Given $\operatorname{Im}(f) \subset \operatorname{ker}(x^*)$ does that imply if $ f∈End_A(M)$ is not a right divisor of zero then f is surjective?

Question

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 $\operatorname{Im}(f) \subset \operatorname{ker}(x^*)$.

Answer 1

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 $gf=0$.