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Questions.

(0) Is there a usual technical term in ring theory for the following kind of module?

$M$ is a free $R$-module over a commutative ring $R$, and for each $R$-basis $B$ of $M$, each $b\in B$, and each unit $r$ of $R$, we have $r\cdot b=b$

(1) Is there a classification of such modules, or a non-classifiability result in the literature?

Remarks.

--This is not a homework problem. I need to know more about this, yet it is not central to what I am doing currently, and I hope this well-known and documented.

--Trivial examples are

  • the $R$-module $\{0\}$ whose only basis is $\{\}$, for which the condition is vacuously true,

  • $R:=\mathbb{Z}/2\mathbb{Z}$, $\kappa:=$some cardinal, and $M:=\prod_{i\in\kappa} R$,

  • $R:=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$, $\kappa:=$some cardinal, and $M:=\prod_{i\in\kappa} R$,

  • $\kappa_0,\kappa_1$ arbitrary cardinals, $R:=\prod_{i\in\kappa_0} \mathbb{Z}/2\mathbb{Z}$, and $M:=\prod_{i\in\kappa_1} R$.

EDIT: beware, the following is evidently not a free $R$-module; I keep it here since with the warning, it seems instructive * $R:=\mathbb{Z}/4\mathbb{Z}$, whose units are $1+(4)_R$ and $3+(4)_R$, and $\kappa:=$some cardinal, and $M:=\prod_{i\in\kappa}\mathbb{Z}/2\mathbb{Z}$,

--Since $R$ need not be a principal ideal domain, or even a domain, no classification theorem known to me applies here.

Peter Heinig
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    In your example with $R=\mathbb{Z}/4\mathbb{Z}$, your module $M$ is not a free module. Also, if $\kappa_0$ is infinite, then $R=\bigoplus_{i\in\kappa_0} \mathbb{Z}/2\mathbb{Z}$ is not a ring at all (I assume you want your rings to be unital, in order to talk about units). – Eric Wofsey Sep 23 '17 at 17:20
  • @EricWofsey: thanks for setting the record straight. I appreciate this very much. The writing was indeed very hasty are flawed. For the record: Eric Wofsey is right in saying that $\bigoplus_\kappa\mathbb{Z}/2\mathbb{Z}$ is not a unital ring if $\kappa$ is infinite. The direct product $\prod_\kappa\mathbb{Z}/2\mathbb{Z}$ is a unital ring, and, being a $\mathbb{F}2$-vector space, is free. (But be careful: it is a well-known that e.g. the direct product $\prod{i\in\omega}\mathbb{Z}$ is not a free $\mathbb{Z}$-module (=free abelian group). – Peter Heinig Sep 23 '17 at 17:43

2 Answers2

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There are some answers to the question here.

Basically, you are asking that the group of units be trivial. In the case that there are only finitely many maximal ideals in the ring, it seems that the answer is known, and the ring consists of products of $\mathbb Z/2\mathbb Z$, so it looks like for a large class of rings, your trivial examples are all of them.

However, after sifting around a bit, it does not look like there is an easy classification.

Andres Mejia
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To elaborate on Andres Mejia's answer, I think you are getting a bit confused thinking about bases here. If $b$ is an element of a basis of a free module $M$ and $r\cdot b=b$, that immediately implies $r=1$, since $(r-1)b=0$ is a linear dependence among elements of the basis and that can only happen if $r-1=0$.

So the module $M$ itself is pretty much completely irrelevant here. If $R$ is a ring such that every unit of $R$ is equal to $1$, then every free module over $R$ will have this property. And if $R$ is a ring such that not every unit of $R$ is equal to $1$, then the only free module over $R$ with this property is the trivial module $0$ (since there cannot exist any element of a basis).

Eric Wofsey
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