Classification of the algebraic affine smooth group schemes of dimension 1

Question

How to prove that all algebraic affine smooth group schemes of dimension 1 are the additive group and the multiplicative one ?

Answer 1

I think the following might be of interest to the OP

Theorem: Let $k$ be a perfect field and let $G$ be a finite type connected geometrically reduced group scheme over $k$. Then, $G$ is either an elliptic curve, $G$ is $\mathbb{G}_{a,k}$, or $G$ is a one-dimensional torus (i.e. $G_{\overline{k}}\cong \mathbb{G}_{m,\overline{k}}$). Moreover, one can explicitly parameterize all of these families.

For a proof of this one can see blost post [1] (DISCLAIMER: THIS IS MY BLOG--THIS IS ALSO A WELL-KNOWN RESULT WHICH IS NOT MINE).

Let me give the main ingredients for the case when one assumes that $G$ is affine and that $k$ is algebraically closed. One wants to show that $G$ is either $\mathbb{G}_{m,k}$ or $\mathbb{G}_{a,k}$.

Step 1: Note that if one lets $C$ denote the unique smooth projective compactification of $G$ then one has an injection

$$G(k)\hookrightarrow \mathrm{Aut}(C,Z)$$

where $Z=\{p_1,\ldots,p_m\}$ denotes the closed subscheme $C-G$ (given the reduced structure) and $\mathrm{Aut}(C,Z)$ denotes the group of automorphisms of $C$ as a $k$-scheme which preserves the subscheme $Z$.

Step 2: But, since $Z$ is finite and $G(k)$ is infinite it's easy to see that this implies that the group $\mathrm{Aut}(C,p_1,\ldots,p_m)$, the subgroup of $\mathrm{Aut}(G,Z)$ which fixes each $p_i$ pointwise, is also infinite.

Step 3: Now, since curves of genus at least $2$ have only finitely many automorphisms (e.g. see [2] for a nice proof by Dan Litt) we easily deduce that $g(C)\leqslant 1$.

Step 4: But, if $g(C)=1$ then for any choice of $p_i$ in $Z$ we see that $\mathrm{Aut}(C,p_i)$ is infinite. But, $\mathrm{Aut}(C,p_i)$ is just the group of automorphisms of elliptic curve $(C,p_i)$ which is finite (e.g. see [3, Theorem 10.1]). Thus, $g(C)=0$.

Step 5: Finally, since $\mathrm{Aut}(C,p_1,\ldots,p_m)$ is trivial for $m\geqslant 3$ we see that $m\in\{1,2\}$. This implies that $C=\mathbb{A}^1_k$ or $C=\mathbb{A}^1_k-\{0\}$ as schemes.

Step 6: The final step is to verify that the only group structures on the schemes $\mathbb{A}^1_k$ and $\mathbb{A}^1_k-\{0\}$, up to isomorphism are $\mathbb{G}_{a,k}$ and $\mathbb{G}_{m,k}$ respectively.

References:

[1] Youcis, Alex. Classifying one dimensional groups (II). 2019. https://ayoucis.wordpress.com/2019/11/19/classifying-one-dimensional-groups-ii/

[2] Litt, Daniel. Curves have finite automorphism groups. http://math.columbia.edu/~dlitt/briefnotes/notes/curve-automorphisms.pdf

[3] Silverman, J.H., 2009. The arithmetic of elliptic curves (Vol. 106). Springer Science & Business Media.