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Timeline for Surjectivity of frobenius

Current License: CC BY-SA 3.0

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Jul 20, 2013 at 16:41 vote accept Lan
Jul 20, 2013 at 13:58 comment added user36938 Very good. Strictly speaking, this gives $x = {y'}^p + p^{1-1/p}z$, but surjectivity of Frobenius on $\overline{R}/(p^{\varepsilon})$ for some $0 < \varepsilon \le 1$ implies the same for $\overline{R}/(p)$ after finitely many iterations (depending on $\varepsilon$).
Jul 19, 2013 at 19:58 review First posts
Jul 19, 2013 at 20:00
Jul 19, 2013 at 19:41 history answered almostuser CC BY-SA 3.0