Let $k$ be any field and $G$ a semi-abelian variety over $k$, i.e., an algebraic group that fits into an exact sequence

$$ 1 \to T \to G \to A \to 1$$

of algebraic groups, where $T$ is an algebraic torus and $A$ is an abelian variety. I have heard somewhere that, given an algebraic group $G$, if $G$ is semi-abelian then it is so in a unique way, meaning that $T$, $A$ and even the maps in the short exact sequence are uniquely determined by $G$. But I have failed to find a reference for this in full generality (or counterexamples).

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    $\begingroup$ The maps are unique but modulo automorphisms. $\endgroup$ – Xarles Jul 16 at 12:33
  • $\begingroup$ Any map from a torus to an abelian variety is trivial (for example, because an abelian variety can not contain any rational curves). This implies that $T$ is the unique maximal torus in $A$, and proves the statement. $\endgroup$ – Angelo Jul 16 at 12:47
  • $\begingroup$ @Angelo That's exactly what I was going to explain in an answer... $\endgroup$ – Xarles Jul 16 at 12:58
  • $\begingroup$ @Angelo Do you mean "the unique maximal torus in $G$"? Just to make sure that I haven't misunderstood your statement $\endgroup$ – 57Jimmy Jul 16 at 14:07
  • $\begingroup$ Yes, I meant the unique maximal torus in $G$. $\endgroup$ – Angelo Jul 16 at 16:38

First of all, there are no non-trivial homomorphisms from a torus $T$ to an abelian variety $A$ (also true from additive group $\mathbb{G}_a$ to $A$, or other unipotent like Witt groups). This is because there are no non-constant rational maps from $\mathbb{A}^1$ to $A$ (see for example Milne's book on abelian varieties, proposition I.3.9).

Hence, any homomorphism $f$ from a semiabelian variety $$ 0\to T \to G \to A \to 0$$ to another semiabelian variety $$ 0\to T' \to G' \to A' \to 0$$ gives homomorphisms $g:T\to T'$ and $h:A\to A'$. The homomorphism $f$ is an isomorphism if and only if both $h$ and $g$ are.


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