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Mar 13 at 20:16 history edited Sándor Kovács CC BY-SA 4.0
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Apr 13, 2017 at 12:58 history edited CommunityBot
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Oct 16, 2012 at 21:52 comment added aglearner Dear Sandor, thank you for adding this reasoning, this is very helpful. I realised though that I have one more question about the same paragraph in your answer. Do you know an example of a variety $X$ that satisfies $R1$ but not $S_2$ and has a function that vanishes on a set $Z$ of co-dimension $2$ or larger in $X$? I have a problem to imagine such an example (I don't think that such example exists among affine varieties over $\mathbb C$...)
Oct 13, 2012 at 1:38 comment added Sándor Kovács @aglearner: I added a statement to explain that. I believe it tells you that in fact you can define property S2 this way. See also mathoverflow.net/questions/45347/… and mathoverflow.net/questions/45347/…
Oct 13, 2012 at 1:36 history edited Sándor Kovács CC BY-SA 3.0
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Oct 11, 2012 at 21:04 comment added aglearner Dear Sandor, when you say "...this is essentially equivalent..." is this really equivalent? Is it true that if for every regular function $f$ its zero set has codimension $1$ then the variety has property $S_2$? Can property $S_2$ be formulated in such a language, or this is just implied by $S_2$? (maybe I miss-understand what you wrote).
Jun 5, 2011 at 15:07 vote accept Jesus Martinez Garcia
Nov 22, 2010 at 20:14 history edited Sándor Kovács CC BY-SA 2.5
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Nov 20, 2010 at 19:07 history edited Sándor Kovács CC BY-SA 2.5
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Nov 20, 2010 at 8:46 history edited Sándor Kovács CC BY-SA 2.5
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Nov 20, 2010 at 0:01 history edited Sándor Kovács CC BY-SA 2.5
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Nov 19, 2010 at 22:00 comment added Eric Zaslow This was a pleasure to read, Sandor! I love your top-down reasoning. It makes the abstruce accessible. Thanks.
Nov 19, 2010 at 21:38 history edited Sándor Kovács CC BY-SA 2.5
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Nov 19, 2010 at 18:23 history answered Sándor Kovács CC BY-SA 2.5