Let $R$ be a Dedekind domain, $K$ the field of fractions of $R$, $\mathfrak{m}$ be a fractional ideal of $R$, i.e. a non-zero finitely generated $R$-submodule of $K$. We define
$$\mathfrak{m}^{-1}:=\{y\in K: y\mathfrak{m}\subseteq R\}$$
I want to prove that $\mathfrak{m}^{-1}$ is a fractional ideal.
For every $y\in\mathfrak{m}^{-1}, r\in R$, we have $ry\in\mathfrak{m}^{-1}$, which ensures that $\mathfrak{m}^{-1}$ is an $R$-submodule of $K$.
Let $x\in\mathfrak{m},x\neq 0$. Then $\mathfrak{m}^{-1}x\subseteq R$, by definition of $\mathfrak{m}^{-1}$, so that $\mathfrak{m}^{-1}\subseteq x^{-1}R$, but $x^{-1}R$ is a finitely generated $R$-module, $R$ is noetherian, hence $x^{-1}R$ is noetherian (as a module), so $\mathfrak{m}^{-1}$ is finitely generated, as a sub-module of a noetherian module.
In order to have $\mathfrak{m}^{-1}$ fractional ideal, i need $\mathfrak{m}^{-1}\neq 0$. How can i prove this?
Best Answer
Since $\mathfrak m$ is finitely generated, you can find a generating family $\frac {a_1}{b_1}, \ldots, \frac {a_n}{b_n} \in \mathfrak m$. Now, let $c = b_1b_2 \ldots b_n$. Forall $i$, $c\frac {a_i}{b_i} \in R$, and so forall $m \in \mathfrak m, cm \in R$. Hence $c \in \mathfrak m^{-1}$