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S Dec 25, 2024 at 10:59 history bounty ended user267839
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Dec 25, 2024 at 10:59 vote accept user267839
Dec 24, 2024 at 17:16 answer added SeanC timeline score: 2
Dec 24, 2024 at 12:24 comment added user267839 @SeanC: Yeah, that was surely part of my confusion in order to develop better intuition how to "transfer" appropriately the "picture" from diffgeo to algeo. Could you maybe sketch to idea how to prove that as you wrote if $f:X \to S$ ($S$ conncted), proper smooth and it has one irreducible geometric fibre, then all geometric fibres are irreducible? (That seems to be exactly what I was looking for)
Dec 23, 2024 at 21:40 comment added SeanC Flatness is not the algebro-geometric analogue of being submersive - that would be smoothness. For example, a degeneration of a family of elliptic curves to a nodal curve is flat but does not induce a submersion on $\mathbf{C}$-points. In general, if $f\colon X \to S$ is a smooth and proper morphism of schemes with $S$ connected, then irreducibility of a single geometric fiber of $f$ implies irreducibility of all geometric fibers. In particular, there's no need to work over $\mathbf{C}$, or to assume any properties of $S$ in that case. Is this the kind of answer you're looking for?
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S Dec 17, 2024 at 19:00 history bounty started user267839
S Dec 17, 2024 at 19:00 history notice added user267839 Canonical answer required
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Dec 14, 2024 at 16:39 comment added Jason Starr Sorry, replace "flat" by "proper".
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Dec 14, 2024 at 15:49 comment added user267839 The point which confuses me a bit is that the presented argument via Ehresmann - if I not made a mistake - seemingly even imply that then all fibres are irreducible under given assumpions, that's about what ZCT doesn't say something. Are there "algebraic" versions of this? So something, which imposes even stronger assumtions than ZCT + that general fibre is irreducible, and implies that every fibre is irreducible? (...I'm happy here to work only over $\Bbb C$)
Dec 14, 2024 at 15:44 comment added user267839 @Jason Starr: Yes, that's with flatness assumption is not a supprise, as flatness may be regarded as a reasonable "algebraic" replacement of beeing submersive, eg Hartshorne's AG, III, Prop. 10.4. The crucial point is that it seems that although argueing via Ehresmann as sketched above imposes stronger assumptions to get connectedness, so ZCT gives fiber connectedness under much weaker assumptions, it makes no statement about irredubility of every fibre assuming there exist a dense open where all fibres are irreducible. And indeed there are counterexamples for this with assumptions of ZCT.
Dec 14, 2024 at 15:22 comment added Jason Starr If you assume proper and submersive, then of course you can argue in this manner. The main significance of the Connectedness Theorem is that the morphism need only be flat and surjective with normal target. This gives many consequences in the "not-necessarily-smooth" setting, e.g., the Connectedness Theorem of Enriques-Severi-Zariski, the Bertini Connectedness Theorem, etc.
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Dec 14, 2024 at 14:56 history asked user267839 CC BY-SA 4.0