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Sep 28, 2012 at 16:58 comment added Jérémy Blanc Thanks for editing, anyway the good point was to describe the isomorphism and action on $\mathbb{P}^{n-1}$. But it is funny that the kernel is quite strange and depends in fact on the roots of unity in $\mathbb{K}^*$.
Sep 28, 2012 at 14:02 history edited rita CC BY-SA 3.0
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Sep 28, 2012 at 14:00 comment added rita You are right, I had been too hasty. Thanks for pointing this out. I'll edit my answer.
Sep 28, 2012 at 12:25 comment added Jérémy Blanc Pay attention that the kernel of the map $G\to \mathrm{PGL}(n,\mathbb{K})$ is not equal to $\mathbb{K}^*\times \mathbb{K}^{N+1}$. In fact is it not abelian: for example, the elements $(x_1:\dots:x_n:y)\mapsto(x_1:\dots:x_n:\alpha y)$ and $(x_1:\dots:x_n:y)\mapsto (x_1:\dots:x_n:y+(x_1)^k)$ do not commute in general. See below.
Sep 24, 2012 at 22:18 history edited rita CC BY-SA 3.0
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Sep 24, 2012 at 15:46 vote accept Stanley Yao Xiao
Sep 24, 2012 at 7:31 history edited rita CC BY-SA 3.0
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Sep 24, 2012 at 7:12 history answered rita CC BY-SA 3.0