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I'm currently battling with the following question:

Over an $R$-module $M$, for a given non-zero $m \in M$, is the map $F : R \rightarrow M$ given by $F(r) = rm$ necessarily non-zero? i.e. is there some element $r \in R$ such that $rm \neq 0$? ($R$ here is assumed to be a non-trivial commutative ring with unity).

Also, given instead the map $F_s : M \rightarrow M$ for some non-zero $s \in R$ given by $F_s (m) = sm$ for all $m \in M$, is this map also non-zero?

This has led me to investigate the following:

In an $R$-module $M$ does $rm = 0$ with ($m \in R$ non-zero) necessarily imply $r = 0$?

I'm pretty sure this is false, as otherwise in particular torsion elements wouldn't exist. I'm also fairly certain that the first statement can hold when we strengthen the assumptions a bit - e.g. with the added assumptions that $M$ is torsion-free and $R$ has at least one regular element. Can these assumptions be loosened? In particular, in question I'm looking at has $M$ is a simple module, so is there a way to relate this property to either of these statements?

Any time I've seen people use maps similar to $F$ as above, they always seem to say "clearly $F$ is non-zero" or something to that effect. Am I missing something?

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    $\begingroup$ What is your definition of “torsion free” for rings that sent domains? Aren’t you working with rings having identity? They always have a regular element. $\endgroup$
    – rschwieb
    Nov 18, 2018 at 18:23
  • $\begingroup$ “Sent” should be “aren’t “ $\endgroup$
    – rschwieb
    Nov 18, 2018 at 19:01
  • $\begingroup$ I am working with rings with identity - I hadn't considered that this element must always be regular! So I suppose we only need $M$ torsion-free? $\endgroup$
    – Stuart Green
    Nov 18, 2018 at 19:09
  • $\begingroup$ Again... what do you mean by that? $\endgroup$
    – rschwieb
    Nov 18, 2018 at 19:17
  • $\begingroup$ Sorry... torsion-free meaning containing no torsion elements, i.e. no non-zero elements $m \in M$ such that there is a regular element $r \in R$ (regular meaning a non zero divisor) with $rm = 0$ $\endgroup$
    – Stuart Green
    Nov 18, 2018 at 19:22

2 Answers 2

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For the first question, if the map $F_m$is zero, then $F(1)=1\,m=0$. However $1\,m=m$, so $m=0$.

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  • $\begingroup$ I've accepted the other answer because it came first, but thanks so much for taking the time to answer, much appreciated! $\endgroup$
    – Stuart Green
    Nov 18, 2018 at 18:30
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    $\begingroup$ You're welcom! It's a pleasure ti help. $\endgroup$
    – Bernard
    Nov 18, 2018 at 18:33
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The answer to your first question is yes, take $r=1$. The answer to your second question is yes, take the $\mathbb{Z}$ module $\mathbb{Z}_2$. Then $2 \cdot \overline{1}=\overline{0}.$

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  • $\begingroup$ Oh gosh yes of course - I sort of forgot to ask the more important direction! (I've added an "also..." to my question above. Thanks for this though! $\endgroup$
    – Stuart Green
    Nov 18, 2018 at 18:21
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    $\begingroup$ @Stuartg98 That map is not necessarily nonzero. Take M. Van's example of $\Bbb{Z}$ and $\Bbb{Z}/2\Bbb{Z}$. The map $F_2$ is the zero map from $M\to M$. $\endgroup$
    – jgon
    Nov 18, 2018 at 18:23
  • $\begingroup$ Brilliant, so basically the people I've seen say "clearly" it is true were wrong (or maybe working under broader assumptions). I was tearing my hair out! Thanks so much both of you. $\endgroup$
    – Stuart Green
    Nov 18, 2018 at 18:25

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