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 $\mathcal{p} \cap \mathbb{Z} = 0$.
[Math] Residue class field of ring of integers is finite
algebraic-number-theory
Related Solutions
There are lots of facts of this kind that are well-known to algebraic number theorists, but may not be in standard textbook accounts. The general theme of which your question is an instance is that of "global realization of a given local situation", namely can a given local situation (e.g. extension of local fields, or something analogous) be obtained from an appropriate global context (e.g. an extension of number fields) by completing.
In your particular case, there are probably many ways to answer the question, but here is one approach, which is fairly flexible.
Start with your finite extension $\mathbb F_q$ over $\mathbb F_p$. Now this can be obtained from an extension of local fields, i.e. we can find an extension $K$ of $\mathbb Q_p$ whose residue field equals $\mathbb F_q$. E.g. if $q = p^f$ then we can take $K$ to be the unramified extension of $\mathbb Q_p$ of degree $f$.
Now if we write $K = \mathbb Q_p(\alpha),$ let $f(x)$ be the minimal polynomial of $\alpha$ over $\mathbb Q_p$. Since $\mathbb Q$ is dense in $\mathbb Q_p$, we may approximate $f(x)$ as closely as we like by an element $g(x) \in \mathbb Q[x]$. If we assume that $g$ is close enough to $f$, then $g$ will also have a root $\beta$ in $K$ which is very close to $\alpha$ (in particular, so that $K = \mathbb Q_p(\beta)$.) (This is a version of Krasner's lemma; in fact it can be deduced pretty directly from Hensel's lemma.)
Now if let $F = \mathbb Q(\beta)$ then $F$ is a number field (since $\beta$ is the root of a polynomial in $\mathbb Q$), and the given embedding $F = \mathbb Q(\beta)\hookrightarrow K = \mathbb Q_p(\beta)$, which as dense image, corresponds to a prime ideal $P$ of $\mathcal O_F$ such that $F_P = K$. In particular, the residue field of $P$ equals the residue field of $K$, i.e. $\mathbb F_q$.
If we do choose $K$ to be unramified over $\mathbb Q_p$ of degree $f$, then we can choose $\alpha = \zeta$, a $(p^f-1)$st root of $1$, which is already algebraic over $\mathbb Q$. Thus we can take $\alpha = \beta = \zeta$, and so we can set $F = \mathbb Q(\zeta)$, the field generated by $(p^f-1)$st roots of unity. (Thus we get essentially the same answer as that indicated by Cam McLeman.)
Note, though, that the above result proves something stronger --- namely not just that any finite extension of $\mathbb F_p$ can be obtained as the residue field of a prime ideal in a number field, but that any finite extension of $\mathbb Q_p$ can be obtained as the completion at a prime ideal of a number field.
Furhermore, these Krasner-type arguments are very flexible and can be generalized in various directions. Suitably strengthened versions play an important role in current research related to Galois representations and automorphic forms.
$2 \in \mathcal{O}_K$. $\frac{1}{2}$ is not an algebraic integer, and therefore not in $\mathcal{O}_K$.
Best Answer
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 $\mathfrak{o} / I$.