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EDIT: I learned with this question (and this observation answers it) that we can detect whether a fractional ideal is principal or not looking at its prime decomposition $\prod\mathfrak{p}^{v_\mathfrak{p}}$ and answering whether or not there is an element $a\in K^{\times}$ with $v_{\mathfrak{p}}(a)=v_{\mathfrak{p}}$ for all these $\mathfrak{p}$. If yes, then yes; If not, then no.

I'm learning idèles through Neukirch's Algebraic Number Theory and got stuck in a claim.

Let $I_{K}$ be the idèle group of $K$ and $J_{K}$ the group of fractional ideals of $K$.

Let $(\:)\colon I_{K}\to J_{K}$ be the homomorphism taking an idèle $\alpha=(\alpha_{\mathfrak{p}})$ and gives the fractional ideal \begin{equation} (\alpha)=\prod_{\mathfrak{p}\: \text{finite}}\mathfrak{p}^{v_{\mathfrak{p}}(\alpha_{\mathfrak{p}})}. \end{equation} It is obviously surjective and has kernel consisting of $I_{K}^{S_\infty}$, which is just the product of the various $\mathcal{O}_{\mathfrak{p}}^{\times}$, but with no restriction in the archimedean places.

The claim is that this induces a surjective homomorphism $C_{K}\to C_{l_{K}}$, where $C_{K}=I_{K}/K^{\times}$ is the idèle class group of $K$ and $C_{l_{K}}$ is the usual class group of $K$.

Now, how is this so?

For this to be true we should have that the image of $K^{\times}$ via this map $(\:)$ is contained in the subgroup of principal ideals $P_{K}$ of $J_{K}$, but I don't think this is the case. I mean, for me is not obvious that we can detect whether or not some fractional ideal is principal only by looking at its prime decomposition. The equality $(\alpha)=\prod\mathfrak{p}^{v_{\mathfrak{p}}}$ seems like a definition, which may be taken also for elements of $\alpha\in K^{\times}$.

While writing this question just now, I realized it could be that the above equality is a 'piece-wise' definition, considering the principal ideal in the case where the element belongs to $K^{\times}$, but then doesn't it becomes trickier and not so straightforward to show that it is a homomorphism?

Thanks for any help.

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If you have a fractional principal ideal $(\alpha)$ then it's thought of as an idele $(\alpha_{\mathfrak p})$ by letting $\alpha_{\mathfrak p}=\alpha$ at each place. As an ideal, then $$\langle\alpha\rangle=\prod_{\mathfrak p}\mathfrak p^{v_{\mathfrak p}(\alpha)} =\prod_{\mathfrak p}\mathfrak p^{v_{\mathfrak p}(\alpha_{\mathfrak p})}$$ which is the image of the idele $(\alpha_\mathfrak p)$ under the $(\ )$-map. So elements of $K^\times$ map to principal ideals.

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  • $\begingroup$ So I guess precisely what I've said on the last paragraph. Thanks!! $\endgroup$ – Shoutre Apr 25 '17 at 15:38

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