A number field $K$ is a finite extension of $\mathbb Q$, they well always be of the form $\mathbb Q[X]/(p(X))$. The ring of integers $\mathcal O_K$ of a number field is the subset of $K$ for which the elements satisfy a monic irreducible polynomial with integer coefficients.
One of the fundamental theorems about this is that the prime ideals are maximal, in other words if $\mathfrak q$ is a prime ideal then $\mathcal O_K/\mathfrak q$ is a field, just like $\mathbb Z/(p)$ is.
Lemma (Primes of a ring of integers lie above rational primes) If $\mathfrak q$ is a prime ideal then $\mathfrak q \cap \mathbb Z$ is a nonzero prime ideal in $\mathbb Z$.
The inverse image of prime ideal through a homomorphism is again a prime ideal and we have the embedding $\mathbb Z \longrightarrow \mathcal O_K$ so $q \cap \mathbb Z$ is a prime ideal, it just remains to check that it's nonzero.
Lemma $\mathcal O_K/\mathfrak q$ is an integral extension of $\mathbb Z/(p)$.
First of all $\mathcal O_K/\mathfrak q$ is a $\mathbb Z/(p)$-module because there is a ring homomorphism $\mathbb Z/(p) \longrightarrow \mathcal O_K/\mathfrak q$. Secondly any element $\alpha \in \mathcal O_K$ is the root of a polynomial: $p(\alpha) = 0$. Reduce this mod $\mathfrak q$ to see that $\bar \alpha$ is integral over $\mathbb Z/(p)$.
Lemma An integral extension of a field is a field.
Since $\alpha^n + c_{n-1} \alpha^{n-1} + \cdots + c_0 = 0$, we have $$ \alpha \cdot (- c_0^{-1}) (\alpha^{n-1} + c_{n-1} \alpha^{n-2} + \cdots + c_1) = 1. $$
Question How many elements does $\mathcal O_K/\mathfrak q$ have? I think this is related to the ramification "efg" theorem - how $\mathfrak q$ appears in the factorization of $\mathfrak q \cap \mathbb Z$.
This is me trying to understand KConrads note <http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/idealfactor.pdf>. I've read through these notes several times over the years. Incredible how ignorant I was in the past, skipping so many important details oblivous.
To prove the theorem about unique factorization we need existence and uniqueness, the uniqueness part relies on a lot of theory about ideals which is proved in terms of fractional ideals. The ultimate goal is to prove a cancellation lemma $\mathfrak p \mathfrak a = \mathfrak p \mathfrak b$ implies $\mathfrak a = \mathfrak b$. This would follow immediately from multiplying by an "inverse" of $\mathfrak p$ but to do that we need to generalize to fractional ideals and define inverses.
Definition $\mathfrak p$ is a prime ideal of $R$ when: $ab \in \mathfrak p$ implies $a \in \mathfrak p$ or $b \in \mathfrak p$.
Theorem For any ideals $\mathfrak a$, $\mathfrak b$, $\mathfrak p \supseteq \mathfrak a \mathfrak b$ implies $\mathfrak p \supseteq \mathfrak a$ or $\mathfrak p \supseteq \mathfrak b$.
Theorem If $\mathfrak p \supseteq \prod_i \mathfrak p_i$ then $\mathfrak p = \mathfrak p_j$ (some $j$).
By the previous we have $\mathfrak p \supseteq \mathfrak p_j$ (some $j$), but prime ideals are maximal (proved as a lemma at the start) so we have equality.