Are linear reductive algebraic groups closed under extensions? Say we have a ses of algebraic groups
$1 \to A \to B \to C \to 1$
where $A,C$ are linear reductive algebraic groups. 

Does it follow that $B$ is also a linear reductive algebraic group? In other words, are linear reductive algebraic groups extension-closed?


Let $$1 \rightarrow A \rightarrow B \rightarrow C \rightarrow 1$$
be an exact sequence of algebraic groups where $A,C$ are linear algebraic groups.
Claim : $B$ is also a linear algebraic group.
If not, let $A' \subset B$ be the unique normal linear algebraic group contained in $B$ such that $B/A'$ is an Abelian variety(Chevalley's Theorem for Algebraic Groups). It follows that $A \subset A'$ by uniqueness.(otherwise one may consider $A.A'$ the group generated by $A$ and $A'$ and the quotient $B/A.A'$ is a quotient of $B/A'$ and hence an Abelian variety)
Since $A \subset A'$, we get a map $C \rightarrow B/A'$ surjective on $\overline{k}-$points, but the latter is an abelian variety, hence it is a constant map. Thus $B/A' = \lbrace e \rbrace$. That is $B$ is actually linear.
It follows from Corollary 14.11 in Borel's Linear algebraic groups, that $R_uB \twoheadrightarrow R_uC = \lbrace e \rbrace$, if $C$ is reductive. Thus $R_uB \subset A$. Hence $R_uB \subset R_uA = \lbrace e \rbrace$. Thus $R_uB = \lbrace e \rbrace$. We get $B$ is reductive if $A$ and $C$ are.
