Let $T\_1$ and $T\_2$ be algebraic tori over a field of characteristic 0. Let $T$ be an extension of $T\_1$ by $T\_2$, namely
$$
1\longrightarrow T\_1\longrightarrow T\longrightarrow T\_2\longrightarrow 1.
$$ Is $T$ necessarily an algebraic torus? If so, is there any simple proof?



An extension of algebraic torus
@MartinSkilleter has posted an [answer](https://mathoverflow.net/questions/413179/an-extension-of-algebraic-torus#comment1059067_413179) in the comments. I'll summarise here an elementary proof; it is almost the same as in @MartinSkilleter's link [Extensions of tori by tori are tori](https://blog.jpolak.org/?p=1125), just written slightly differently.


Let $k$ be the field of definition. Since $k$ has characteristic $0$, the group scheme $T$ is smooth. (In general, we could observe that $0 \to \operatorname{Lie}(T\_1) \to \operatorname{Lie}(T) \to \operatorname{Lie}(T\_2)$ is exact, so that $\dim \operatorname{Lie}(T)$ is at most $\dim \operatorname{Lie}(T\_1) + \dim \operatorname{Lie}(T\_2) = \dim T$, whence $\dim \operatorname{Lie}(T)$ equals $\dim T$ and so we have equality.) Since $T$ is connected, it suffices to show that all points of $T(\overline k)$ are semisimple. Consider $g \in T(\overline k)$. Then the unipotent part $g\_u$ of $g$ maps to the unipotent part of the image of $g$ in $T\_2(\overline k)$, hence is trivial because $T\_2$ is a torus; so $g\_u$ belongs to $T\_1(\overline k)$, hence is trivial because $T\_1$ is a torus.


