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Let $M$ be an infinite $R$-module, for a commutative ring $R$. Let $F \subseteq M$ be the set of elements which have a trivial annihilator: that is, for all $f \in F$, $$ rf = 0_M \iff r = 0_R.$$ Suppose that $T: M \to M$ preserves $F$: that is, $f \in F \implies Tf \in F$. Does it follow that $T$ is left-invertible? Does this depend on whether $R$ is finite?

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Please do not add stuff to questions which radically changes them after you have answers. You could post a new question with the new stipulation, and it would not be counted as a duplicate because it is completely different from the original question. – rschwieb Mar 21 '14 at 13:03
@rschwieb: Sorry --- I had forgotten to specify conditions which I was taking for granted. I've rolled back my revision, and I think I have found an approach for my revised question anyway, thanks. – Niel de Beaudrap Mar 21 '14 at 13:06
@nieldebeudrap Thanks! And by the way, asking in comments for another example with more conditions is perfectly fine, it's just that changes like that to the original question will sometimes get you in trouble with people who worked a long time writing solutions for the other question. I guess you already know this though :) – rschwieb Mar 21 '14 at 13:09
@rschwieb: I had not been thinking of integral domains, so when I saw you invoking them in your answer I knew at once that I had missed conditions... but also (perhaps presumptuously, but maybe not so much given your rep score) supposed that you hadn't needed much time to formulate your answer. Otherwise I would have been more hesitant to modify my question. – Niel de Beaudrap Mar 21 '14 at 13:15
oh no, don't worry: you caused me no major inconvenience! But I thought I should say something, or else you might actually run into that situation in the future :) What new example did you settle on, by the way? I was thinking that you should be able to make $\Bbb Z/(n)$ into a $\Bbb Z/(m)$ module somehow, and then use the second half of my answer again. Artinian rings all have the property about unit and zero divisors that you are asking about. – rschwieb Mar 21 '14 at 13:16

Let R be an integral domain that is not a field and x be a nonzero element of R.

The map $T:r\mapsto rx$ maps F into F for the left R module R. If T were left invertible, that would mean there is a y such that $xy=1$. Since you can choose an element x that isn't a unit, it's possible for T to be noninvertible.

This same example does not survive if R is finite, though, since a finite domain is a field.

Here's a different example: let $M=\Bbb Z/(n)$ be viewed as a $\Bbb Z$ module. Then $F$, being empty, is obviously preserved by whatever module endomorphism you choose, and there are many noninvertible endomorphisms of this module. This one shows finiteness is not much help.

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Finiteness of the module, that is. – Niel de Beaudrap Mar 21 '14 at 13:17
@NieldeBeaudrap That's true, I looked at the wrong thing being finite :) But perhaps you can swap $\Bbb Z$ for another quotient of $\Bbb Z$ and make a similar example with the ring finite too. – rschwieb Mar 21 '14 at 13:23

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