Timeline for Advances and difficulties in effective version of Thue-Roth-Siegel Theorem
Current License: CC BY-SA 2.5
8 events
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| Nov 23, 2018 at 4:40 | comment | added | user19475 | Is there a version of abc implying the uniform Mordell conjecture? | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 4, 2011 at 1:12 | comment | added | Joe Silverman | @Felipe: Thanks for supplying the reference where Bombieri and Gubler say explicitly that one needs effective $ABC$ to get effective Roth. For those who don't have [BG], here's the exact quote from page 497: "Remark 14.4.20. The proof we have given actually shows that an effective version of the $K$-rational $abc$-conjecture implies effective versions of Roth's and Faltings's theorems." | |
| Jul 3, 2011 at 11:40 | comment | added | Felipe Voloch | @Joe: And theorem 12.2.9 of op.cit. | |
| Jul 3, 2011 at 11:35 | comment | added | Felipe Voloch | @Joe: Yes, it all depends on what is meant by abc conjecture. See remarks 14.4.18 and 14.4.20 in the book of Bombieri and Gubler. | |
| Jul 3, 2011 at 2:42 | comment | added | Joe Silverman | @Felipe: Do you mean that an effective ABC conjecture implies an effective version of Roth's theorem? If one only knows, say, that $\max(|a|,|b|,|c|) < K*N(abc)^2$, for some constant $K$, but no bound for $K$ is effectively known, can you deduce an effective version of Roth's theorem? One can certainly conceive of a Diophantine approximation style proof of the ABC conjecture that would lead to an ineffective constant $K$. | |
| Mar 19, 2011 at 4:44 | comment | added | Junkie | The techniques are often some sort of Padé approximation, or can be put in a framework of that. The work of the Chudnovskys had a linear diff equation in the background. This gives some cases that "work" I think, but I am no expert. springerlink.com/content/j02t25105r35g171 | |
| Mar 19, 2011 at 0:59 | history | answered | Felipe Voloch | CC BY-SA 2.5 |