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Is it easy to give an example of a function field $K$ and a smooth proper variety $X$ over $K$ that does not extend to a flat scheme over $B$, where $B$ is a smooth proper variety with function field $K$?

The transcendence degree of $K$ has to be at least two for this to exist.

Note that I want $B$ to be proper. You can always extend $X$ to a flat $U$-scheme, where $U$ is some open subscheme of $B$ (maybe up to an etale cover?).

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    $\begingroup$ You want the extension to be $B$-proper. If $B$ is any reduced and irreducible noetherian scheme and $X$ is a proper scheme over the function field $K$ then $X$ extends to a proper $B$-scheme $\overline{X}$ by the Nagata compactification theorem. This is flat over some dense open $U \subset B$. Raynaud--Gruson "flattening by blow-up" (Theorem 5.2.2 in their Inventiones paper) provides a blow-up $B' \rightarrow B$ away from $U$ such that the strict transform of $\overline{X}$ relative to the blow-up is $B'$-flat. So at the expense of blowing up $B$, a proper flat extension always exists. $\endgroup$ – Marguax Nov 5 '13 at 16:57
  • $\begingroup$ In the projective case you can use closure in a Hilbert scheme as an elementary substitute to appealing to the Raynaud-Gruson theorem. $\endgroup$ – Marguax Nov 5 '13 at 16:58
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    $\begingroup$ If $B$ is fixed: take $X$ be the spectrum of some finite separable extension $L/K$. The only possible proper extension of $X$ over $B$ is the normalization of $B$ in $L$. Now chose $L/K$ in such a way that this normalization is not flat (this requires that $\dim B\ge 3$). $\endgroup$ – Cantlog Nov 5 '13 at 22:46
  • $\begingroup$ @Cantlog: Perhaps one should assume $X$ is geometrically connected over $K$, to keep things "interesting". :) $\endgroup$ – Marguax Nov 6 '13 at 0:31
  • $\begingroup$ @Marguax: anyway my construction is deadly wrong. There are finite morphism different from the normalization. $\endgroup$ – Cantlog Nov 6 '13 at 6:25
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Cantlog gives a nice example, but perhaps the OP wants an example where the geometric generic fiber is irreducible: it depends on your interpretation of "variety". Let $B$ be $\mathbb{A}^3$ with coordinates $(x,y,z)$. Let $U$ be $\mathbb{A}^3\setminus\{0\}$. Let $\mathcal{E}$ be the cokernel of the following map of locally free $\mathcal{O}_B$-modules, $$ \mathcal{O}_B \xrightarrow{(x,y,z)} \mathcal{O}_B^{\oplus 3}.$$ The point is that for every integer $d$, $\textbf{Sym}^d_{\mathcal{O}_B}(\mathcal{E})$ is coherent and reflexive, yet not locally free at $0$.

Now take $\overline{X}$ to be $\textbf{Proj}_B(\textbf{Sym}^\bullet_{\mathcal{O}_B}(\mathcal{E}))$ with its natural projection $\overline{\pi}:\overline{X}\to B$. Define $X$ to be the inverse image of $U$. This is flat over $U$, in fact just a $\mathbb{P}^1$-bundle. However, I claim that there is no flat extension of $X$ over $B$.

It is "well known" that every flat, proper morphism with $1$-dimensional fibers is projective, at least étale locally over the base. You can find this explicitly in some writings of Brian Conrad (I think also Max Lieblich, probably many other sources as well). So if $X/U$ extended to a flat family over the base, then, after passing to the Henselization of $\mathcal{O}_{B,0}$, the family would be projective. The relative Picard of $X/U$ is just the constant fppf sheaf $\mathbb{Z}$ with generator the Serre twisting sheaf $\mathcal{O}(1)$ naturally associated to the Proj construction. Thus, on the flat model, the very ample invertible sheaf restricts to $\mathcal{O}(d)$ over the punctured Henselization for some positive integer $d$.

Because the morphism is flat and projective, for a sufficiently high twist of the very ample invertible sheaf, i.e., after replacing $d$ by a sufficiently positive multiple, there are no higher direct images and the pushforward to the base is locally free. But, over the punctured Henselization, the pushforward is just the pullback of $\textbf{Sym}^d_{\mathcal{O}_B}(\mathcal{E})$. Thus, there exists an extension of this sheaf to a locally free sheaf over the entire Henselization. But this is absurd: already $\textbf{Sym}^d_{\mathcal{O}_B}(\mathcal{E})$ is one reflexive extension over the entire Henselization, and reflexive extensions are unique up to unique isomorphism (using the S2 property). Since $\textbf{Sym}^d_{\mathcal{O}_B}(\mathcal{E})$ is not locally free at $0$, the base change to the Henselization is also not locally free.

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In the special case where $X$ is a point, this question is essentially asking for resolution of singularities: there is no reason such a $B$ should exist otherwise.

Conversely, if you assume resolutions, then the answer is yes: by Raynaud-Gruson, for any proper variety $B_0$ with function field $K$, there is some blowup $B \to B_0$ such that $X$ extends to a flat family over $B$. By resolutions, one can do a further blowup to assume $B$ is smooth and proper, and this clearly does not affect flatness.

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    $\begingroup$ My interpretation of the question is that $B$ is given at the outset; but perhaps the OP should clarify. $\endgroup$ – Daniel Litt Nov 5 '13 at 17:05
  • $\begingroup$ @Litt: Even with the assumption, I doubt one can prove much without resolutions as one needs to handle the blowup arising in the R-G argument. $\endgroup$ – flatness Nov 5 '13 at 20:43

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