Let $S$ be a regular Noetherian scheme, and let $U\subset S$ be the complement of a divisor. If $A\to B$ is an isogeny of abelian schemes over $U$, and $A$ extends to a semi-abelian scheme over $S$, does $B$ also extend to a semi-abelian scheme over $S$?


I'm happy to assume that the degree of the isogeny is invertible in $\mathcal{O}\_S$.



Extensions of (semi-)abelian schemes
Under the hypothesis that the degree of the isogeny is invertible in $S$, the answer is yes; I attempted to sketch a proof below under slightly more general assumptions on the base. [On the other hand, if the degree of the isogeny is not invertible in $S$, the answer is no. There are counterexamples over regular rings of dimension $2$. You can look at page 192 of the book by Faltings-Chai for a counterexample in equal characteristic $p$; or to chapter 6 of De Jong-Oort's "On extending families of curves" for an example in mixed characteristic.]


Statement: let $S$ be a normal noetherian integral scheme, $U\subset S$ a non-empty open, $f\colon A\to B$ an isogeny of $U$-abelian schemes, of degree $m$ invertible on $S$. Suppose $A$ extends to a semiabelian scheme $\mathcal A/S$; then $B$ extends to a semiabelian scheme $\mathcal B/S$, and $f$ extends to an isogeny $\mathcal A\to \mathcal B$ (i.e. a fibrewise finite surjective homomorphism).


Proof: the $m$-torsion $\mathcal A[m]$ is a quasi-finite etale group scheme over $S$, and the inclusion $\mathcal A[m]\to \mathcal A$ is a closed immersion, as semiabelian schemes are separated.


Let $K/U$ be the kernel of $f$, a finite etale group scheme. The closed immersion $K\to A[m]$ is etale hence an open immersion. We can therefore write $A[m]$ as a disjoint union of schemes $K\sqcup Z$.


Claim: the schematic closure of $K$ inside $\mathcal A[m]$ (or $\mathcal A$), is etale over $S$. Proof of claim: Write $\mathcal A[m]$ as the disjoint union $Y\_1\sqcup Y\_2\sqcup \ldots Y\_n$ of its connected components. As $S$ is geometrically unibranch, so is $\mathcal A[m]$, hence each $Y\_i$ is also an irreducible component. In particular, the restrictions $Y\_{1 ,U},\ldots,Y\_{n,U}$ are the irreducible (and connected) components of $A[m]$, and by reordering the indexes we can write $K=Y\_{1,U}\sqcup \ldots \sqcup Y\_{k,U}$, $Z=Y\_{k+1,U}\sqcup\ldots\sqcup Y\_{n,U}$. It follows that $\mathcal A[m]$ is the **disjoint** union of $\overline K$ and $\overline Z$, the schematic closures of $K$ and $Z$. This proves the claim.


It is clear that the multiplication map of $\mathcal A$ maps $\overline K\times \overline K$ inside $\overline K$, hence $\overline K$ is a subgroup of $\mathcal A$. The sheaf quotient $\mathcal B=\mathcal A/\overline K$ is a smooth group algebraic space extending $B$; it is separated because quotient of a separated group scheme by a closed subgroup; one argues easily that the fibres are semiabelian varieties. It remains to show that it is a scheme; here you can apply theorem XI 1.13 from Raynaud's "Faisceaux amples sur les schemas en groupes et les espaces homogenes" which shows that $\mathcal B/S$ is quasi-projective. It is here that the hypothesis that $S$ is normal, noetherian, integral is used.


