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The following question is related to this post: Example of a an endomorphism which is not a right divisor of zero and not onto

I was told there is a simple motivation for this problem from a linear algebra textbook such as Hofmann and Kunze but I have yet to find a clear explanation. In particular I do not see why existence of one invertible element tells us about the behavior of injective endomorphisms... I think this problem is related to exercises in Algebra Volume II by Bourbaki but the notation is probably a little different.

Suppose that $M$ is a module over a commutative ring $R$ with identity and that there $\exists x' \in M^*$ a linear form and $\exists x \in M$ such that $\langle x , x' \rangle$ is invertible.

  1. How do we show that an element which is not a left divisor of zero in $End_R (M)$ is an injective endomorphism?

(Actually the notes say in parituclar if $M$ is free but I am even confused about this statement.. All we need to show is 1. right and the free case follows?)

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    You're welcome. Suggestion: use the `@Pierre-Yves` if you want to ping me. I'm busy right now, but I'll get back to that soon.2011-09-25

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If $f\in\text{End}(M)$ is not injective, there is a nonzero $y$ in $M$ such that $f(y)=0$. Define $g\in\text{End}(M)$ by g(z):=\langle z , x' \rangle y. Then $g(x)\not=0$ and $f\circ g=0$, and $f$ is a left divisor of zero.