I'm trying to solve this exercise, but I can't...
Let S be a multiplicatively closed subset of A, M an A-module, N a submodule of M and $\phi_M:M\longrightarrow S^{-1}M$ the homomorphism given by $\phi_M(m)=\frac{m}{1}$. The following are equivalent:
- There exists an $S^{-1}A$-submodule $P$ of $S^{-1}M$ such that $N=\phi_M^{-1}(P)$.
- $\forall s \in S,x \in M $ such that $sx \in N$, then $x \in N$.
This is what I've done:
=> 2. If $sx \in N=\phi^{-1}_M(P)$ then $\phi_M(sx)=\frac{sx}{1}\in P$. As $s\in S$, $\frac{sx}{s}=\frac{x}{1}\in P$. Then, $\phi^{-1}_M(\frac{x}{1})=x\in \phi^{-1}_M(P)=N$. Is it proved?
=> 1. I can't start because I don't know if $P=\phi_M(N)$ .
If $N=\phi^{-1}(P)$ then take $s\in S, x\in M$ such that $sx\in N$. Then $\frac{sx}{1}\in P$, so $\frac{1}{s}\cdot\frac{sx}{1}=\frac{x}{1}\in P$ and so $x\in \phi^{-1}(P)=N$.
If the second property holds, define $P=<\phi(N)>$. Notice that $P$ will in general be bigger than $\phi(N)$ because $\phi$ is an homomorphism of $A$-modules but not $S^{-1}A$ modules (that wouldn't even make sense because $M$ is not an $S^{-1}A$ module). In fact $P$ is the set of the elements of the form $\frac{x}{s}$ where $s\in S$ and $x\in N$: those are certainly there, and viceversa a general element of $P$ is $\sum \frac{a_i}{x_i}{s}$ and by making a common denominator you get an element of the simpler form. Now we prove that $N=\phi^{-1}(P)$. Of course if $x\in N$, then $\phi(x)\in \phi(N)\subseteq P$. On the other hand, take $\frac{x}{1}\in P$. Then we know that $\frac{x}{1}=\frac{y}{s}$ for $y\in N$, that is there exists $t\in S$ such that $sx=ty$, that is $sx\in N$, that is $x\in N$ by the hypothesis.