Let $\mathcal{o}_K$ be a ring of algebraic integers. I have a proof for the fact that $\mathcal{o}_K$ is a free module of finite rank over $\mathbb{Z}$. Now, let $\mathcal{p}$ be a prime ideal of $\mathcal{o}_K$ (in fact $\mathcal{p}$ is a maximal ideal). I would like to proof that $\mathcal{o}_K / \mathcal{p}$ is finite. It seems quite simple but I'm stuck when $\mathcal{p} \cap \mathbb{Z} = 0$.

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    $\begingroup$ how can p intersects Z = 0? take any element in p, the norm of that element is in p, and in Z. $\endgroup$
    – Long
    Mar 31 '14 at 22:30

The cases that $\mathfrak{p} \cap \mathbf{Z} = 0$ are indeed a counterexample to your conjecture. However, it turns out there is only one solution to this equation amongst prime ideals (or even amongst all ideals): the zero ideal.

An easy way to see this is that for any algebraic integer $\alpha$, $\mathbf{N}_{\mathfrak{o}/\mathbf{Z}}(\alpha)$ is a rational integer and it is a multiple of $\alpha$.

A different method would be to observe that every nonzero ideal is a free module over $\mathbf{Z}$ of the same rank as $\mathfrak{o}$: if $\beta_i$ is an integral basis for $\mathfrak{o}$, then $\alpha \beta_i$ is a set of linearly independent elements of $(\alpha)$. Since every ideal contains a principal ideal, every ideal must be a submodule of $\mathfrak{o}$ of full rank, and thus the quotient module must be finite.

If you take a basis matrix for a nonzero ideal $I$ (i.e. its rows are the coordinates of a set of basis vectors), then it turns out that its determinant is the number of elements in $\mathfrak{o} / I$.


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