In Vakil's FOAG, it is an exercise to show that if $K$ is a number field and $\mathfrak a\subseteq K$ is a fractional ideal, then $\mathfrak a$ determines an invertible sheaf of $\operatorname{Spec}\mathcal O_K$, where $\mathcal O_K$ denotes the ring of integers of $K$.

My first thought is to just take $\mathcal L=\widetilde{\mathfrak a}$, but this doesn't seem to be locally trivial. I would need to show that there exists a generating set $\{\alpha_i\}$ of $\mathcal O_{K}$ such that $\mathfrak a_{\alpha_i}\cong(\mathcal O_K)_{\alpha_i}$ as $(\mathcal O_K)_{\alpha_i}$-modules. The latter condition is equivalent to there existing some $n\ge0$ such that $\alpha_i^n\in\mathfrak a$.

None of what I'm saying is in any way using the defining property of fractional ideals, i.e. that there exists some $\alpha\in\mathcal O_K$ such that $\alpha\mathfrak a\subseteq\mathcal O_K$. It seems like this problem should be simple, but I don't see how to do it. Can anybody possibly suggest something to guide me in the right direction?

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    $\begingroup$ The condition that there is some $n$ with $\alpha_i^n\in\mathfrak{a}$ is equivalent to the condition that the inclusion $\mathfrak{a}\hookrightarrow\mathcal{O}_K$ induces an isomorphism $\mathfrak{a}_{\alpha_i}\cong(\mathcal{O}_K)_{\alpha_i}$. To conclude $\mathfrak{a}$ is locally trivial, we just need there to be some isomorphism (not necessarily induced by inclusion). $\endgroup$ – Julian Rosen Jul 18 '17 at 18:40
  • $\begingroup$ @JulianRosen ah, yeah I see the difference now. So really what I need to show is that $\mathfrak a_{\alpha_i}$ is generated over $(\mathcal O_K)_{\alpha_i}$ by a single element if I make the proper choice of $\alpha_i$. $\endgroup$ – Alex Mathers Jul 18 '17 at 18:43

The fractional part is meaningless, since multiplication by a scalar is an isomorphism onto its image, so every fractional ideal is isomorphic as a module to some ideal. To me, the most geometric way to think about this is to introduce the notion of "order of vanishing" of an element at a prime ideal (geometrically, of a function at a point), which amounts to showing that the local rings are discrete valuation rings. Then you can get your $\alpha_i$ as follows: given a prime $\mathfrak{p}$, let $f$ be an element of your ideal with minimal order of vanishing at $\mathfrak{p}$.

Exercise: There is some $\alpha$ not in $\mathfrak{p}$ such that after inverting $\alpha$, $f$ generates your ideal (just throw away the other points where $f$ vanishes).


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