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.


  • 2
    $\begingroup$ 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. $\endgroup$ Apr 21, 2021 at 18:56
  • $\begingroup$ @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. $\endgroup$ Apr 21, 2021 at 19:22
  • 1
    $\begingroup$ "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 $\geq 2$. $\endgroup$ Apr 21, 2021 at 19:30
  • $\begingroup$ @R.vanDobbendeBruyn Thanks for that clarification! I will update the question to ask about Mochizuki's proof technique $\endgroup$ Apr 21, 2021 at 19:33

1 Answer 1


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. $\square$

  • $\begingroup$ Thanks for the wonderful answer! I was having some trouble parsing what is going on in the second to last paragraph but I think I've got it now. $\endgroup$ Apr 21, 2021 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.