Timeline for algebraic subgroups of $\mathbb{G}_m$
Current License: CC BY-SA 4.0
7 events
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| Jul 26, 2021 at 18:04 | comment | added | user164898 | 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 $n$. | |
| Jul 26, 2021 at 13:46 | history | edited | Mikhail Borovoi |
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| Jul 26, 2021 at 13:40 | review | Close votes | |||
| Aug 4, 2021 at 21:53 | |||||
| Jul 26, 2021 at 13:21 | review | First posts | |||
| Jul 26, 2021 at 15:41 | |||||
| Jul 26, 2021 at 13:21 | comment | added | Jason Starr | 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 $n$. | |
| Jul 26, 2021 at 13:21 | comment | added | YCor | The closed subsets (of any 1-dimensional irreducible variety) are finite except the whole variety. So the proper algebraic subgroups are finite. | |
| Jul 26, 2021 at 13:17 | history | asked | roots99 | CC BY-SA 4.0 |