# Uniqueness of presentation for semi-abelian varieties

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).

• The maps are unique but modulo automorphisms. – Xarles Jul 16 at 12:33
• 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. – Angelo Jul 16 at 12:47
• @Angelo That's exactly what I was going to explain in an answer... – Xarles Jul 16 at 12:58
• @Angelo Do you mean "the unique maximal torus in $G$"? Just to make sure that I haven't misunderstood your statement – 57Jimmy Jul 16 at 14:07
• Yes, I meant the unique maximal torus in $G$. – 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.