# If a PID has no nonzero divisible elements, then is the same true of its finitely-generated modules?

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?

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

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.