Is it true that every connected Affine Algebraic Group has a subgroup isomorphic to $\mathbb{G}_a$ or $\mathbb{G}_m$? If so- why?
1 Answers
I think this is true for any affine algebraic group of strictly positive dimension. Let $G$ be an affine algebraic group and let $U$ be the unipotent radical of $G$. If $G=U$, then $G$ is a unipotent group. In this case, we can embed $U\subseteq\operatorname{GL}(V)$ as a group of unipotent matrices. Take some element $u\in U$ with $\log(u)\ne 0$. Then, the morphism \begin{align*} \lambda: \mathbb G_a &\longrightarrow U \\ t &\longmapsto \exp(tn) \end{align*} is an injective group homomorphism, an inverse is given by the restriction of $\log$ to the image of $\lambda$. Hence, $U$ contains a $\mathbb G_a$. A reference for the above statements would be the book by Tauvel & Yu, Lie Algebras and Algebraic Groups, item 22.3.3.
If $G\ne U$, then $G/U$ is reductive and contains a maximal torus, which contains a $\mathbb G_m$. This $\mathbb G_m$ can be lifted to $G$ in this case, which is actually in this answer to a question I once had.
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If you're working over $\bar{k}$, then this is equivalent to "does every affine algebraic group contain an integral smooth subgroup of dimension $1$". Somehow, this seems even harder though! – Alex Youcis Mar 12 '15 at 07:56