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Oct 7, 2020 at 18:37 history edited LSpice CC BY-SA 4.0
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Oct 7, 2020 at 17:35 answer added Tim Campion timeline score: 1
Oct 7, 2020 at 6:55 comment added YCor A further remark: (1) for $R$ commutative, the condition that some f.g. module has a divisible element is equivalent to: some nonzero cyclic module $M$ satisfies $nM=M$ for all $n\ge 1$, and is still equivalent to: some residual field of $R$ ($R/I$ for max ideal $I$) has characteristic zero. (2) for $R$ commutative, the condition that $1$ is not divisible is equivalent to: some residual field of $R$ has finite characteristic. (3) for a PID, $1$ is not divisible iff no nonzero element is divisible. So Question 1(=2) amounts to finding one PID with residual fields of char both 0 and finite.
Oct 6, 2020 at 23:14 history edited Tim Campion CC BY-SA 4.0
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Oct 6, 2020 at 23:06 comment added Tim Campion I think I want to be assuming that $R$ is a PID.
Oct 6, 2020 at 22:59 comment added Tim Campion @YCor Thanks. At least I'm not still trying to convince myself that this is true! Regarding your second comment, I edited a couple of minutes ago when I realized that a your observation about cyclic modules (which I had earlier edited to place at the very end of the question) implied this. Sorry for the flux of edits!
Oct 6, 2020 at 22:56 comment added YCor Q1 and Q2 are equivalent. Actually if some f.g. $R$-module has a nonzero divisible element, then some quotient $R/I$ does. (Take $x_0$ divisible, choose $x_n$ with $n!x_n=x_{n-1}$, then $(Rx_n)$ stabilizes, so for large $n$, $Rx_n$ is a cyclic module with a divisible element.)
Oct 6, 2020 at 22:54 comment added YCor Q1: no: take $R=\mathbf{Z}_p[t]$ and $M=R/(1-pt)R\simeq\mathbf{Q}_p$
Oct 6, 2020 at 22:51 history edited Tim Campion CC BY-SA 4.0
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Oct 6, 2020 at 22:44 history edited Tim Campion CC BY-SA 4.0
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Oct 6, 2020 at 22:30 history asked Tim Campion CC BY-SA 4.0