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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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    Hi! Can I ask you how you found this exercise?2011-09-24
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    It part of a huge pool of problems that have been given in the past for graduate level qualifying exams in linear algebra.2011-09-24
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    Is the last sentence supposed to have $\mathrm{End}_R(M)$ in both? If not, what is $A$?2011-09-24
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    thank you I had the wrong letter for the base ring it should be $R$2011-09-24
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    Dear user7980: Thanks! The sentence "every endomorphsim in $\text{End}_R (M)$ is not a left divisor of zero in $\text{End}_R (M)$" is not clear to me. Does it mean (1) "if $f\in\text{End}_R (M)$ then $f$ is not a left divisor of zero"? Or (2) "at least one $f\in\text{End}_R (M)$ is not a left divisor of zero"? Or something else? Clearly (2) is true: take the identity. (1) is false in general: Take $R\not=0$, $M$ free of rank $\ge2$, $f\not=0$ nilpotent.2011-09-25
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    I think I have been misreading the problem. Thanks for pointing out the issues. I have been using a previous students handwritten notes to study for a qualifying exam and I think I completely misread this problem. I will update the problem shortly. Thanks again for your time and sorry to have misstated things.2011-09-25
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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.