Skip to main content
9 events
when toggle format what by license comment
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
added 40 characters in body
Sep 5, 2016 at 20:40 history edited Sándor Kovács CC BY-SA 3.0
added 1695 characters in body
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