An extension of algebraic torus 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?
 A: @MartinSkilleter has posted an answer 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, 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.
