Clarifying maps in algebraic number theory

Question

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!

Answer 1

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 $\mathcal{O}_K/\mathfrak{m}$.

Answer 2

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 $K$.