Timeline for Non existence of cyclic infinite linear algebraic groups
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 31, 2014 at 10:05 | comment | added | David Stewart | Over fields of characteristic $p$ it isn't true that the all connected abelian groups are products of copies of $G_a$ and $G_m$. You also have Witt groups. See Chapter VII, \S2 of Serre's book Algebraic Groups and Class Fields. | |
| Mar 30, 2014 at 6:50 | comment | added | Jérémy Blanc | Yes, I agree with Jason Starr, I still do not understand the details. | |
| Mar 29, 2014 at 11:51 | comment | added | Jason Starr | @Aakumadula: It sounds to me like you do indeed have a proof, yet the rest of us do not understand it the way that you explained it. Could you explain the proof carefully in the special case that $G_K$ is isomorphic to $\mathbb{G}_{m,K}$? | |
| Mar 29, 2014 at 11:36 | comment | added | Venkataramana | @Starr: another thing. The norm map is not like a valuation map. It rarely specializes, at the level of $L^*$ into $k^*$, as a map of $T'(k)$ onto $\mathbb Z$ | |
| Mar 29, 2014 at 11:02 | comment | added | Venkataramana | @Starr; No, that is not what I am doing. There is a finite kernel map from $T'$ into the anisotropic torus $T$ and a surjection from the anisotropic torus to $R^1({\mathbb G}_m=T'$; this implies that the group of norm one elements (if we assume $G(k)=\mathbb Z$) in $l^*$ is also $\mathbb Z$ up to finite kernel and cokernel; that is impossible. | |
| Mar 29, 2014 at 9:41 | comment | added | Jason Starr | @Aakumadula: I guess you are trying to say that the norm map will be a homomorphism from $G(K)$ to $G(k)$, and you are trying to rule this out. However, the group $K^*$ can, indeed, admit homomorphisms to $\mathbb{Z}$, e.g., discrete valuations of $K$ give such homomorphisms. | |
| Mar 29, 2014 at 9:38 | comment | added | Venkataramana | @Cornulier, Any such torus maps onto a torus f the form $T'=R^1_{l/k}({\mathbb G}_m$, and contains such an anisotropic torus $T'$. We need only prove it when $T'=T$, and this is really like the case $k^*$ (with a little more work. | |
| Mar 29, 2014 at 7:56 | comment | added | YCor | @Aakumadula: could you explain why $T(k)$ can't be infinite cyclic when $T$ is an anisotropic torus? Your argument does not seem to include this case. | |
| Mar 29, 2014 at 7:47 | history | edited | Venkataramana | CC BY-SA 3.0 |
added 603 characters in body
|
| Mar 29, 2014 at 7:11 | comment | added | Jérémy Blanc | Thanks for having added details. Do you use something like "there is no form of $(\mathbb{G}_a)^n$"? Which assertions on the field do you need for this? What happens for $(\mathbb{G}_a)^n\times (\mathbb{G}_m)^l$ ? | |
| Mar 29, 2014 at 1:30 | comment | added | Venkataramana | @Jason,thank you. All I am saying is that over $K$, the group is a product of mult and add groups (not over the smaller field $k$). The group of $k$ rational points, in case add factors are involved, cannot be $\mathbb Z$; in case only mult factors are involved, we will have to use that such a group, after multiplying by a suitable product of $R_{l/k}({\mathbb G }_m$, where $R$ iw thw Weil restriction of scalars. | |
| Mar 28, 2014 at 15:55 | comment | added | Jason Starr | @Aakumadula: Explicitly the issue being raised by Jeremy at the end of his first comment is that not every Abelian linear algebraic group over $k$ is of the form an extension of copies of $\mathbb{G}_a$ and copies of $\mathbb{G}_m$. | |
| Mar 28, 2014 at 14:10 | history | edited | Venkataramana | CC BY-SA 3.0 |
edited body
|
| Mar 28, 2014 at 13:58 | history | edited | Venkataramana | CC BY-SA 3.0 |
gave some more detailsof the proof
|
| Mar 28, 2014 at 13:35 | comment | added | Jason Starr | @JérémyBlanc: Just to restate: first replace $G$ by the centralizer $Z_G(g)$, and then replace $Z_G(g)$ by its center $Z(Z_G(g))$, which is an Abelian algebraic group that contains $g$. | |
| Mar 28, 2014 at 13:19 | comment | added | Jérémy Blanc | I still do not see how one can deduce that $K$ or $K^{*}$ has to be cyclic from the fact that $G$ is abelian. Sorry if this is easy. | |
| Mar 28, 2014 at 13:17 | comment | added | Jérémy Blanc | In fact after thinking a bit, I understand now why it is abelian; it follows from the fact that $G$ was replaced with the closure of $G(k)$. The centralizer of an element is closed, so the centralizer of $G(k)$ contains $G$. Then, the intersection of all centralizers of all elements of $G$ is again closed and contains $G(k)$, so $G$ is abelian. | |
| Mar 28, 2014 at 13:03 | comment | added | Jason Starr | @Aurel: "how about replacing $G$ with the centralizer of ... $g$" That seems like a good idea, but the centralizer need not be commutative, e.g., for $g$ in the center of $G$, the centralizer of $g$ is all of $G$. | |
| Mar 28, 2014 at 12:53 | comment | added | Aurel | For the first part, how about replacing $G$ with the centralizer of a generator $g$ of $G(k)$? | |
| Mar 28, 2014 at 11:42 | comment | added | Jérémy Blanc | Thanks for the answer, but I do not get all details. Why is $G$ abelian if $G(k)$ is? Why "the assumptions mean that $K$ or $K^*$ must be $\mathbb{Z}$" ? A priori, I expected that we can have some forms of tori for instance. How do you deal with these? | |
| Mar 28, 2014 at 11:34 | history | answered | Venkataramana | CC BY-SA 3.0 |