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Jun 17, 2019 at 21:55 history edited aglearner CC BY-SA 4.0
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Jun 17, 2019 at 20:03 comment added aglearner That is an interesting conjecture
Jun 17, 2019 at 13:52 comment added Piotr Achinger It seems reasonable to conjecture that they are all obtained from the one above by pullback along an endomorphism of P1 totally ramified at 0.
Jun 17, 2019 at 11:20 history edited aglearner CC BY-SA 4.0
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Jun 17, 2019 at 11:07 comment added aglearner Thanks for this comment! Indeed the first half of my question seems to be answered in the post you refer to. There is still a second part... What can be said in general about this type of three-folds? Is there a "simplest" among them?
Jun 17, 2019 at 10:44 comment added Piotr Achinger Seems that your question is almost a duplicate of mathoverflow.net/questions/80989/…
Jun 17, 2019 at 10:10 comment added aglearner Thanks Piotr. Could you point me to a nice reference for this standard degeneration, so that by reading it I would be able to fully understand your comment?
Jun 17, 2019 at 10:08 history edited aglearner CC BY-SA 4.0
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Jun 17, 2019 at 9:50 comment added Piotr Achinger Of course. Cover $\mathbf{P}^1$ by two copies of $\mathbf{A}^1$. Over one of those, construct the standard degeneration of $\mathbf{P}^1\times \mathbf{P}^1$ to $F_2$ (by degenerating the Euler sequence on $\mathbf{P}^1$ to a split sequence and taking the projective bundle). On the other copy, take the trivial family $\mathbf{P}^1\times \mathbf{P}^1 \times \mathbf{A}^1$. Identify the two families on the overlap.
Jun 17, 2019 at 9:15 history asked aglearner CC BY-SA 4.0