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

This question is a follow up to this post: Question about an endomorphsim of modules $N \subset M$ given $\exists f \neq 0$ an endomorphism such that $f(M) \subset N$

I have been working through a bunch of similar exercises and after I had the proof to the above problem explained to me I thought I was in good shape to attack the problem below but I keep getting stuck on choosing the correct submodule.

Let $A$ be a commutative ring with identity and let $M$ be an $A$ module. Suppose for every submodule $N \neq M$ with $N \subset M$ there exits a linear form $x^{*} \in M^{*}$ which is zero on $N$ and surjective,

How do we show that if $ f \in End_A(M)$ is not a right divisor of zero then f is a surjective endomorphism?

I thought the proof would be very similar to the previous problem cited above. But when I consider $N = \operatorname{Im}(f)$ and assume $\operatorname{Im}(f) \neq M$ then I am having trouble getting a contradiction out of the behavior of the linear form on $N$. The next step in the argument I thought would give us a contradiction by considering the fact that $\operatorname{Im}(f) \subset \operatorname{ker}(x^*)$.

share|cite|improve this question
If the title and the question are related, you might want to explain how. – Did Sep 26 '11 at 1:52
What is $E$ in your third paragraph? What is $R$ in your fourth paragraph? – Mariano Suárez-Alvarez Sep 26 '11 at 3:50
up vote 1 down vote accepted

If $f:M\to M$ is not surjective, its image $f(M)$ is a proper submodule of $M$ and, according to your hypothesis, there exists a surjective linear map $\phi:M\to A$ such that $\phi|_{f(M)}=0$. Let $m_0\in M$ be non-zero, and let $g:m\in M\mapsto \phi(m)m_0\in M$. This is a non-zero endomorphism (because there exists $m_1\in M$ such that $\phi(m_1)=1$, so that $g(m_1)=m_0\neq0$) and $gf=0$.

share|cite|improve this answer
Thanks for your help. This makes perfect sense after seeing your construction of $g$. – user7980 Sep 26 '11 at 5:09

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.