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Oct 8, 2012 at 11:14 vote accept Nina
Oct 3, 2012 at 12:53 history edited Sándor Kovács CC BY-SA 3.0
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Oct 3, 2012 at 8:40 vote accept Nina
Oct 3, 2012 at 9:03
Oct 3, 2012 at 4:02 comment added Sándor Kovács ps: I edited the answer to make the proof simpler not even needing any assumption about $N$.
Oct 3, 2012 at 4:01 history edited Sándor Kovács CC BY-SA 3.0
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Oct 3, 2012 at 2:36 comment added Sándor Kovács Sasha, I meant that $N$ is defined locally by a single equation which is the definition of a Cartier divisor. $X$ is neirther normal nor irreducible, so the usual concept of Weil divisors don't work. I think this is equivalent to $M\cap N$ being Cartier in $M$ and that is indeed what's needed. It doesn't have to be Cartier in $N$.
Oct 2, 2012 at 17:17 comment added Sasha Sandor, what do you mean by saying that $N$ is a divisor in $X$. Both have the same dimension! Did you mean that the intersection $M \cap N$ is a Cartier divisor in both $M$ and $N$?
Oct 2, 2012 at 16:03 history edited Sándor Kovács CC BY-SA 3.0
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Oct 2, 2012 at 15:06 history answered Sándor Kovács CC BY-SA 3.0