Timeline for How does normalization behave on closed subschemes?
Current License: CC BY-SA 3.0
9 events
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Sep 6, 2016 at 3:59 | vote | accept | stupid_question_bot | ||
Sep 5, 2016 at 21:27 | comment | added | Sándor Kovács | I don't think you actually need that, but it is because $Z$ is irreducible, $\mathbb A^1_k$ is regular of dim 1 and $g$ is dominant. | |
Sep 5, 2016 at 21:20 | comment | added | stupid_question_bot | Thanks for your example! How do you deduce that $g : Z\rightarrow\mathbb{A}_k^1$ is flat? | |
Sep 5, 2016 at 20:53 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
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Sep 5, 2016 at 20:40 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
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Sep 5, 2016 at 20:18 | comment | added | Sándor Kovács | I've just realized that I've been thinking about this for $A[t]$. I don't think that changes the outcome, but that certainly explains why you are puzzled by my using the words "nearby fibers". Sorry about that. In any case, it might be worth thinking about examples that come from examples above $A[t]$. | |
Sep 5, 2016 at 20:12 | comment | added | Sándor Kovács | Dear @rtz, I don't think on this site one needs to address others as "Professor". For instance, I can't tell whether you are one, so I can't reciprocate.... :) Anyway, yes, by "nearby fibers" I meant fibers above other primes. I'll try to add an example. | |
Sep 5, 2016 at 19:56 | comment | added | stupid_question_bot | Dear Professor Kovacs, In your second to last paragraph, what do you mean by "nearby fibers"? Are you referring to the preimage of Z above primes other than $(t)$? Also, is there a simple example where normalization separates some components of the preimage above $t = 0$ but leaves others intersecting? | |
Sep 5, 2016 at 19:45 | history | answered | Sándor Kovács | CC BY-SA 3.0 |