More confusion in showing ring of integers is a Dedekind Domain I am having some trouble understanding the proof we did in class in showing a ring of integers is a Dedekind Domain. We proved the following: Let $R$ be a Dedekind Domain with field of fractions $K$. Let $L/K$ be a finite separable extension. Then the integral closure of $S$ in $R$ in $L$ is a Dedekind Domain.
Specifically, I have a bunch of questions in our proof of prime ideals being maximal, which I will outline below. We only proved this part for when $R = \mathbb Z$ and $K = \mathbb Q$. The parts I don't get are bolded.
Take $R = \mathbb Z$, $S = \mathcal{O}_L$. Then $S$ is a free module of rank $n = [L:K]$ so we can write $$S = \oplus_{i = 1}^n\mathbb Z \omega_i.$$ Now let $P$ be a nonzero prime ideal of $\mathcal{O}_L$. We know that $$P \cap \mathbb Z = \begin{cases} p\mathbb Z \\
0
\end{cases}.$$ Let $x \in P\setminus \{0\}$. $\textbf{Then we see that}$ $0 \neq N_{L /\mathbb Q}(x) \in P \cap \mathbb Z.$ (So the norm is just the product of the Galois conjugates so I can see why it's nonzero here. But why must it be in $P \cap \mathbb Z$?).
Therefore we must have that $P \cap \mathbb Z = p\mathbb Z$.
Now we come to the next part I don't get. I'm not sure if this is obvious, but we come to the conclusion that $\textbf{pS is contained in P}.$
From here we note that there's a surjective homomorphism $S/pS \to S/P.$ But apparently $\textbf{S/pS is an n dimensional vector space over}$ $\mathbb F_p$. I don't really see how we obtained this result. But this tells us that $S/P$ is a finite integral domain and therefore a field.
Can anyone clarify the doubts I have in this proof? I would appreciate it very much. Thanks!
 A: Addressing your questions in order:
For knowing that the norm lives in $\mathbb{Z}\cap P$, note that it lives in $P$ since one of its factors lives in $P$, and $P$ is an ideal, and lives in $Z$ since its an algebraic integer, that lives in the image of the norm map, which is $\mathbb{Q}$ (one way to see this is to view the norm via matrices, we are taking the determinant to the underlying field).
The claim that $pS$ is contained in $P$ follows since we know that $P\cap \mathbb{Z}=p\mathbb{Z}$, so in particular, $p$ is in $P$, so since $P$ is an ideal of $S$, $pS$ is in $P$.
Then finally, the claim that $S/pS$ is $n$ dimensional is because we knew $S$ is free of rank $n$ as a $\mathbb{Z}$ module, and we can check that any basis descends to a basis once we reduce mod $p$. Perhaps easier is to convince yourself that any generating set of $S$ over $\mathbb{Z}$ will generate $S/pS$ and $\mathbb{Z}/p$ module, and then worry about the exact dimension of this vector space (since we really only needed finiteness to get the conclusion).
I would reccomend chapter $3$ of Atiyah Macdonald if you want to go over this, this is general module theory rather than anything particularly arithmetic.
