Timeline for 1-Parameter subgroups
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2011 at 22:50 | comment | added | George McNinch | I think the OP means that $\lambda$ is a "$k$-cocharacter" of $G$ -- homomorphism $\mathbf{G}_m \to G$ defined over $k$. Then indeed the statement follows from the Zariski density of the $k$-points of the multiplicative group. | |
| Feb 2, 2011 at 15:55 | comment | added | Bugs Bunny | Having said that, if your $\lambda$ is what you think it is (split+multiplicative), the statement is obviously true as $k^\times$ is Zariski dense in $\overline{k}^\times$. Please, clarify. | |
| Feb 2, 2011 at 15:53 | comment | added | Bugs Bunny | $\lambda (k^\times)$ is a horrible notation as your $\lambda$ is neither split, nor even multiplicative... | |
| Feb 2, 2011 at 11:19 | comment | added | darij grinberg | Read $\times$ for $\otimes$ please. Deformation professionelle of a Hopf algebraist... | |
| Feb 2, 2011 at 11:19 | comment | added | darij grinberg | The above is written from the above affine-groups-are-made-of-points viewpoint. From the scheme-theoretic viewpoint, you will probably have to do something like this: the preimage $\lambda^{-1}\left(Z\left(G\right)\right)$ is Zariski-closed (being the preimage of a Zariski-closed set) but contains the Zariski dense subset $k^{\otimes}$ of $\overline{k}^{\otimes}$, thus must be the whole $\overline{k}^{\otimes}$. I hope the Zariski topology makes sense here. | |
| Feb 2, 2011 at 11:16 | comment | added | darij grinberg | Anyway, nothing changes if it's infinitely many polynomial identities... | |
| Feb 2, 2011 at 11:15 | comment | added | darij grinberg | The field $k$ is infinite (since $\operatorname*{char}k=0$), so $k^{\times}$ is infinite. On the other hand, the property of a $x\in k^{\times}$ to satisfy $\lambda\left(x\right)\in Z\left(G\right)$ is an AND-combination of finitely many polynomial identities (I hope it's really "finitely many"), and we should be able to apply the standard Zariski argument (if a polynomial identity holds for an infinite subset of $k$, then it should also hold for all $k$). | |
| Feb 2, 2011 at 10:13 | history | asked | Ana | CC BY-SA 2.5 |