# Cancellation in quotient of fractional ideals

When reading about fractional ideals of rings of integers, I came upon the following footnote:

For fractional ideals $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ with $\mathfrak{a} \supset \mathfrak{b}$, $$\displaystyle ^{\mathfrak{a}\mathfrak{c}}/_{\mathfrak{b}\mathfrak{c}} \simeq \ ^{\mathfrak{a}}/_{\mathfrak{b}}$$ as $\mathcal{O}_K$-modules.

This was not obvious to me, so I tried to prove it, however did not succeed. I think it must be connected to the unique product decomposition in Dedekind domains. I also found this question, where someone was also not sure how to prove this isomorphism, but did not succeed either.

Any help is greatly appreciated! Thanks in advance!

• This answer could be helpful. – Watson Dec 23 '17 at 16:45
• @Watson It was helpful, sorry and thanks. The question is over a year old, so it kind of slipped off my radar in the last weeks. – johnnycrab Jan 3 '18 at 10:22
• – Watson Nov 25 '18 at 17:31

Here is the main statement:

Proposition: Let $R$ be a Dedekind domain (for instance the ring of integers $R=\mathcal O_K$ of some finite extension $K$ of $\Bbb Q$). Let $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ be non-zero fractional ideals of $R$ with with $\mathfrak{a} \supset \mathfrak{b}$. Then there is an isomorphism of $R$-modules $$\dfrac{\mathfrak{a}\mathfrak{c}}{\mathfrak{b}\mathfrak{c}} \cong \dfrac{\mathfrak{a}}{\mathfrak{b}}.$$

We first need a little result:

Lemma: let $I,J$ be non-zero integral ideals of a Dedekind domain $R$. Then there is $x \in \mathrm{Frac}(R)$ such that $xI+J=R$.

This lemma is proved here ; we just take $x:=b/a \in \mathrm{Frac}(R)$ from the proof there.

Proof of proposition.

— Since $\frak a, b, c$ are fractional ideals, there are elements $\alpha, \beta, \gamma \in \mathrm{Frac}(R)$ such that $\alpha \frak a, \beta \frak b, \gamma \frak c$ are all integral ideals of $R$. We clearly have isomorphisms of $R$-modules $$\dfrac{\frak a}{\frak b} \cong \dfrac{\alpha \beta \frak a}{\alpha \beta \frak b}, \qquad \dfrac{\frak ac}{\frak bc} \cong \dfrac{\alpha \beta \gamma \frak ac}{\alpha \beta \gamma \frak bc}$$

We may therefore assume that all of $\frak a, b, c$ are integral ideals (i.e. contained in $R$) : replace them by $\alpha \beta \frak a, \alpha \beta \frak b, \gamma \frak c$ respectively.

— Then, by the lemma, there is $x \in \mathrm{Frac}(R)$ such that $x \mathfrak{c}^{-1} + \mathfrak{ba}^{-1} = R$, i.e. $$x \frak{a + bc = ac}.$$

Consider the morphism $f : \frak{a} \to \frak{ac/bc}$ be defined by $f(r) = [xr]$. It is clearly surjective. Moreover, the kernel of $f$ is $$\mathfrak{a} \cap x^{-1}\mathfrak{bc} = (\mathfrak{c}^{-1} \cap x^{-1}\mathfrak{ba}^{-1}) \mathfrak{ac} = x^{-1}(x\mathfrak{c}^{-1} \mathfrak{ba}^{-1})\mathfrak{ac} = \frak b$$

The first equality comes from (13) in this answer and the second comes from (14), using $x \mathfrak{c}^{-1} + \mathfrak{ba}^{-1} = R$ (notice that the answer holds in our setting with fractional ideals of the Dedekind domain $R$, not only for integral ideals in $R$).

As a conclusion, $f$ induces an isomorphism of $R$-modules $\frak{a/b \cong ac/bc}$ as claimed. $\hspace{1.5cm}\blacksquare$

• The claim that two modules are isomorphic whenever their localizations at each prime are isomorphic is false in general. It's true when the local isomorphisms are induced from some global map. To correct the above proof, you can restrict the argument to the case where $\mathfrak{c}$ is either prime or inverse of prime, in which case it's easy to write down a map from $\mathfrak{ac}$ to $\mathfrak{a}$ or $\mathfrak{a}$ to $\mathfrak{ac}$. Then inductively apply this result to get the theorem for an arbitrary fractional ideal $\mathfrak{c}$. – Ashvin Swaminathan Feb 13 '18 at 22:23
• @AshvinSwaminathan : thank you for pointing this out. There is a counterexample here. But what would be a suitable map $\mathfrak{ac} \to \mathfrak a$ when $\mathfrak c$ is prime? Then I can correct my answer. – Watson Feb 14 '18 at 10:50
• When $\mathfrak{c}$ is prime, note that $\mathfrak{c} \subset R$, so $\mathfrak{c} \mathfrak{a} \subset R \mathfrak{a} = \mathfrak{a}$. It's just the inclusion map. In the other case, when $\mathfrak{c}$ is the inverse of a prime, note that $\mathfrak{c} \ni 1$, so you get an inclusion of $\mathfrak{a}$ into $\mathfrak{ac}$. – Ashvin Swaminathan Feb 14 '18 at 14:41
• My apologies, please refer to the following answer, which defines the correct map: math.stackexchange.com/a/1404927/259604. – Ashvin Swaminathan Feb 15 '18 at 14:19
• @AshvinSwaminathan : you might see my edited answer. Localization doesn't seem to be the right tool here, actually. It is possibly to use it, though, as in this answer (which uses local isomorphisms which don't come from a global morphism!). – Watson Feb 16 '18 at 10:52