Let $f:X \to S$ be a smooth morphism and $S$ the spectrum of a discrete valuation ring. If the generic fiber of $f$ is rationally (chain) connected then is the special fiber of $f$ also rationally (chain) connected? We know this to be true if $S$ is over a field of characteristic zero, due to Kollar, Miyaoka and Mori. If not true in general, is there any known special cases when this holds true? Any hint/reference will be most welcome.
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1$\begingroup$ For rational chain connected: yes. Essentially what happens is that rational curves could break up into chains under specialisation, and if the special fibre is also in characteristic zero this means it is also rationally connected. In char p it is open whether RCC and RC differ though. Over a more general base one has to worry about countably many components of the Chow scheme, and pathologies do occur, see eg Kollar's (Rational curves) book: IV.1.8.7 and IV.3.5. $\endgroup$– FrankCommented Apr 17, 2018 at 13:54
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$\begingroup$ @Frank Thank you. Is there a reference for the statement on rationally chain connected varieties? $\endgroup$– user45397Commented Apr 17, 2018 at 14:09
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$\begingroup$ Yes IV.3.5 from above! $\endgroup$– FrankCommented Apr 17, 2018 at 14:18
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1$\begingroup$ In the reference you mention $S$ is over a field of characteristic zero. I am more interested when $S$ is not a field of characteristic zero. $\endgroup$– user45397Commented Apr 17, 2018 at 14:34
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1$\begingroup$ IV.3.5.2 is stated (and proven) over arbitrary base $S$ - the book is written carefully so as to distinguish between these cases. In IV.3.5.3 the result is strengthened in the case where $S$ is over characteristic zero $\endgroup$– FrankCommented Apr 17, 2018 at 17:30
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In characteristic $ p>0 $, RCC and SRC (separable rational connectedness) do differ (see Janos Kollar's (Rational curves) book: V.5.19, I know that Janos Kollar refers to this even though I do not have the book.). In Higher Dimensional Varieties and Rational Points pg. 41 and pg. 43, Carolina Araujo and Janos Kollar show that SRC is deformation invariant.
Namely they refer to Kollar IV.3.11 and Kollar, Miyaoka and Mori to show that the existence of a very free curve is an open condition (condition 6) and the existence of a chain of rational curves connecting any two points (condition 1+) is a closed condition. I do not have access to these sources, so you will have to check that the proof that condition 6 is an open condition does not use properties that only hold for varieties over a field of characteristic zero.
Assuming that there are no problems with condition 6 being an open condition, then since $ \operatorname{Spec}(R) $ is connected, if $ \mathcal{X} $ is a scheme over $ \operatorname{Spec}(R) $ and the general fibre of $ \mathcal{X} $ is SRC, then the special fibre is also SRC.
I don't know about whether RC is deformation invariant in positive characteristic, so I cannot answer that part of the question per se.
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$\begingroup$ I am not sure I follow the argument. In characteristic 0, it is important that conditions 1+ and 6 are both equivalent to being RC. Then you get a nonempty clopen subset of the base which must therefore be everything. But in characteristic $p>0$, condition 1+ is not equivalent to SRC. $\endgroup$ Commented May 16, 2023 at 10:09