Let $M$ be an $R$-module and $N$ a submodule of $M$. $N$ is said to be a fully invariant submodule of $M$ if $f(N)\subseteq N$ for all $f\in \text{End}_R(M)$. We call an $R$-module $M$ invariant if each of its submodules is fully invariant.

In Lemma 1.1 of the paper (Duo Modules by A. C. Ozcan, A. Harmanci and P. F. Smith), it is proved that a right $R$-module $M$ is an invariant module if and only if for each endomorphism $f$ of $M$ and each element $m$ of $M$ there exists $r$ in $R$ such that $f(m) = mr$. Here it means that any $m\in M$ determines an element of $R$ such that $f(m)$ is right multiplication by $r$.

My question here is: How does the whole image $f(M)$ and the kernel $\ker(f)$ of $M$ under $f$ look like in terms of elements of $R$? I know $f(M)$ is also right multiplication by some element of $R$ but which one and how? Is it $f(M)=Ma$ where $a=(r_1,r_2,\ldots,r_n)$ for some sequence of elements in $R$?

For the kernel, I think the following: $$\ker(f)=\{m_i\in M:f(m_i)=0\}=\{m_i\in M:m_ia_i=0~\text{for some}~a_i\in R\}=\{m_i\in M:m_i\in\text{Ann}(a_i)~\text{for some}~a_i\in R\}=\bigcup_{a_i\in R}\text{Ann}(a_i),$$ where $\text{Ann}(a_i)$ is the left annihilator of $a_i$ in $M$.

I need some help to understand on how the entire images and kernels are determined by the right multiplication by the ring element?

  • $\begingroup$ I din't know! Let me edit. $\endgroup$
    – mariam
    Jan 30 at 10:20
  • $\begingroup$ Also, note $m0 =0$ for any $m\in M$, thus the second equality does not hold $\endgroup$
    – Astyx
    Jan 30 at 10:21


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