If $\{N_i\}_{i\in I}$ is a totally ordered set of prime submodules then ${\bigcup}_{i\in I} N_i$ is a prime submodule

Question

In a finitely generated $R$-module $M$, let $\{N_i\}, i\in I$ be a totally ordered prime submodules family. I have tried that definition $\Rightarrow$(There is a prime submodule that includes all proper submodules of $M$). How can I use that definition?

Answer 1

Right, from your title I guess you understand that the only thing you need to show is that $\bigcup_{i \in I} N_i$ is a prime submodule.

Let's recall definitions. A module $M$ is prime if $\operatorname{Ann}(M) = \operatorname{Ann}(M')$ for every $0 \not= M' \subseteq M$. The annihilator of a prime module is a prime ideal, but the converse need not hold.

A submodule $N \subseteq M$ is a prime submodule of $M$ if $M/N$ is a prime module, i.e. if $\operatorname{Ann}(M/N) = \operatorname{Ann}(M'+N/N)$ for every submodule $M' \subseteq M$ such that $M' \nsubseteq N$. Equivalently, a submodule $N$ is prime if $(N : M) = (N : M')$ for every $N \subsetneq M' \subseteq M$, where $(N : M) := \{r \in R \mid rM \subseteq N \}$ denote the colon ideal. Note that it suffices to check primeness of the submodule $N \subseteq M$ on the cyclic submodules $mR \subseteq M$ for $m \in M \setminus N$, i.e. $(N : mR) = (N : M)$ for every $m \in R$ implies that $(N : M') = (N : M)$ for every $N \subsetneq M' \subseteq M$.

If I understand your question, it is asking why the union of a chain of prime submodules is a prime submodule.

Let $N_i$ be a chain of prime submodules of $M$ and set $N = \bigcup_i N_i$.
Let $m \in M \setminus N$. So $m \notin N_i$ for all $i$. We want to check that $(N : M) = (N : mR)$. As I mentioned above, it suffices to show that $rm \in N$ implies $rM \subseteq N$. That $rm \in N$ implies $rm \in N_i$ for some $i$, and since $N_i$ is a prime submodule of $M$ and $m \in M \setminus N_i$, we conclude $rM \subseteq M$ by definition.