Let $K$ be a quadratic imaginary number field. I wonder why, something which seems to be standard (yet by no means clear for me) is a natural map:

$$\mathbf{Z}/\mathfrak{m} \cap \mathbf{Z} \longrightarrow \mathcal{O}_K/\mathfrak{m}$$

Do we know explicitly what this map is? I am more precisely interested in determining whether or not a given element is in the image, but I feel totally lost. It is often mentionned the following exact sequence:

$$1 \to \mathbf{C}^\times \times \hat{\mathcal{O}}^\times \to \mathbf{A}(K)^\times/K^\times \to Cl(K) \to 1$$

But I do not see the relation... Do someone has any idea or source? It will be of great help!


This is just a general fact about modules over commutative rings. Note that $\mathcal{O}_K/\mathfrak{m}$ is a $\Bbb Z/\mathfrak{m}\cap\Bbb Z$ module--the action is just multiply by $n\in\Bbb Z$, and clearly it is trivial on $\mathfrak{m}\cap\Bbb Z$, so you can map the latter in the former by looking at $\Bbb Z/\mathfrak{m}\cap\Bbb Z\cdot 1$, as a cyclic $\Bbb Z/\mathfrak{m}\cap\Bbb Z$ submodule of $\mathcal{O}_K/\mathfrak{m}$.

  • $\begingroup$ Thank you, I haven't understood that. But I remain not grasping what is the image exactly. I got prime-to-$\mathfrak{m}$ ideals of the form $\varpi_\mathfrak{p}^e$ (with $\varpi$ and uniformizer), and I am really lost when I wonder if I can know if they are in the image or not... do we know an explicit/computable/reachable form for the elements in the image ? $\endgroup$ – Wolker Dec 8 '16 at 19:43
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    $\begingroup$ @Wolker see my edit, it's simpler to observe the image as a cyclic $\Bbb Z/\mathfrak{m}\cap\Bbb Z$ module. $\endgroup$ – Adam Hughes Dec 8 '16 at 19:43
  • $\begingroup$ @AdamHugues Ok I will try it, thank you for the help and enlightening formulation ;) $\endgroup$ – Wolker Dec 8 '16 at 19:52
  • $\begingroup$ @Wolker my pleasure. Cheers! $\endgroup$ – Adam Hughes Dec 8 '16 at 19:55
  • $\begingroup$ @AdamHugues It enligthened me a lot, but I begin to understand where I get lost each time : the elements I am trying to determine to be or not in this image are of the form $\varpi_\mathfrak{p}^e$ or product of those... but aren't they ideles ? How can I view them as classes of integers ? Maybe something relying on the same kind of identifications, for instance $\hat{O}^\times/(1+\mathfrak{m}\hat{O}^\times) \cong O_K/\mathfrak{m}$ ? $\endgroup$ – Wolker Dec 8 '16 at 20:24

I'd think of it like this.

There is a natural embedding $\mathbb Z\hookrightarrow\mathcal O_K$, and hence a projection $$\mathbb Z\hookrightarrow\mathcal O_K\to\mathcal O_K/\mathfrak m.$$

The kernel of this map is exactly the elements of $\mathfrak m$ which are also in $\mathbb Z$ - i.e. $\mathbb Z\cap\mathfrak m$. We deduce that there is an injection $$\mathbb Z/\mathbb Z\cap \mathfrak m\hookrightarrow\mathcal O_K/\mathfrak m.$$

In the case where $K$ is an imaginary quadratic number field, these two fields are just the residue fields of $\mathbb Z$ and $\mathcal O_K$ with respect to $\mathbb Z\cap \mathfrak m$ and $\mathfrak m$ respectively. Therefore, the map will be surjective if and only if the two residue fields are equal, which happens if and only if $\mathbb Z\cap\mathfrak m$ ramifies or splits in $K$.

  • $\begingroup$ It's not going to be a field unless $\mathfrak{m}$ is prime. Also I would avoid "hence a projection," the projection exists regardless of any inclusion. And you mean the kernel of the map restricted to the subring $\Bbb Z$. $\endgroup$ – Adam Hughes Dec 9 '16 at 13:20
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    $\begingroup$ I assumed $\mathfrak m$ referred to a maximal ideal, but I think my meaning in the last paragraph is clear enough. I think it's reasonably clear too that "this map" refers to the composite map $\mathbb Z\to\mathcal O_K/\mathfrak m$. $\endgroup$ – Mathmo123 Dec 10 '16 at 18:06

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