I have been recently studying about Pure submodules of a module in a commutative algebra course. I came up with the following two questions, please help me as I am completely stuck.
The definition of a pure submodule of a module is as follows:
$a)$ $\textbf{Definition.}$ Let $R$ be a commutative ring with unity and let $M$ be an $R$-module. A submodule $N$ of $M$ is called a pure submodule if the sequence $0 \rightarrow L \otimes_{R} N \xrightarrow{1 \otimes i} L \otimes_{R} M$ is exact for all $R$-module $L$, where $i: N \rightarrow M$ is the inclusion.
Another definition goes like this:
$b)$ $\textbf{Definition.}$ Let $R$ be a commutative ring with unity and let $M$ be an $R$-module. A submodule $N$ of $M$ is called a pure submodule if $IM \cap N = IN$ for all ideals $I$ of $R$.
Are these two definitions equivalent? however, I have shown $a) \Rightarrow b)$. I can not understand how to proceed for $b) \Rightarrow a)$.
Next, I have come across the following well-known theorem:
$\textbf{Theorem.}$ Let $M$ be a finitely generated module over a Principal ideal domain $R$. If $N$ is a pure submodule of $M$ and $M/N$ is a direct sum of cyclic modules, then $N$ is a direct summand of $M$.
I have solved this problem but I can not understand why finite generation of $M$ is needed in that theorem. I have also seen the proof of this theorem here: Question about pure submodules over a P.I.D., and there also I can not see the importance of finite generation of $M$ or $M/N$.
I am completely stuck with these two problem, please please help me.
Best Answer
Edit: @Takumi Murayama has pointed out an inaccuracy in my answer to Question 1. The two definitions are in fact not equivalent! Though this answer was accepted, it would be best to look at the Takumi's correct response.
Question 1: Are the two definitions equivalent?The OP has shown (a)$\Rightarrow$(b) so let's show the other direction. Let $L$ be an $R$-module and consider the map $L\otimes_R N\stackrel{1\otimes i}{\to} L\otimes_R M$ where $i:N\hookrightarrow M$. The idea the shines to me is a similar fact that a module $M$ is flat iff $I\otimes_R M\to R\otimes_R M$ is injective for any finitely generated ideal $I$ of $R$: see this post for a proof and the statement.
Let me consider the sequence $0\to N\to M\to M/N\to 0$ and tensor by $L$ to get $$ \dots\to Tor_1^R(L,M/N)\to L\otimes_R N\to L\otimes_R M\to L\otimes_R M/N\to 0. $$ It suffices to show that first Tor-group vanishes in this case. Now the approach from that post more or less applies. The following line is an inaccuracy caught by @Takumi Murayama: Every $R$-module $L$ is a filtered colimit of its cyclic submodules which are all isomorphic to $R/I$. So we get $$ Tor_1^R(L,M/N)\cong \varinjlim_{R/I\subseteq L} Tor_1^R(R/I,M/N). $$ Now the $Tor_1^R (R/I,M/N)$ groups vanish because if we take $0\to N\to M\to M/N\to 0$ and tensor by $R/I$, $$ R/I\otimes_R N\to R/I\otimes_R M\to R/I\otimes_R M/N\to 0 $$ the definition (b) says this is left exact. Indeed, this exact sequence is the same as $$ N/IN\to M/IM\to R/I\otimes_R M/N\to 0 $$ and the kernel of the first map is $IM\cap N=IN$ from the definition.Question 2: Why the finite generation hypothesis?
I answered this in the comments. It comes from the application of the structure theorem for finitely generated modules over a PID.