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Feb 1, 2016 at 18:14 comment added Tyler Lawson @mathvader Very well, the same type of analysis will be effective. In the case of $\Bbb P(1,4)$ there is only one $k$-point with a nontrivial automorphism group, and it is the group of fourth roots of unity. In the case of $\Bbb P(2,3)$ there are exactly two $k$-points with nontrivial automorphism groups--the second roots of unity and the third roots of unity respectively.
Feb 1, 2016 at 18:05 comment added mathvader Actually I do not want to consider cases like $\mathbb{P}(2,2)$. I have in mind cases like $\mathbb{P}(1,4)$ and $\mathbb{P}(2,3)$
Feb 1, 2016 at 17:23 history answered Tyler Lawson CC BY-SA 3.0