Let $k$ be an algebraically closed field. $\Bbb G_a$ is the Linear algebraic group with the underlying variety being $\Bbb A = k$ with the additive group operation . $\Bbb G_m$ is the Linear algebraic group with the underlying variety being $\Bbb A^*$ with the multiplicative group operation.
The question is :
Does there exist an injective homomorphism $\phi : \Bbb G_a \to \Bbb G_m$ ?
My attempt :
No!
Let $\phi \in Hom(\Bbb G_a,\Bbb G_m)$ .
Then that will induce a $k$-algebra homomorphism from the Coordinate ring of $\Bbb G_m$ to the Coordinate ring of $\Bbb G_a$ i.e. $\phi ^* : \frac{k[X,Y]}{(XY-1)} \to k[T]$ i.e. $\phi ^* : k[X,X^{-1}] \to k[T]$ .
$X$ is a unit in $ k[X,X^{-1}]$ , hence $\phi^*(X)$ must be a unit in $k[T]$ . But ${k[T]}^* = k^*$ thus $\phi^*(X)=c \text{ (say)}\in k^*$ , but that will fail the injectivity of $\phi^*$ since $c \in k[X,X^{-1}]$ and $\phi^*(c)=c$ .
Are my arguments valid? Please point out mistakes if any.
Just to get this off the unanswered list: no. There do not exist injective homomorphisms $\mathbb{G}_a\to\mathbb{G}_m$ as your concrete calculation shows. In fact, there exists no non-trivial homomorphisms $\mathbb{G}_a\to \mathbb{G}_m$ since the image would then be isomorphic to $\mathbb{G}_a$ (since all quotients of $\mathbb{G}_a$ are isomorphic to itself) and thus we would have produced an injective homomorphism $\mathbb{G}_a\to\mathbb{G}_m$.
Two more general comments:
If $f:G\to H$ is an injective homomorphism of algebraic groups, then it's actually a closed embedding (e.g. see [Mil, Corollary 3.3.5]) and so if $G$ and $H$ are of the same dimension (and $H$ is reduced), then it's an isomorphism!
There are no non-trivial homomorphism $\mathbb{G}_a\to\mathbb{G}_m$ over any ring $R$ that is reduced, but if $R$ contains a non-zero element $x$ such that $x^2=0$ then there are non-trivial homomorphisms $\mathbb{G}_a\to\mathbb{G}_m$ (try to find them!).