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Mar 3, 2014 at 12:00 vote accept Heitor
Nov 29, 2013 at 17:37 history edited Heitor CC BY-SA 3.0
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Nov 29, 2013 at 17:37 comment added Heitor Thanks @WillSawin. You also made me notice the terrible mistake I made by mixing notation for $X$ and $Y$ (now $Z$ and $W$). Let me edit and fix that.
Nov 29, 2013 at 17:30 comment added Will Sawin Any nontrivial relation between $X$ and $Y$ would, by substitution, immediately give a nontrivial relation between $z$ and $w$, which does not exist.
Nov 29, 2013 at 17:12 answer added abx timeline score: 1
Nov 29, 2013 at 16:40 history edited Heitor CC BY-SA 3.0
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Nov 29, 2013 at 16:11 comment added Francesco Polizzi $\mathbb{C}[z^2, w] \cong \mathbb{C}[z^2][w]$. It is obvious that $\mathbb{C}[z^2]$ is a polynomial ring, so $\mathbb{C}[z^2, w]$ is a polynomial ring, too. Hence $\textrm{Spec}\,(\mathbb{C}[z^2, w])$ is smooth.
Nov 29, 2013 at 16:02 history asked Heitor CC BY-SA 3.0