Fully invariant submodule and maximal submodule Let $M$ as a $R$-module and $N\subset M$ a fully invariant submodule of $M$. Why the exists a maximal submodule of $M$ such as $L$ that $L\cap N=\{0\}$. 
Definition: We refer $N$ as a fully invariant submodule of $M$ if for every homomorphim $\varphi : M\to M$ we have $\varphi(N)\subset N$.
 A: Zorn's Lemma: define
$$C:=\left\{\,K\le_R M\,\;/\;K\cap N=\{0\}\right\}$$
since $\;\{0\}\in C\;$ we have that $\;C\neq\emptyset\;$ , and it's easy (do it) to see $\;C\;$ is partially ordered by set inclusion and fulfills the conditions on any chain in it , so it has a maximal element...
A: I interpret the wording of the question as saying: If $N$ is a fully invariant submodule of $M$ then there is a maximal submodule of $M$, call it $L$, such that $L\cap N=\{0\}$.
This claim is false:
Consider $M:=\mathbb{Z}$ as a module over itself ($R:=\mathbb{Z}$). Any $R$-endomorphism $\varphi$ of $M$ is determined by the image of $1$, since if $\varphi(1)=n$ then $\varphi(m)=\varphi(m\cdot 1)=m\varphi(1)=mn$. Therefore
$$\varphi(\mathbb{Z})=n\mathbb{Z}$$
and
$$\varphi(k\mathbb{Z})=kn\mathbb{Z}\subset k\mathbb{Z}.$$
Thus all ideals of $\mathbb{Z}$, which are precisely its $R$-submodules, are fully invariant. On the other hand we have
$$k_1\mathbb{Z}\cap k_2\mathbb{Z}=\text{lcm}(k_1,k_2)\mathbb{Z},$$
which is nonzero unless $k_1=0$ or $k_2=0$. In particular, all nonzero fully invariant submodules have nontrivial intersection with all maximal submodules (which are those $p\mathbb{Z}$ with $p$ prime).
