# Question about an endomorphsim of modules $N \subset M$ given $\exists f \neq 0$ an endomorphism such that $f(M) \subset N$

Hi the following problem is related to the post Example of a an endomorphism which is not a right divisor of zero and not onto.

Below is one of the many facts I am having trouble dealing with in a multi part example centered around standard notions about endomorphism modules of commutative rings with identity. I think the original source for these type of problems comes from one of the Bourbaki books on Algebra.

Suppose $M$ is a left $R$-module such that for every submodule $N \neq 0$ with $N \subset M$ there exits an endomorphsim $f \neq 0$ with $f(M) \subset N$. How do you show an element which is not a left divisor of zero in $End_R(M)$ is injective.

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How about this: suppose that $g \in \text{End}_R(M)$ is not a left-zero divisor and look at $\text{ker}(g)$, which is a submodule of $M$. If $g$ is not injective, then $\text{ker}(g) \neq 0$ and we have a non-trivial endomorphism $f$ with $f(M) \subset \text{ker}(g)$. By definition, this means that $gf=0$. But $g$ is not a left-zero divisor, so we must have $f=0$, which contradicts its non-triviality. Thus our assumption that $\text{ker}(g) \neq 0$ was wrong, and so $\text{ker}(g) = 0$ which implies that $g$ is injective.