# Existence of “zero-divisors” over a module?

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?

• 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. – rschwieb Nov 18 '18 at 18:23
• “Sent” should be “aren’t “ – rschwieb Nov 18 '18 at 19:01
• 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? – Stuartg98 Nov 18 '18 at 19:09
• Again... what do you mean by that? – rschwieb Nov 18 '18 at 19:17
• 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 $rm = 0$ – Stuartg98 Nov 18 '18 at 19:22

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 $$2 \cdot \overline{1}=\overline{0}.$$

• 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! – Stuartg98 Nov 18 '18 at 18:21
• @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 $M\to M$. – jgon Nov 18 '18 at 18:23
• 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. – Stuartg98 Nov 18 '18 at 18:25

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

• I've accepted the other answer because it came first, but thanks so much for taking the time to answer, much appreciated! – Stuartg98 Nov 18 '18 at 18:30
• You're welcom! It's a pleasure ti help. – Bernard Nov 18 '18 at 18:33