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EDIT: The question was originally about general Noetherian rings instead of PID's. Thanks to YCor for pointing out how wrong this was in the comments below (1 2 3).

Question 1: Let $R$ be a PID. Suppose that some finitely-generated $R$-module $M$ contains a nonzero $\mathbb Z$-divisible element. Then does $R$ contain a nonzero $\mathbb Z$-divisible element?

Here I say that $x$ is $\mathbb Z$-divisible if, for every $0 \neq n \in \mathbb Z$, there is $y$ such that $ny = x$. Since this is the only kind of "divisibility" I'm interested in, I'll say "divisible" instead of "$\mathbb Z$-divisible" from now on.

My expectation is that the answer is "yes" -- my feeling is that in order to produce a divisible element of some module, some sort of localization must be performed, which is a sort of infinitary construction.

A relevant observation is that in a Noetherian module $M$, if $x \in M$ is divisible, then the submodule $xM \subseteq M$ generated by $x$ is a divisible submodule (i.e. all of the elements of $xM$ are divisible in $xM$). It follows that the following is an equivalent formulation of the question:

Question 2: Let $R$ be a PID. Suppose that some quotient ring $R/I$ contains a nonzero divisible element. Then does $R$ contain a nonzero divisible element?

Note that if $R$ is a ring and some quotient ring $R/I$ contains a nonzero divisible element, then we may assume that $R/I$ is a field of characteristic 0. So an equivalent form of Question 2 would be: if $R$ is a PID surjecting onto a field of characteristic 0, then must $R$ contain a divisible element?

Restricting Question 2 to the case where $R$ is $p$-local for some prime $p \in \mathbb Z$, there is also the following formulation:

Question 3: Let $R$ be a $p$-local PID. If $p$ does not lie in the Jacobson radical of $R$, then must $R$ contain a nonzero divisible element?

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    $\begingroup$ Q1: no: take $R=\mathbf{Z}_p[t]$ and $M=R/(1-pt)R\simeq\mathbf{Q}_p$ $\endgroup$
    – YCor
    Oct 6, 2020 at 22:54
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    $\begingroup$ 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.) $\endgroup$
    – YCor
    Oct 6, 2020 at 22:56
  • $\begingroup$ @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! $\endgroup$
    – Tim Campion
    Oct 6, 2020 at 22:59
  • $\begingroup$ I think I want to be assuming that $R$ is a PID. $\endgroup$
    – Tim Campion
    Oct 6, 2020 at 23:06
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Oct 7, 2020 at 6:55

1 Answer 1

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Thanks to YCor's reduction in the comments (1 2 3), we know that if $R$ is a PID with a residue field of characteristic 0 and a residue field of finite characteristic, then $R$ answers Question 1 in the negative. In fact Heitmann has constructed PID’s with specified residue fields in a very general way. So the answer to each question is no.

These PID's still seem very exotic to me — I wish there were an easy example.

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  • $\begingroup$ I learned of these examples from here. $\endgroup$
    – Tim Campion
    Oct 7, 2020 at 17:47

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