1
$\begingroup$

Let $\mathcal{o}$ be a Dedekind domain with fraction field $K$, let $L/K$ be finite separable and let $\mathcal{O}$ denote the integral closure of $\mathcal{o}$ in $L$, denote by $Tr$ the trace $Tr_K^L$.

In $\textit{Neukirch's Algebraic Number Theory}$ on page 195 he defines the inverse different $$ \mathfrak{C}_{\mathcal{O}\vert \mathcal{o}}=\{x\in L \vert Tr(x\mathcal{O})\subset \mathcal{o}\} $$ and the different $\mathfrak{D}_{\mathcal{O}\vert \mathcal{o}}$ to be the inverse of $\mathfrak{C}_{\mathcal{O}\vert \mathcal{o}}$.

In Proposition (2.2) he then says that it is trivial to prove that for $S\subset \mathcal{o}$ a multiplicative set, $S^{-1}\mathfrak{D}_{\mathcal{O}\vert \mathcal{o}}=\mathfrak{D}_{S^{-1}\mathcal{O}\vert S^{-1}\mathcal{o}}$.

It is certainly enough to prove this for the inverse different. The direction I am having trouble with is $\mathfrak{C}_{S^{-1}\mathcal{O}\vert S^{-1}\mathcal{o}} \subset S^{-1}\mathfrak{C}_{\mathcal{O}\vert \mathcal{o}}$, i.e. given $x\in L$ such that $Tr(x S^{-1}\mathcal{O})\subset S^{-1}\mathcal{o}$, we want to conclude that $Tr((sx)\mathcal{O})\subset \mathcal{o}$ for some $s\in S$.

I have no idea how this is obvious, I am grateful for any kind of help.

1 Answers 1

1

$ \mathcal O $ is finitely generated as an $ o $-module, say by $ y_1, y_2, \ldots, y_n $. For each $ y_i $, we may find $ s_i \in S $, $ z_i \in o $ such that

$$ \textrm{Tr}(x y_1) = z_i / s_i $$

so that $ \textrm{Tr}((s_i x) y_i) \in o $. Now, take $ s = \prod s_i $.