Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.