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Sep 25, 2020 at 21:03 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Aug 26, 2020 at 19:20 history edited YCor
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Aug 26, 2020 at 18:00 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jul 27, 2020 at 17:31 answer added Badam Baplan timeline score: 1
Jul 21, 2020 at 6:19 history edited leo monsaingeon CC BY-SA 4.0
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Jul 21, 2020 at 5:12 comment added Pavel Čoupek This does not exist. The fact that a valuation ring has dimension $1$ is equivalent to the value group being "rank $1$", i.e. archimedean, which in turns translates to the property of the valuation ring: For any two nonzero elements $a, b$, there is $n$ such that $(b^n) \subseteq (a)$. The condition on ideals you impose would in fact mean that the valuation is of "infinite rank", which translates to $R$ having infinite dimension.
Jul 21, 2020 at 5:00 review First posts
Jul 21, 2020 at 6:19
Jul 21, 2020 at 4:53 history asked S. T. Stanly CC BY-SA 4.0