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Jul 13, 2019 at 1:12 history edited Fan Zheng CC BY-SA 4.0
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Feb 7, 2019 at 18:35 vote accept Will Chen
Feb 7, 2019 at 17:54 comment added YCor Just for exposition, the middle argument simplifies as follows: $v(1+at)=0$ for all $a\in K$; assuming by contradiction $v_0\neq 0$ and choosing $a$ with $v_0(a)<-v(t)$, we have $v(at)<0$ while $v(1+at)=v(1)=0$, contradicting the ultrametric axiom on $v$. So $v_0=0$.
Feb 7, 2019 at 16:31 history answered Will Sawin CC BY-SA 4.0