problem has been discussed already (Hartshorne ex III.10.2 on smooth morphisms) but I cannot understand the solution and since that questions is 8 years old or so I decided to make a new question rather than ask for clarifications in the comments.

The problem is as follows: We are given a flat and proper morphism $$f:X\to Y$$ of varieties over a field $$k$$ such that there is a point $$y\in Y$$ for which the restriction $$X_y\to k(y)$$ is smooth. Then we should prove that there is a neighborhood $$U$$ of $$y$$ such that $$f^{-1}(U)\to U$$ is smooth. I think I have an argument that shows that $$\Omega_{X/Y,x}\otimes k(x)$$ has dimension $$n:=\dim X-\dim Y$$ for any $$x\in X_y$$. This means that $$\Omega_{X/Y,x}\otimes k(x)$$ has dimension at most $$n$$ in a neighborhood containing $$X_y$$. I would like to show that there is a neighborhood containing $$X_y$$ on which $$\Omega_{X/Y,x}\otimes k(x)$$ is exactly $$n$$ but I haven't had any success.

In the thread linked above they use that the set of points $$x$$ where $$\Omega_{X/Y,x}$$ is free is open. My problem here is I cannot show that $$\Omega_{X/Y,x}$$ is free at points of $$X_y$$. If anyone could point me in the right direction it would be much appreciated.

$$\def\Spec{\operatorname{Spec}}\def\W{\Omega}$$
First, $$f$$ is open by exercise III.9.1 and closed by the definition of proper, so it's image is a clopen subset of $$Y$$. Since $$Y$$ is integral, this means that $$f(X)=Y$$ and therefore the generic point of $$X$$ maps to the generic point of $$Y$$ and we get an extension of fields $$k(Y)\subset k(X)$$. Since transcendence degree adds over towers of extensions, we see that $$k(X)$$ has transcendence degree $$n=\dim X-\dim Y$$ over $$k(Y)$$, and hence $$\W_{X/Y}$$ has rank at least $$\dim X-\dim Y$$ at the generic point of $$X$$ by the compatibility of $$\W_{X/Y}$$ with localization (proposition II.8.2A). Thus by exercise II.5.8(a), we have that the closed set of points $$x\in X$$ where $$\W_{X/Y}$$ is of rank at least $$n$$ is all of $$X$$ because it contains the generic point.
Next, by corollary III.9.6, we have that every irreducible component of $$X_y$$ is of dimension $$\dim X-\dim Y$$ for any $$y$$, so by the definition of a smooth morphism we have that $$\W_{X_y/\{y\}}$$ is of rank $$n$$ for every $$x\in X_y$$. Now I claim that the rank of $$\W_{X/Y}$$ is exactly $$n$$ for every $$x\in X$$ mapping to $$y$$: writing $$\W_{X/Y}\otimes k(x)$$ as the pullback of $$\W_{X/Y}$$ along the map $$\Spec k(x)\to X$$, we can factor this map as $$\Spec k(x)\to X_y\to X$$, and as $$\W_{X_y/\{y\}}$$ is the pullback of $$\W_{X/Y}$$ along $$X_y\to X$$ by proposition II.8.10, we have that the rank of $$\W_{X/Y}$$ at $$x$$ is the same as the rank of $$\W_{X_y/\{y\}}$$ at $$x$$. Therefore by exercise II.5.8(a), the set $$Z$$ of points where $$\W_{X/Y}$$ has rank $$>n$$ is closed and does not contain $$X_y$$. So $$f(Z)$$ is a closed subset of $$Y$$ not containing $$y$$, and letting $$U=Y\setminus f(Z)$$ we have found our $$U$$: condition (1) in the definition of 'smooth of relative dimension $$n$$' holds as flatness is stable under base change, condition (2) holds by our previous discussion involving corollary III.9.6, and condition (3) holds by construction of $$U$$.