I'm currently working my way through Matsumura's "Commutative Ring Theory", but I'm stuck on what I feel should be an easy proof of the following theorem:
Theorem Let $R$ be a commutative ring with identity, with $\mathbb{M}$ an $R$-module and $N\subseteq\mathbb{M}$. Then for all $r,r'\in R$, $$rN-r'N=(r-r')N\iff N\ \text{is a submodule of}\ \mathbb{M}.$$
This is not directly in the book, but I've reduced an implicit theorem in there to this simpler one. (he claims that $N:N'$ is an ideal iff $N$ and $N'$ are submodules on p.6; this is part of showing that claim to be valid).
It is clear that $(r-r')N\subseteq rN-r'N$ regardless of the structure of $N$, since $$x\in(r-r')N\implies\exists n\in N\big[x=(r-r')n\big]$$ $$\therefore x=(r-r')n=rn-r'n\in rN-r'N.$$
The reverse inclusion is not so clear to me, since $$y\in rN-r'N\implies\exists n,n'\in N\big[y=rn-r'n'\big],$$ but I cannot see how to decompose $rn-r'n'$ into something with a coefficient of $r-r'$. (usually adding in an appropriate preimage of $0$ under multiplication and addition works, but it doesn't seem to here)
What does $N_1-N_2$ even mean for submodules? The naive thing would be that it means $\{a−b∣a∈N_1,b∈N_2\}$. But if that's the case here, this is obviously wrong, say, in the case $r=r′$. I think your "implicit theorem" might be a tad off...
Suppose $x,y\in (N:N')$.
Then $xN'\subseteq N$ and $yN'\subseteq N$.
If $N$ is an additive group, $-xN'\subseteq -N=N$. So $-x\in (N:N')$
Also if $N$ is an additive group, $(x+y)N'\subseteq xN'+yN'\subseteq N$. So $x+y\in (N:N')$
At this point we've established that $(N:N')$ is an additive subgroup of $R$.
If $N'$ is a submodule of $N$, then $xrN'=xN'r\subseteq xN'\subseteq N$. So $xr\in (N:N')$.
I don't see why you'd need the lemma you suggested (even if we defined subtraction between submodules.)