I was studying fractional field from "Algebraic Number fields" Janusz and I am stuck in two problems
First let me give you some definitions:
1) Fractional field: Let $R$ be a Dedekind ring and $K$ be its quotient field. A fractional field of $R$ is a nonzero finitely generated $R$-submodule of $K$.
2) If $\mathfrak M$ is a fractional ideal of $R$, then $\mathfrak M^{-1}$ is the set $\{x \in K : x\mathfrak M \subseteq R\}$. We call it inverse of $\mathfrak M$.
Now questions are:
1) Let $S$ be a multiplicative set in $R$ define $\mathfrak M_S=\mathfrak MR_S(=S^{-1}\mathfrak M)$ localization of $\mathfrak M$ at S. We have to prove $(\mathfrak M^{-1})_S=(\mathfrak M_S)^{-1}$.
My attempt was: for $x = \frac as \in (\mathfrak M^{-1})_S$ where $a \in \mathfrak M^{-1}$.
Now for any $y=\frac {m}{s_1} \in \mathfrak M_S$. So we have $xy = \frac {am}{ss_1}\in R_S$
Hence, $(\mathfrak M^{-1})_S \subseteq (\mathfrak M _S)^{-1}$
Conversely, for any $z \in (\mathfrak M_S)^{-1}$
then $z(\mathfrak M_S) \subseteq R_S$
Hence, $z \frac ms= \frac {r}{s_1}$ From here how can I prove that $z$ can be written as $\frac {x_3}{s_3}$ such that $x_3 \in \mathfrak M^{-1}$ and $s_3 \in S$.
I was thinking that here I have to use finitely generation of $M$. Please give a detailed answer.
2) If $\mathfrak M, \mathfrak N$ are fractional ideals of $R$ then prove that $(\mathfrak M \mathfrak N)^{-1}=\mathfrak M^{-1} \mathfrak N^{-1}$
In my attempt it is clear to see that $\mathfrak M^{-1} \mathfrak N^{-1} \subseteq (\mathfrak M \mathfrak N)^{-1}$[As the ring is commutative]
But conversely,
Let $x \in (\mathfrak M \mathfrak N)^{-1}$ we have $x(\mathfrak M \mathfrak N) \subseteq R$. Now how to break $x$ down into two parts? Please give detail information please.