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Let $S$ be a scheme and $X$ a smooth separated faithfully flat over $S$.

An $S$-birational group law on $X$ is an $S$-rational map $$m:X\times_S X\dashrightarrow X, (x,y)\mapsto xy$$ such that

a) the $S$-rational maps $$\Phi:X\times_S X\dashrightarrow X\times_S X, (x,y)\mapsto(x,xy)$$ $$\Psi:X\times_S X\dashrightarrow X\times_S X, (x,y)\mapsto(xy,y)$$ are $S$-birational, and

b) m is associative; i.e., (xy)z=x(yz) whenever both sides are defined.

Now, assume that $S$ is a valuation ring and that the generic and special fibers of $X$ have birational group laws (for instance if they are group schemes).

When and how is it possible to get an $S$-birational group law on $X$?

Context:

Let $K$ be a valuation field with valuation ring $R$. Let $H$ be a (qc, separated, integral but not necessarily of finite type) group scheme over $R$. Assume that the generic fiber $H_K$ is an algebraic group (i.e. of finite type). We may write $H=\varprojlim_i H_i$ for integral, separated, of finite type $H_i$ over $R$. They may not be group schemes over $R$. But $(H_i)_K$ is an algebraic group. Assume that for large enough $i$, the inverse limit is birational on the special fiber, so for large enough $i$, $(H_i)_k$ has a $k$-birational group law.

Putting the smoothness conditions aside, does $H_i$ have an $R$-birational group law for large enough $i$?

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  • $\begingroup$ A birational group law on the generic fiber is an $S$-birational group law. $\endgroup$ – Jason Starr Nov 10 '16 at 10:58
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    $\begingroup$ @JasonStarr: The OP must be reading in the book "Neron Models", and in there the definition of $S$-rational map (and likewise $S$-birational map) requires the crucial condition that the open domain of definition is $S$-dense in the sense of meeting each fiber over $S$ in a dense open subset. So for integral $S$, a birational group law on the generic fiber is a different concept from an $S$-birational group law in general (and extending to the latter is non-obvious). $\endgroup$ – nfdc23 Nov 10 '16 at 14:54
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    $\begingroup$ I assume you mean "discrete valuation ring" rather than "valuation ring". What is the motivating context where you would have unrelated birational group laws on the two fibers and want to splice them into a single $S$-birational group law? Or is this a question of idle curiosity? $\endgroup$ – nfdc23 Nov 10 '16 at 14:57
  • $\begingroup$ In the "Context", did you intend "but not necessarily" rather than "but necessarily"? Is $R$ a general valuation ring, not necessarily a discrete valuation ring? By Corollary 3.4.7 in Part I of "Criteres de platitude et de projectivite" by Raynaud and Gruson, flatness of finite type integral schemes over a valuation ring implies they are locally of finitely presentation, so finitely presented if also quasi-separated. Did you intend "finitely presented" rather than "finite type", or have in mind that Raynaud-Gruson result? And have you looked in Perrin's 1976 paper in Bull. SMF 104? $\endgroup$ – nfdc23 Nov 13 '16 at 12:58
  • $\begingroup$ @nfdc23, I changed to "not necessarily". R is a general valuation ring. Afterwards I meant "of finite type" since the existence of an R-point says that it is faithfully flat but then "of finite type"="finitely presented" by "Finitely generated rings over a valuation ring" (Nagata), no? Also, isn't Perrin's paper for qc group schemes over fields? Is there a way to generalize for valuation rings? $\endgroup$ – user4231 Nov 13 '16 at 13:44
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Thanks to nfdc23 for explaining the definition. There are counterexamples. For instance, begin with $S=\text{Spec}(R)$ for a discrete valuation ring $(R,\mathfrak{m})$. Let $\nu:\text{Spec}(\widetilde{R})\to R$ be an étale double cover with $R$-involution $\iota:\text{Spec}(\widetilde{R}) \to \text{Spec}(\widetilde{R})$. Assume that $\widetilde{R}/\mathfrak{m}\widetilde{R}$ is $R$-isomorphic to $R/\mathfrak{m} \times R/\mathfrak{m},$ and assume that $\widetilde{R}\otimes_R \text{Frac}(R)$ is a field. Let $\overline{X}$ be $\text{Spec}(\widetilde{R}) \sqcup \text{Spec}(R)$, and extend $\nu$ to $\overline{X}$ by the identity morphism on $\text{Spec}(R)$.

There is a unique $S$-group law $\overline{m}$ on $\overline{X}$. It has the property that $\text{Spec}(R)\subset \overline{X}$ is the identity section, that $\iota$ is the restriction to $\text{Spec}(\widetilde{R})$ of the group inverse morphism, and that the composition of $\Delta:\text{Spec}(\widetilde{R}) \to \overline{X}\times_S \overline{X}$ and $\overline{m}$ equals $\iota$. This is a $3$-torsion group scheme.

The closed fiber of $\overline{X}$ is a disjoint union of three copies of $\text{Spec}(R/\mathfrak{m})$, two of which are points of order $3$ in the group of $R/\mathfrak{m}$-rational points. Choose one of these order $3$ points, and define $X$ to be the complement of that closed point in $\overline{X}$. There is a unique group structure on the closed fiber of $X$ compatible with the identity section, and that group structure is $2$-torsion.

There is no $S$-rational group law on $X$. Since $\overline{X}$ is proper over $S$, every $S$-rational transformation from $X$ to $\overline{X}$ extends to all of $\overline{X}$ by the valuative criterion. So every $S$-rational group law on $X$ gives an "honest" $S$-group law on all of $\overline{X}$. But the unique $S$-group law on $\overline{X}$ is not an $S$-rational transformation from $X$ to $X$.

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  • $\begingroup$ Thanks, I added above the context of the question. It may help to focus my question. $\endgroup$ – user4231 Nov 13 '16 at 7:42

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