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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.

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$\def\Spec{\operatorname{Spec}}\def\W{\Omega}$

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 $U$.

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  • $\begingroup$ Thank you very much KReiser! $\endgroup$
    – budwarrior
    Jun 7 at 19:53

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