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Oct 9, 2012 at 4:08 vote accept Henry Yuen
Oct 3, 2012 at 15:03 vote accept Henry Yuen
Oct 8, 2012 at 18:00
Oct 3, 2012 at 7:28 comment added Qing Liu Without non-degeneracy condition, the answer is no. Consider $R=\mathbb C[t]$, $Q(X)=X^2-t$ (in one variable), and $I=tR$. There is no lifting mod $I^2$ of the zero solution mod $I$. However, there is a weaker condition (non-degeneracy when tensoring with the total ring of fractions of $R$) which insures that there is a constant $c$ such that for any $d>0$, any solution mod $I^d$ is congruent mod $I^{d-c}$ to a solution in $R$ (see Renée Elkik's paper in Ann. Sci. ENS in the 70's).
Oct 3, 2012 at 5:43 answer added Will Sawin timeline score: 6
Oct 3, 2012 at 3:57 history edited Henry Yuen CC BY-SA 3.0
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Oct 3, 2012 at 3:42 history edited Henry Yuen CC BY-SA 3.0
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Oct 3, 2012 at 3:27 history asked Henry Yuen CC BY-SA 3.0