Take the 2-minute tour ×  As a character of an algebraic group
is defined to be any morphism (of algebraic groups) from this
group to $\mathbb{G}_m$, how to prove:  $\mathbb{G}_a$ has no
nontrivial characters.  If I take the ground field
$K=\mathbb{C}$, and define a map $\phi: \mathbb{G}_a \rightarrow
\mathbb{G}_m$, $a+bi \mapsto $exp$(a+bi)$, why isn't $\phi$ a
character of $\mathbb{G}_a$? (Why isn't it a morphism of
algebraic groups?)  Many thanks.  share|improve this question
What exactly is $\mathbb{G_m}$? –  Thelonius Sep 16 '12 at 2:29
4 Answers 4  up vote 9 down vote accepted  One way to establish
that there is no nontrivial homomorphism of algebraic groups
from $\mathbb{G}_a$ to $\mathbb{G}_m$ is via the Jordan
decomposition: every element $x$ in a linear algebraic group may
uniquely be written as the product of a semisimple element $x_s$
and a unipotent element $x_u$, and this decomposition is
preserved by homomorphisms of algebraic groups. See for instance
Theorem 6 of these notes of Ryan Reich.  N