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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.
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.
Now we can realize $\mathbb{G}_a$ as the subgroup of matrices $\{ \left[ \begin{array}{cc} 1 & a \\ 0 & 1 \end{array} \right] , \ a \in K \}$,
which shows that every element of $\mathbb{G}_a$ is unipotent. On the other hand, every element of $\mathbb{G}_m$ is semisimple. Since an element which is both unipotent and semisimple is the identity, the only homomorphism from $\mathbb{G}_a$ to $\mathbb{G}_m$ is the trivial one.
As for your example with the exponential map: that's not an algebraic function! Remember that in dealing with homomorphisms of algebraic groups, we restrict to maps which are locally given by polynomials.
You might be interested in this answer of mine from MO. Over a base ring $R$ which contains $\mathbb{Q}$, I show that the hom's from $\mathbb{G}_a$ to $\mathbb{G}_m$ are in canonical bijection with the nilpotent elements of $R$. In particular, if $R$ is itself a field, then the only nilpotent is $0$, giving the trivial hom.
Since these are affine algebraic groups, there is an equivalent ring-theoretic statement. Namely, a character over a commutative ring $R$ is an $R$-Hopf algebra map from $R[x,x^{-1}]$ to $R[y]$, where the comultiplication is $x \mapsto x \otimes x$ on the left and $y \mapsto y \otimes 1 + 1 \otimes y$ on the right.
To be an $R$-algebra map, $x$ has to go to an invertible element of $R[y]$. We will write it as $f(y) = \sum_{i=0}^n r_i y^i$, where $r_0 \in R^\times$. Taking tensor square is a functor on algebras, so we get a map $$R[x,x^{-1}] \otimes_R R[x,x^{-1}] \to R[y] \otimes_R R[y] = R[y,z]$$ that in particular yields the assignment: $$x \otimes x \mapsto f(y) f(z) = \sum_{j=0}^n \sum_{k=0}^n r_j r_k y^j z^k.$$ To have a Hopf algebra map, it is necessary and sufficient that this be equal to the image of $f(y)$ under the comultiplication map, which is $$\sum_{i=0}^n \sum_{j=0}^i \binom{i}{j} r_i y^j z^{i-j}.$$ Comparing the coefficients of $y^j z^k$ yields the system of equations $r_jr_k = \binom{j+k}{j} r_{j+k}$, where $r_m = 0$ for $m > n$.
We find that $r_0 = 1$ (since $r_0$ is an invertible idempotent), and all $r_i$ for $i \geq 1$ are nilpotent. Since the complex numbers have no nonzero nilpotents, all characters defined over $\mathbb{C}$ are trivial.
A little more work shows that there is a functor from schemes to sets that takes the spectrum of a ring $R$ to the character group of the additive group over $R$. This functor is formally represented by the multiplicative formal group $\widehat{\mathbb{G}_m}$, whose reduced subscheme is $\operatorname{Spec} \mathbb{Z}$. This is an example of Cartier duality where one has to allow formal schemes to take duals of commutative affine groups.
Regarding the exponential function, it does not yield a map from $R[x,x^{-1}]$ to $R[y]$, but in characteristic zero, there is a formal exponential map to the completion $R[[y]]$, which is the coordinate ring of the formal additive group $\widehat{\mathbb{G}_a}$. This is given by sending $x$ to $\exp(y) = \sum_{i = 0}^\infty \frac{y^i}{i!}$, and it is straightforward to check that the assignment $r_i = 1/i!$ satisfies the system of equations $r_j r_k = \binom{j+k}{j} r_{j+k}$. In positive characteristic, you can't divide by $i!$, but there is a formal exponential to a formal divided power series algebra, which yields a sort of PD-additive group. In other words, the exponential does yield a homomorphism when restricted to an infinitesimal neighborhood of the identity, but that homomorphism cannot be extended to a homomorphism (in group schemes) on the whole additive group.
If ${\Bbb G}_a$ had a non-trivial character, then there would be a systematic (in fact, functorial) way to produce invertible elements in any ring $R$ (or maybe in any $R_0$-algebra, where $R_0$ is the ring of definition of the character) from its additive structure.
This would be quite strange........
