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Let $(G,+)$ be an abelian group . The first Prufer theorem (https://en.wikipedia.org/wiki/Pr%C3%BCfer_theorems) states that if $nG=\{0\}$ for some non-zero $n \in \mathbb Z$ then $G$ is a direct sum of cyclic groups .

Is there any similar result for special kind of non-faithful (i.e. $rM=\{0\}$ for some $0\ne r \in R$ ) modules over any class of commutative rings ?

I add the "non-faithful" condition as this seems like the right generalization of the concept of " bounded exponent" , of groups , for modules . Any reference or partial results are highly appreciated .

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The same result holds for modules over any Dedekind domain $R$: if $rM=0$ for some nonzero $r\in R$ then $M$ is a direct sum of cyclic modules. I don't know a reference but the proof is basically the same as for $\mathbb{Z}$. First, $M$ is the direct sum of its localizations at each of the finitely many prime ideals containing $(r)$. So we can localize and assume $R$ is a DVR, with maximal ideal $(p)$ and $r=p^n$ for some $n$. Now show that any element of $M$ of maximal order generates a direct summand (for instance, by proving that $R/(p^n)$ is injective as a module over itself via Baer's criterion). Peeling off summands one by one by transfinite induction, we find that $M$ is a direct sum of cyclic modules.

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  • $\begingroup$ Nice (+1)! It seems that this is pretty close to the largest class of rings that will work. $\endgroup$ – Stephen Aug 23 '17 at 17:09
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You definitely need to restrict the class of rings in some dramatic way: modules over $R$ with $rM=0$ are the same as $R/(r)$ modules, so you are asking for rings $R$ such that $R/(r)$ has some very well-understood representation theory for every $r \in R \setminus \{0 \}$. I suppose the most direct way to ask a specific, well-defined question sharpening yours is

For what commutative rings $R$ is it true that for all non-zero $r \in R$, every $R/(r)$-module is a direct sum of cyclic modules?

Thinking instead about $S=R/(r)$, we obtain that every indecomposable $S$-module is cyclic. This is a very strong condition: it implies that for every minimal prime $p$ of $S$, the field of fractions of the quotient $S/p$ is a cyclic $S/p$-module. This implies that in fact $S/p$ is a field, or in other words every minimal prime is also maximal. Thus $S=R/(r)$ is of Krull dimension zero for every non-zero $r$, or in other words $R$ is at most one dimensional. Thus for commutative rings, at least, Eric Wofsey's answer is fairly close to being the complete answer: you can say something only when the ring $R$ is of Krull dimension (at most) one, and when it is actually a Dedekind domain the answer is essentially the same as for the integers.

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