Ramification Indices
We keep our previous notation, that is, $K, L$ are number fields with $K \subset L$ and $R = \mathbb{A}\cap K, S =\mathbb{A}\cap L$. Suppose $P$ is prime in $R$.
The primes lying over $P$ are those that occur in the prime decomposition of $P S$. Their corresponding exponents are called the ramification indices. If $Q^e$ is the exact power of $Q$ dividing $P S$, then $e$ is the ramification index of $Q$ over $P$. We denote this by $e(QP)$.
Example: Let $R = \mathbb{Z}, S = \mathbb{Z}[i]$. Then $(1i)$ lies over $2$ (which we use as shorthand for $2\mathbb{Z}$). It can be checked that $(1i)$ is prime, and hence $e((1i)2) = 2$ since $2S = (1i)^2$. Also, we have $e(Qp) = 1$ for $p\ne 2$ and any $Q$ lying over $p$.
More generally, if $R = \mathbb{Z}, S=\mathbb{Z}[\omega]$ for $\omega=e^{2 \pi i/m}$ where $m = p^r$ for some prime $p\in\mathbb{Z}$ then the principal ideal $(1\omega)$ is $S$ is a prime lying over $p$, and $e((1\omega)p)=\phi(m)=p^{r1}(p1)$ (TODO: proof). Also $e(Qq)=1$ whenever $q\ne p$ and $Q$ lies over $q$, which will follow from a theorem we shall prove later.
Notice $R/P, S/Q$ are fields since $P,Q$ are maximal. Moreover, there is a natural way to view $R/P$ as a subfield of $S/Q$. Since $R \subset S$, there is a ring homomorphism from $R\rightarrow S/Q$ whose kernel is $R\cap Q$. Now $R\cap Q=P$ hence we have an embedding $R/P\rightarrow S/Q$. The fields $R/P, S/Q$ are called the residue fields associated with $P,Q$. The residue fields of a number ring are finite as was shown when proving every number field is a Dedekind domain, hence $S/Q$ is a finite extension of $R/P$. Let $f$ be the degree of this extension. We call $f$ the inertial degree of $Q$ over $P$, and denote it by $f(QP)$.
Example: Let $R=\mathbb{Z}, S=\mathbb{Z}[i]$. Recall 2 in $\mathbb{Z}$ lies under the prime $(1i)$ in $\mathbb{Z}[i]$. Now $S/2S = 4$ and $(1i)$ properly contains $2S$ so $S/(1i)$ must be 2. Then $R/P = S/Q = 2$, so $f = 1$ in this case. On the other hand, $3S$ is prime in $S$, and $S/3S=0$. Thus $f(3S3)=2$.
It can be easily verified that $e,f$ are multiplicative in towers. That is, if $P\subset Q\subset U$ are primes in $R\subset S\subset T$ then
In general if $Q$ is a prime in $S$, then $Q$ lies over a unique prime $p \in \mathbb{Z}$. Thus $S/Q$ is a field of order $p^f$ where $f = f(Qp)$. Since $p S \subset Q$ we have $p^f \le S/p S= p^n$ where $n = [L:\mathbb{Q}]$. Thus $f\le n$ when the ground field is $\mathbb{Q}$. In fact, we can show more.
For an ideal $I$ of $R$, define $\I\$ to be $R/I$. We shall prove the following two theorems simultaneously:
Theorem: Let $n = [L:K]$. Let $Q_1,...,Q_r$ be primes of $S$ lying over a prime $P$ of $R$. Let $e_1,...,e_r$ and $f_1,...f_r$ be the ramification indices and inertial degrees. Then $\sum_{i=1}^r e_i f_i = n$.
Theorem: Let $n = [L:K]$. Then

For ideals $I,J$ in $R$, $\I J\ = \I\ \J\$.

Let $I$ be an ideal of $R$. Then $\I S\ = \I\^n$.

Let $\alpha\in R$ be nonzero. Then $\(\alpha)\ = N^K_\mathbb{Q}(\alpha)$.
Proof of (1): We first prove for the case when $I, J$ are coprime, and then show $\P^m\ = \P\^m$ for all primes $P$. The result will then follow from uniqfue factorization of ideals.
Assume $I, J$ are coprime, so that $I +J=R, I\cap J = I J$. Then by the Chinese Remainder Theorem we have an isomorphism
hence $\I J\ = \I\ \J\$.
Now consider $\P^m\$ where $P$ is prime. We have the chain $R\supset P \supset P^2 \supset ... \supset P^m$. We shall show that for all $k$, $\P\ = P^k / P^{k+1}$ where the $P^k$ are viewed as additive groups. In fact we shall exhibit a group isomorphism $R/P \rightarrow P^k / P^k/P^{k+1}$.
Take any $\alpha\in P^k  P^{k+1}$. We have the isomorphism $R/P \rightarrow \alpha R / \alpha P$. The inclusion $\alpha R \subset P^k$ induces the homomorphism $\alpha R \rightarrow P^k /P^{k+1}$, with kernel $(\alpha R) \cap P^{k+1}$ and image $((\alpha R) + P^{k+1} / P^{k+1}$. It remains to show that $(\alpha R)\cap P^{k+1} = \alpha P$ and $(\alpha R) + P^{k+1} = P^k$, which can be done easily by viewing these as the lcm and gcd of $\alpha R, P^{k+1}$, and noting that $P^k$ is the exact power of $P$ dividing $\alpha R$.
Proof of first theorem, special case We prove the first theorem for $K=\mathbb{Q}$. Thus $P =p \mathbb{Z}$ for some prime $p\in\mathbb{Z}$. We have $p S = \prod_{i=1}^r Q_i^{e_i}$, hence
Since we know $\p S\ = p^n$, we have proved the theorem for $K=\mathbb{Q}$.
Proof of (2): Since (1) has been proved, we need only show this fact for prime ideals $P$. It can be checked that $S / P S$ is a vector space over $R / P$, and in fact $S/P S$ is a ring containing $R/P$. We shall show that the dimension is $n$.
First take any $\alpha_1,...,\alpha_{n+1}\in S$. We wish to show that the corresponding elements in $S/P S$ are linearly dependent. They are linearly dependent over $K$ and hence also over $R$. Hence $\alpha_1 \beta_1 + ... + \alpha_{n+1} \beta_{n+1} = 0$ for some $\beta_1,...,\beta_{n+1} \in R$, not all zero. We shall show that not all the $\beta_i$ lie in $P$, so after reducing modulo $P$ they do not all become zero. First we need the following generalization of a previous lemma:
Lemma: Let $A,B$ be nonzero ideals in a Dedekind domain $R$, with $B \subset A$ and $A \ne R$. Then there exists $\gamma \in K$ with $\gamma B \subset R, \gamma B \nsubset A$.
Proof: Recall there exists a nonzero ideal $C$ such that $B C$ is principal, say $B C = \langle \alpha \rangle$. Then $B C \nsubset \alpha A$. Fix any $\beta \in C$ such that $\beta B \nsubset \alpha A$ and set $\gamma = \beta/\alpha$.∎
Apply the lemma with $A = P$ and $B=\langle \beta_1,...,\beta_{n+1}\rangle$, Thus $S/P S$ is at most $n$dimensional over $R/P$.
To establish equality, let $P\cap \mathbb{Z} = p\mathbb{Z}$ and consider all primes $P_i$ lying over $p$. We know $S/P_i S$ is vector space over $R/P_i$ of dimension $n_i \le n$. We show equality holds for all $i$ and in particular $P_i = P$. Set $e_i = e(P_i p), f_i = f(P_ip)$. Then from above $\sum_i e_i f_i = m$ where $m = [K:\mathbb{Q}]$.
Thus $p R = \prod P_i^{e_i}$ hence $p S = \prod (P_i S)^{e_i}$, whence
Also we have $\p S\ = p^{m n}$ so $m n = \sum_i f_i n_i e_i$. Since all $n_i \le n$ and $\sum_i e_i f_i = m$ and we have $n = n_i$ for all $i$.∎
Proof of first theorem, general case: We have $P S = \prod Q_i^{e_i}$, hence
by (1) and the definition of $f_i$. By (2) we have $\P S\ = \P\^n$. Thus $n = \sum_i e_i f_i$.∎
Proof of (3): Extend $K$ to a normal extension $M$ of $\mathbb{Q}$ and let $T = \mathbb{A} \cap M$. For each embedding $\sigma$ of $K$ in $\mathbb{C}$ we have $\\sigma(\alpha)T\ = \\alpha T\$. This is because $\sigma(T) = T$ when extended to an automorphism of $M$. Set $N = N^K(\alpha)$. Then by (1) we have
Clearly $\N T\ = N^{m n}$ where $m = [M:K]$ and by (2) we have $\\alpha T\ = \\alpha R\^m$. Thus $\\alpha R\ = N$.∎
Example: We can use these theorems to show that $\langle 1\omega \rangle$ is prime in $\mathbb{Z}[\omega]$ where $\omega = e^{2 \pi i /m }, m = p^r$: since $\langle 1\omega\rangle^n = p \mathbb{Z}[\omega]$ where $n=\phi(m)$, any further splitting of $\langle 1\omega\rangle$ into primes would contradict the first theorem since $\mathbb{Q}[\omega]:\mathbb{Q} = n$.
Alternatively we could have used the first part of the second theorem:
hence $\\langle 1\omega\rangle\ = p$. Since is $p$ is prime we have that $\langle 1  \omega\rangle$ prime.
Example:
TODO
If $L$ is a normal extension of $K$, and $P$ is a prime of $R = \mathbb{Z}_K$, then the Galois group $G=Gal(L/K)$ permutes the primes lying over $P$. In fact, we have:
Theorem: Let $L$ be a normal extension of $K$ with corresponding number rings $S$ and $R$. Let $Q, Q'$ be primes of $S$ lying over the same prime $P$ of $R$. Then $\sigma(Q) = Q'$ for some $\sigma$ in the Galois group $G = Gal(L/K)$.
Proof: Suppose $\sigma(Q)\ne Q'$ for all $\sigma \in G$. Then by the Chinese Remainder Theorem there is a solution the the system of congruences
Let $\alpha \in S$ be such a solution. Then
since one of the factors of $N^L_K(\alpha)$ is $\alpha \in Q'$. Since $\alpha \notin \sigma(Q)$ for all $\sigma$ we have $\sigma^{1}(\alpha)\notin Q$. As $N^L_K(\alpha)$ is the product of $\sigma^{1}(\alpha)$ for all $\sigma$ we see that $N^L_K(\alpha) \notin Q$, a contradiction since we have just shown $N^L_K(\alpha) \in P \subset Q$.∎
Corollary: If $L$ is normal over $K$, and $Q,Q'$ are two primes lying over $P$ then $e(QP) = e(Q'P)$ and $f(QP)=f(Q'P)$.
Proof: We have the unique factorization $P S = Q_1^{e_1} ... Q_r^{e_r}$. For any $Q_i$, there exists $\sigma\in G$ that maps $Q_1$ to $Q_i$. We have
hence by unique factorization we must have $e_1 = e_i$ for all $i$.
The theorem implies there is an isomorphism between $S/Q$ and $S/Q'$, which establishes the result for the inertial degrees.∎
Hence for normal extensions, a prime $P$ of $R$ splits into $(Q_1 ... Q_r)^e$ in $S$ where the $Q_i$ are distinct primes in $S$ of inertial degree $f$ over $P$. By a previous theorem we must have $ r e f = [L:K]$.
If $e(QP)\gt 1$ then we say $P$ is ramified in $S$. (In other words, $P S$ is not squarefree.)
Theorem: Let $p$ be a prime in $\mathbb{Z}$. Suppose $p$ is ramified in a number ring $R$. Then $pdisc(R)$.
Proof: Let $P$ be a prime of $R$ lying over $p$ with $e(Pp) \gt 1$. For some ideal $I$ we have $p R = P I$. Note $I$ is divisible by all primes of $R$ lying over $p$. Let $\sigma_1,...,\sigma_n$ be the embeddings of the corresponding number field $K$ in $\mathbb{C}$.