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Let $G$ be an affine reductive group defined over a field of characteristic zero $k$, denote by $\bar{k}$ the algebraic closure of $k$, and by $k^{\times}$ the multiplicative group of $k$. Let $Z(G)$ be the center of $G$. Let $\lambda$ be a 1-parameter subgroup of $G$ defined over $k$. Suppose that $\lambda(k^{\times})\subset Z(G)$, is it true that $\lambda(\bar{k}^{\times})\subset Z(G)$ or can you give me a counterexample?

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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 $k$). – darij grinberg Feb 2 2011 at 11:15
Anyway, nothing changes if it's infinitely many polynomial identities... – darij grinberg Feb 2 2011 at 11:16
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 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. – darij grinberg Feb 2 2011 at 11:19
Read $\times$ for $\otimes$ please. Deformation professionelle of a Hopf algebraist... – darij grinberg Feb 2 2011 at 11:19
$\lambda (k^\times)$ is a horrible notation as your $\lambda$ is neither split, nor even multiplicative... – Bugs Bunny Feb 2 2011 at 15:53
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