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.

  • $\begingroup$ For the first part, choose finitely many generators, and clear denominators to show $z$ times (some element of $S$) sends every generator of $M$ into $R$. $\endgroup$
    – Steve D
    Aug 26, 2017 at 7:28
  • $\begingroup$ For the second part, the last inclusion shows $xM\subset N^{-1}$. So $xMM^{-1}\subset N^{-1}M^{-1}$. $\endgroup$
    – Steve D
    Aug 26, 2017 at 7:34
  • $\begingroup$ I have understood that now if $M $ is invertible then $MM^{-1}=R$ otherwise we don't know. So what will be the next step? $\endgroup$
    – Ri-Li
    Aug 26, 2017 at 19:36


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