Top modules were defined in this article. Let $R$ be a ring and $M$ an $R$-module, throughout for any submodule $N$ of $M$, $V(N)$ denotes the set $\{P\in Spec(M)|N\subseteq P\}$, and $\mathcal{V}(M)$ denotes the set $\{V(N)|N \leq M\}$. Clearly, $V(0)=Spec(M)$, $V(M)=\emptyset$ and for any family of submodules $N_i$ of $M$, $\bigcap\limits_{i\in I}V(N_i)=V(\sum\limits_{i\in I}N_i)$. But $\mathcal{V}(E)$ is not closed under finite union in general.
I want to fined a module $M$ which is not top module since $V(N)\cup V(L)\neq V(E)$ where $N,~L,~E$ are submodules of $M$.
Thanks in advance.
The following theorem is useful for constructing examples:
Example 1: The $\mathbb{Z}$-module $\mathbb{Z}\oplus\mathbb{Z}$ is not top.
Example 2: It is rutin to check that the $\mathbb{Z}$-module $\mathbb{Z}\oplus \mathbb{Z}_p$ is not a top module, where $p$ is a prime number(without usnig obove theorem).
By definition, a perfect ring is a type of ring in which all left modules have projective covers. Since every vector space is projective, every field is perfect. Thus by above theorem a vector space $F$ over a field $K$ is top if and only $Dim_K(F)=1$.