Timeline for More on Vojta's exceptional set for a more general abc conjecture

8 events

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Dec 27, 2014 at 18:23	comment	added	joro		@FelipeVoloch Indeed the question is answered. If you disagree with limsup, check the argument about t^4 vs t^(3+o(1)) which I believe still stands.
Dec 27, 2014 at 18:00	comment	added	Felipe Voloch		$q < 1 + \epsilon$ outside of $Z_{\epsilon}$ so $q \le 1$ outside of the union of all $Z_{\epsilon}$. The original ABC which I guess is $n=3$ in your notation doesn't have an excepcional set, you only need it for $n>3$. Meanwhile Michael answered your question.
Dec 27, 2014 at 12:25	vote	accept	joro
Dec 27, 2014 at 12:09	answer	added	Michael Stoll		timeline score: 4
Dec 27, 2014 at 12:03	comment	added	joro		@FelipeVoloch from the abc records database: math.leidenuniv.nl/~desmit/abc "The ABC conjecture says that the limsup of the quality when we range over all ABC triples, is 1"
Dec 27, 2014 at 12:01	comment	added	joro		@FelipeVoloch Are you sure? I am taking logs(), which kills all epsilons. If in abc you get limsup bigger than one, this will certainly disprove abc.
Dec 27, 2014 at 11:45	comment	added	Felipe Voloch		Your interpretation that $\limsup q = 1$ outside a Zariski closed subset is incorrect. For every $\epsilon > 0$ there is a Zariski closed subset but it depends on $\epsilon$. As $\epsilon \to 0$ the union of these Zariski closed subsets may well be Zariski dense. I haven't looked at your example to see if in fact that's what you are getting.
Dec 27, 2014 at 11:34	history	asked	joro	CC BY-SA 3.0