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Oct 19, 2016 at 18:54 answer added Friedrich Knop timeline score: 3
Oct 19, 2016 at 5:48 history edited Francesco Polizzi CC BY-SA 3.0
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Oct 18, 2016 at 15:11 review Close votes
Oct 18, 2016 at 16:59
Oct 18, 2016 at 13:05 comment added Francesco Polizzi I have posted my comment above as an answer.
Oct 18, 2016 at 13:04 vote accept Miele
Oct 18, 2016 at 13:02 answer added Francesco Polizzi timeline score: 5
Oct 18, 2016 at 12:20 comment added abx In this example $\mathbb{A}^2/G$ is the quadratic cone in $\mathbb{A}^3$ defined by $v^2-uw=0$ (just put $u=x^2$, $v=xy$, $w=y^2$). Thus $Y/G$ is this quadratic cone minus its vertex.
Oct 18, 2016 at 12:04 history edited Miele CC BY-SA 3.0
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Oct 18, 2016 at 12:04 comment added Miele @FrancescoPolizzi Perfect. I should have realized that myself. Could I ask you to post it as an answer? By the way, just out of curiosity, what is $Y/G$ explicitly? It is a smooth quasi-affine variety with Picard group $\mathbb Z/2\mathbb Z$, as far as I can see. Is there a more explicit way of describing $Y/G$ (as an explicit open in some affine variety)?
Oct 18, 2016 at 8:11 comment added Francesco Polizzi What about $\mathbb{Z}/2 \mathbb{Z}$ acting on $\mathbb{A}^2$ by $$(x, \, y) \mapsto (-x, -y)?$$
Oct 18, 2016 at 8:00 review First posts
Oct 18, 2016 at 8:01
Oct 18, 2016 at 7:59 history asked Miele CC BY-SA 3.0