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Nov 6, 2013 at 13:52 answer added Jason Starr timeline score: 3
Nov 6, 2013 at 6:25 comment added Cantlog @Marguax: anyway my construction is deadly wrong. There are finite morphism different from the normalization.
Nov 6, 2013 at 0:31 comment added Marguax @Cantlog: Perhaps one should assume $X$ is geometrically connected over $K$, to keep things "interesting". :)
Nov 5, 2013 at 22:46 comment added Cantlog 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 $\dim B\ge 3$).
Nov 5, 2013 at 17:05 review First posts
Nov 5, 2013 at 17:30
Nov 5, 2013 at 17:01 answer added flatness timeline score: 0
Nov 5, 2013 at 16:58 comment added Marguax In the projective case you can use closure in a Hilbert scheme as an elementary substitute to appealing to the Raynaud-Gruson theorem.
Nov 5, 2013 at 16:57 comment added Marguax 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.
Nov 5, 2013 at 16:48 history asked Louis CC BY-SA 3.0