TITLE: Proposition about localizations of Dedekind domains - Lang: Algebraic Number Theory
QUESTION [1 upvotes]: This proposition is in Lang's book "Algebraic number theory":
"Let $A$ be a Dedekind ring and $S$ a multiplicative subset of $A$. Then $S^{-1}A$ is a Dedekind ring. The map $\mathfrak{a}\mapsto S^{-1}\mathfrak{a}$ is a homomorphism of the group of fractional ideals of $S^{-1}A$, and the kernel consists of those fractional ideals of $A$ which meet $S$."
I am not sure if the statement about the kernel of this map is true. For example, consider $A=\mathbb{Z}$, $S=\{1,3,9,\dotsc\}$ and $\mathfrak{a}=(1/2)\mathbb{Z}=\{n/2\colon n\in\mathbb{Z}\}$. Then $\mathfrak{a}\cap S=S\ne\emptyset$, so $\mathfrak{a}$ meets $S$. If the statement were true, we would have $S^{-1}\mathfrak{a}=S^{-1}\mathbb{Z}$, which is of course false (for instance, $1/2\in S^{-1}\mathfrak{a}\setminus S^{-1}\mathbb{Z}$).
Am I right? Is this statement false? If the answer is yes, then what should be the real kernel?

REPLY [4 votes]: For ordinary ideals $\mathfrak a$ in a Dedekind domain $A$ (or really in an arbitrary commutative ring $A$), we have $\mathfrak a = A$ if and only if $\mathfrak a$ contains a unit of $A$.  But for fractional ideals $\mathfrak a$ of
a Dedekind domain $A$, it is not true that $\mathfrak a = A$ if and only if $\mathfrak a$ contains a unit of $A$. For example, taking $A = \mathbf Z$, a fractional ideal $(1/n)\mathbf Z$ where $n \geq 2$ contains all of $\mathbf Z$ and doesn't equal $\mathbf Z$.
What we can say for fractional ideals $\mathfrak a$ of a Dedekind domain $A$ is that $A \subset \mathfrak a$ if and only if $\mathfrak a$ contains a unit of $A$.
(If $\mathfrak a$ is an ordinary ideal, so $\mathfrak a \subset A$, then the conditions $\mathfrak a = A$ and $A \subset \mathfrak a$ are equivalent.)
In the case of the localization $S^{-1}A$ of a Dedekind domain $A$, fractional ideals $\mathfrak a$ of $A$ lead to fractional ideals $S^{-1}\mathfrak a$ of $S^{-1}A$ and this is multiplicative: $S^{-1}(\mathfrak a\mathfrak b) = (S^{-1}\mathfrak a)(S^{-1}\mathfrak b)$. So by induction, $S^{-1}(\mathfrak a^n) = (S^{-1}\mathfrak a)^n$ for nonnegative integers $n$, and to show this is true for negative $n$ as well we just need to check it for $n=-1$: localization sends inverse to inverses, i.e., $S^{-1}(\mathfrak a^{-1}) = (S^{-1}\mathfrak a)^{-1}$.  That is immediate from the formula $S^{-1}(\mathfrak a\mathfrak b) = (S^{-1}\mathfrak a)(S^{-1}\mathfrak b)$ by using $\mathfrak b = \mathfrak a^{-1}$.
Note that inversion on fractional ideals reverses containments: $\mathfrak a \subset \mathfrak b \Leftrightarrow \mathfrak b^{-1} \subset \mathfrak a^{-1}$ for fractional ideals $\mathfrak a$ and $\mathfrak b$ of the Dedekind domain $A$, and applying that reasoning to the Dedekind domain $S^{-1}A$ we get $S^{-1}\mathfrak a \subset S^{-1}\mathfrak b \Leftrightarrow S^{-1}(\mathfrak b^{-1}) \subset S^{-1}(\mathfrak a^{-1})$ because localizing at $S$ commutes with inversion on fractional ideals.
Now I can answer your question: for which $\mathfrak a$ does $S^{-1}\mathfrak a = S^{-1}A$?  Since $S^{-1}A$ is the fractional ideal of $S^{-1}A$ generated by $1$, we have $S^{-1}A \subset S^{-1}\mathfrak a$ if and only if $1 \in S^{-1}\mathfrak a$, which is equivalent to saying $\mathfrak a$ meets $S$. Thus
\begin{align*}
S^{-1}A = S^{-1}\mathfrak a & \Leftrightarrow S^{-1}A \subset S^{-1}\mathfrak a \text{ and } S^{-1}\mathfrak a \subset S^{-1}A \\ & \Leftrightarrow S^{-1}A \subset S^{-1}\mathfrak a \text{ and } S^{-1}A \subset S^{-1}(\mathfrak a^{-1}).
\end{align*}
Therefore $S^{-1}A = S^{-1}\mathfrak a$ if and only if $\mathfrak a$ and $\mathfrak a^{-1}$ both meet $S$. There is the criterion you were seeking.
Example: $A = \mathbf Z$, $S = \{1,3,9,\ldots\}$, and $\mathfrak a = (1/2)\mathbf Z$.  Then $S^{-1}A = \mathbf Z[1/3]$ and $\mathfrak a$ meets $S$ since $\mathfrak a \supset \mathbf Z$, but $\mathfrak a^{-1} = 2\mathbf Z$ does not meet $S$, and this is consistent with $S^{-1}\mathfrak a = (1/2)\mathbf Z[1/3]$ not being equal to $\mathbf Z[1/3]$.
The criterion that $\mathfrak a$ and $\mathfrak a^{-1}$ both meet $S$ is symmetric in the roles of $\mathfrak a$ and $\mathfrak a^{-1}$, which isn't a surprise since fractional ideals are a group and the condition $S^{-1}A = S^{-1}\mathfrak a$ is  equivalent to $S^{-1}A = S^{-1}(\mathfrak a^{-1})$, so whatever the criterion is should be applicable to $\mathfrak a$ if and only if it is applicable to $\mathfrak a^{-1}$. The rule "$\mathfrak a$ meets $S$" is not equivalent to "$\mathfrak a^{-1}$ meets $S$" when $\mathfrak a$ is a general fractional ideal of $A$. Note, however, that if $\mathfrak a$ is an ordinary ideal of $A$ (meaning $\mathfrak a \subset A$), then the symmetry between the roles of $\mathfrak a$ and $\mathfrak a^{-1}$ is broken: if $\mathfrak a \subset A$ then $A \subset \mathfrak a^{-1}$ so $\mathfrak a^{-1}$ automatically meets $S$ for every multiplicative subset $S$ of $A$, and in that case we recover (for Dedekind domains) the old relation for all ordinary ideals (but not all fractional ideals!) that $S^{-1}A = S^{-1}\mathfrak a$ if and only if $\mathfrak a$ meets $S$.
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