Birational Group Law

Question

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 $i$?

Answer 1

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 $X$.