Proposition 10

Let $M$ be an $A$-module. For every submodule $N'$ of the $S^{-1}A$-module $S^{-1}M$, let $\phi(N')$ be the inverse image of $N'$ under $i_M^S$. Then: (i) $S\phi(N') = N'$.

(ii) For every submodule $N$ of $M$, the submodule $\phi(S^{-1}N)$ of $M$ consists of those $m \in M$ for which there exists $s \in S$ such that $sm \in N$.

(iii) $\phi$ is an isomorphism (for the orderings defined by inclusion) of the set of sub- $S^{-1}A$-modules of $S^{-1}M$ onto the set of submodules $Q$ of $M$ which satisfy the following condition:

(MS) If $sm \in Q$, where $s \in S$, $m \in M$, then $m \in Q$.

Obviously $S^{-1}\phi(N') \subset N'$; conversely, if $n' = m/s \in N'$, then $m/1 \in N'$, hence $m \in \phi(N')$ and consequently $n' \in S^{-1}(\phi(N'))$; whence (i). For an element $m \in M$ to be such that $m \in \phi(S^{-1}N)$, it is necessary and sufficient that $m/1 \in S^{-1}N$, that is that there exist $s \in S$ and $n \in N$ such that $m/1 = n/s$; this means that there exists $s' \in S$ such that $s'sm = sn \in N$, whence (ii). Finally, the relation $sm \in \phi(N')$ is equivalent by definition to $sm/1 \in N'$ and as $s/1$ is invertible in $S^{-1}A$, this implies $m/1 \in N'$, or $m \in \phi(N')$, hence $\phi(N')$ satisfies condition (MS); on the other hand, it follows from (ii) that, if $N$ satisfies (MS), then $\phi(S^{-1}N) = N$, which completes the proof of (iii).

The submodule $\phi(S^{-1}N)$ is called the saturation of $N$ in $M$ with respect to $S$, and the submodules satisfying condition (MS) (and hence equal to their saturations) are said to be saturated with respect to $S$. The submodule $\phi(S^{-1}N)$ is the kernel of the composite homomorphism

where $h$ is the canonical homomorphism, as follows from the commutativity of the diagram

If $S$ is the complement in $A$ of a prime ideal $p$, $\phi(S^{-1}N)$ is also called the saturation of $N$ in $M$ with respect to $p$.