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May 13, 2016 at 4:51 comment added R. van Dobben de Bruyn This argument only deals with the case that $(0,0,\ldots,0) \in V(f,g)$, i.e. $f(0) = g(0) = 0$. You need to say a bit more (cf. Mohan's comment), but it looks like it's ultimately ok.
May 12, 2016 at 22:55 history edited ACL CC BY-SA 3.0
Replaced $h$ by $g$ at 4 places.
May 12, 2016 at 19:45 history answered Friedrich Knop CC BY-SA 3.0