Let $k$ be a field. How do you prove that the algebraic subgroups of $\mathbb{G}_m=\mathrm{Spec}\,k[x, x^{-1}]$ are $\mathbb{G}_m$ itself and the roots of unity $\{x^n=1\}$ for some $n$?
$\begingroup$
$\endgroup$
3
-
2$\begingroup$ The closed subsets (of any 1-dimensional irreducible variety) are finite except the whole variety. So the proper algebraic subgroups are finite. $\endgroup$– YCorCommented Jul 26, 2021 at 13:21
-
4$\begingroup$ Welcome new contributor. Since the multiplicative group is a connected group of dimension $1$, the only subgroup scheme of dimension $1$ equals the entire group scheme. For a finite subgroup scheme of length $n$, the subgroup scheme is contained in the "raising to the power $n$" morphism of group schemes from the multiplicative group scheme to itself, i.e., the closed subgroup scheme $\mu_n$. However, this subgroup scheme already has length $n$. $\endgroup$– Jason StarrCommented Jul 26, 2021 at 13:21
-
2$\begingroup$ Perhaps you can also do this in a pretty naive algebraic way, without knowing much about geometry or alg. groups. Here is what I have in mind: since $G_m$ is represented by $Spec$ of the Hopf algebra $k[x,x^{-1}]$ with $x$ grouplike, classifying algebraic subgroups of $G_m$ amounts to classifying Hopf algebra quotients of $k[x,x^{-1}]$. You proceed by observing that any Hopf algebra quotient of $k[x,x^{-1}]$ has to be obtained by identifying two (or more) grouplikes, and from here you observe that any nontrivial such Hopf algebra quotient must be of the form $k[x]/(x^n - 1)$ for some $n$. $\endgroup$– user164898Commented Jul 26, 2021 at 18:04
Add a comment
|