# Extending rational maps to semi-abelian varieties

Setup. Let $$k$$ be an algebraically closed field of characteristic zero, and let $$G/k$$ be a semi-abelian variety i.e., $$G$$ is a commutative algebraic group which is an extension of an abelian variety $$A/k$$ by a torus $$T/k$$, so it admits the following presentation: $$0 \to T \to G \to A \to 0$$ Let $$Z/k$$ be a smooth, integral variety and let $$U\subset Z$$ be a dense open such that $$\text{codim}(Z\setminus U) \geq 2$$.

Question. Is it true that any morphism of $$k$$-schemes $$U\to G$$ uniquely extends to a morphism $$Z\to G$$?

This result is stated in Lemma A.2 of Mochizuki's Topics in absolute anabelian geometry I: generalities and the proof goes as follows. First, he deduces the result when $$G$$ is proper i.e., $$G$$ is isomorphic to $$A$$. In this case the result is well-known and can be proved in a few different ways; one avenue of proof uses that $$A$$ does not contain any rational curves. Next, he says that we may reduce to the case where $$G$$ is isomorphic to a torus. Again, this result is well-known and follows from an application of Serre's normality criterion.

I understand the proofs of both cases, but I am struggling to see how to combine these to get the result for a general semi-abelian variety $$G/k$$. More precisely, I do not understand how one may reduce to the case where $$G$$ is a torus.

Edit. The answer to the question is yes, the result is true and actually holds in a more general setting by a theorem of Weil (see Bosch–Lütkebohmert–Raynaud's Néron Models, Theorem 4.4.1.). I would still like to understand how Mochizuki proposes to deduce the result for a general semi-abelian variety from only knowing this in the cases of an abelian variety and a torus.

Thanks!

• This is actually true for arbitrary group schemes by a result of Weil. See for example Bosch–Lütkebohmert–Raynaud's Néron Models, Theorem 4.4.1. Apr 21, 2021 at 18:56
• @R.vanDobbendeBruyn Thank you for the comment! I was aware of this result but this requries that the rational map from $Z \dashrightarrow G$ to be defined in codimension $\leq 1$ right? Perhaps I am misunderstanding what defined in codimension $\leq 1$ means. In any event, I would still like to try and understand how one deduces the result for the semi-abelian variety from knowing it for the abelian and toric part. Apr 21, 2021 at 19:22
• "Defined in codimension 1" means that it's defined at all generic points of subschemes of codimension $\leq 1$, i.e. the locus where it's undefined has codimension $\geq 2$. Apr 21, 2021 at 19:30
• @R.vanDobbendeBruyn Thanks for that clarification! I will update the question to ask about Mochizuki's proof technique Apr 21, 2021 at 19:33

What you would like to say is that $$G \cong A \times T$$, which is obviously not quite true. But it is true smooth-locally, and that's enough to conclude.
Indeed, note that the quotient map $$G \to A$$ is faithfully flat with smooth fibres, hence smooth [Tag 01V8]. In fact, $$G \times_A G \cong T \times G$$ via $$(g,h) \mapsto (g-h,g)$$, and likewise $$G \times_A G \times_A G \cong T \times T \times G$$ (in two different ways ― see below).
We can uniquely extend the composition $$U \to G \to A$$ to $$Z \to A$$, which turns $$Z$$ into an $$A$$-scheme and gives a commutative diagram $$\begin{array}{ccc}U & \hookrightarrow & Z \\ \downarrow & & \downarrow \\ G & \to & A.\!\end{array}$$ Pulling back along $$G \to A$$ gives the commutative diagram $$\begin{array}{ccc} U \underset A\times G & \hookrightarrow & Z \underset A\times G \\ \downarrow & & \downarrow \\ G \underset A\times G & \to & G.\!\end{array}$$ Recalling that $$G \times_A G \cong T \times G$$ as schemes over $$G$$ (via the second projection), we get a map $$U \times_A G \to T$$ via the first projection. This uniquely extends to a map $$Z \times_A G \to T$$ since $$U \times_A G \hookrightarrow Z \times_A G$$ is an open immersion of smooth $$k$$-schemes (here we use that we are working smooth-locally and not just fppf-locally!). This gives a diagonal arrow in the diagram above: $$\begin{array}{ccc} U \underset A\times G & \hookrightarrow & Z \underset A\times G \\ \downarrow & \swarrow & \downarrow \\ T \times G & \to & G,\!\end{array}$$ where the top triangle commutes since it does so after composing with the first and second projections $$T \times G \to T$$ (this is the extension property) and $$T \times G \to G$$ (this was given), and the bottom triangle commutes because everything is a morphism over $$G$$.
To see that $$Z \times_A G \to G \times_A G$$ descends to $$Z \to G$$, we need to check the cocycle condition that the two pullbacks along $$(-) \times_A G \times_A G \rightrightarrows (-) \times_A G$$ agree; see [Tag 023Q] for details. But they do so above the dense open $$U \subseteq Z$$ and $$G \times_A G \times_A G \cong T \times T \times G$$ as schemes over $$G$$ (via the last projection), so the uniqueness statement for morphisms to $$T \times T$$ (which just follows since everything is separated) shows that they agree everywhere.
So $$Z \times_A G \to G \times_A G$$ descends to $$Z \to G$$, and by construction the restriction to $$U$$ is the map $$U \to G$$ we started with. $$\square$$