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Let $K$ be a quadratic number field.

I am struggling with some "usual" maps in algebraic number theory, but with which I am not used to, confusing a lot of different settings, as idèles, corresponding ideals and reduced images modulo something. I would like to understand as explicitly as possible the involved maps. In my case:

\begin{equation} \mathrm{Ideals \ of \ } \mathcal{O}_K \to \mathbf{A}_{K,f}^\times/K^\times \to \hat{\mathcal{O}}_K^\times/(1+\mathfrak{m}\hat{\mathcal{O}}_K^\times) \to \mathcal{O}_K/\mathfrak{m} \leftarrow \mathbf{Z}/\mathfrak{m}\cap\mathbf{Z} \end{equation}

Let us be more precise from now on. Take $\mathfrak{a}$ to be an ideal of $\mathcal{O}_K$, the ring of integers of $K$. Write its decomposition into prime ideals:

$$\mathfrak{a} = \prod_i \mathfrak{p}_i^{e_i} \quad \mathrm{\ an \ ideal \ of \ } \mathcal{O}_K$$

We can embed it in the idèles on $K$. For that, I choose local uniformizers $\varpi_{\mathfrak{p}_i}$ in $K_{\mathfrak{p}_i}$ for each $i$, and consider the (is it canonical? could we come back to $\mathfrak{a}$?) image:

$$x = \prod_i \varpi_{\mathfrak{p}_i}^{e_i} \quad \in \mathbf{A}_{K,f}^\times/K^\times$$

Now, we want to "reduce it modulo an ideal $\mathfrak{m}$" prime to $\mathfrak{a}$, that is to say we consider its image modulo $1+\mathfrak{m}\hat{\mathcal{O}}_K^\times$. But here I get lost and do not understand what "reducing modulo $\mathfrak{m}$ can mean for these ideles...:

$$\overline{x} = \prod_i \overline{\varpi_{\mathfrak{p}_i}}^{e_i} \quad \in \hat{\mathcal{O}}_K^\times / (1+\mathfrak{m}\hat{\mathcal{O}}_K^\times) \cong \mathcal{O}_K/\mathfrak{m}$$

Back to $\mathcal{O}_K$, denote $x_0$ the image in the last quotient. Now come my embarrassment: there is a natural map $\mathbf{Z}/\mathfrak{m}\cap\mathbf{Z} \to \mathcal{O}_K/\mathfrak{m}$ coming from $1 \mapsto 1_K$ (that is to say just viewing $\mathcal{O}_K/\mathfrak{m}$ as a $\mathbf{Z}$-module), and here is my question:

I would like to determine whether or not $x_0$ is in its image, depending on $\mathfrak{a}$, i.e. what can be said about $\mathfrak{a}$, the $\mathfrak{p}_i$ and/or the $e_i$ for having $x_0$ in $\mathbf{Z}/\mathfrak{m}\cap\mathbf{Z}$?

And two questions I am not able to answer also:

  • is the association $\mathfrak{a} \mapsto x$ usual? What are its main properties? And are we able to come back?
  • how can we understand the reduction modulo $\mathfrak{m}$ of $x$? ($x$ is an adèle, I'd prefer an ideal...)

Any idea or reference will be really welcome.

(My question could seem in some ways naive or elementary, so I wonder if this question is more suited for MO or MSE, hence I apologize in advance if I am wrong and would be grateful if someone more enlightened transfer me to MSE if it is better.)

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    $\begingroup$ What does O_K-hat-star mean? If it means what I think it means (a profinite completion of O_K, and then take the units) then this does not naturally contain K^* so I think some of what you write does not make sense. $\endgroup$
    – znt
    Commented Dec 12, 2016 at 15:57
  • $\begingroup$ @znt So could it be true without the $K^\times$ ? That is to say, do we have $\hat{O_K}$ isomorphic to $A_K^\times/K^\times$ ? $\endgroup$ Commented Dec 12, 2016 at 16:00
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    $\begingroup$ No -- the kernel will be the global units and the cokernel the class group (at least if you meant to use the finite adeles on the right like you did with the rest of your post). Isn't class field theory wonderful! All these fundamental things linked together via this adelic stuff. $\endgroup$
    – znt
    Commented Dec 12, 2016 at 16:44
  • $\begingroup$ @znt So for the whole adeles we would get $1 \to \mathbf{C}^\times \times \hat{O}^\times \to \mathbf{A}_K/K^\times \to Cl(K) \to 1$, right ? I simplify my question above for it following your comments. What bothers me the most is: we associate the idele $x$ to the ideal $\mathfrak{a}$ (up to this point, everything is fine), and then reduce modulo $\mathfrak{m}$ and wonder if it lies in the image of $\mathbf{Z}/\mathfrak{m}\cap\mathbf{Z}\to \mathcal{O}_K/\mathfrak{m}$ or not, and here I get totally lost. $\endgroup$ Commented Dec 13, 2016 at 8:21
  • $\begingroup$ It is difficult to understand your question because, at the time of writing my comment, your question still has errors in. There is an issue with global units and an issue with the class group, and you might be lost but I cannot help because I still do not understand the question. Can you fix it? I still don't know how you are getting to O_K / m and to be honest I am a little unclear about whether this is possible. $\endgroup$
    – znt
    Commented Dec 13, 2016 at 12:48

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