请求证明：满足循环子模和仍循环的Noetherian左R-模是循环模

Let's work through this proof step by step, leaning on the Noetherian property of the module and the given condition about cyclic submodules:

1. Leverage the Noetherian property to find a maximal cyclic submodule
   Since M is a Noetherian left R-module, every non-empty collection of its submodules has a maximal element (this is one of the core equivalent definitions of Noetherian modules). Consider the collection of all cyclic submodules of M—this collection is non-empty (for any $m \in M$, the submodule $Rm$ generated by $m$ is cyclic). Let $N = Rm_0$ be a maximal cyclic submodule from this collection.

2. Prove this maximal cyclic submodule equals M
   Suppose for contradiction that $N \subsetneq M$. Then there exists some element $x \in M \setminus N$. The submodule $Rx$ is cyclic, so by the problem's given condition, $N + Rx = Rm_0 + Rx$ must also be a cyclic submodule of M. But here's the catch: $N$ is strictly contained in $N + Rx$ (since $x$ is in $N + Rx$ but not in $N$), which directly contradicts the maximality of $N$ (we just found a larger cyclic submodule, which shouldn't exist if $N$ is maximal).

3. Final conclusion
   Our contradiction tells us the assumption $N \subsetneq M$ is false. Therefore, $N = M$, which means M itself is a cyclic left R-module.

内容的提问来源于stack exchange，提问作者siwar