## 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 $$e(Q|P)$$.

Example: Let $$R = \mathbb{Z}, S = \mathbb{Z}[i]$$. Then $$(1-i)$$ lies over $$2$$ (which we use as shorthand for $$2\mathbb{Z}$$). It can be checked that $$(1-i)$$ is prime, and hence $$e((1-i)|2) = 2$$ since $$2S = (1-i)^2$$. Also, we have $$e(Q|p) = 1$$ 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 $$e((1-\omega)|p)=\phi(m)=p^{r-1}(p-1)$$ (TODO: proof). Also $$e(Q|q)=1$$ 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 $$f(Q|P)$$.

Example: Let $$R=\mathbb{Z}, S=\mathbb{Z}[i]$$. Recall 2 in $$\mathbb{Z}$$ lies under the prime $$(1-i)$$ in $$\mathbb{Z}[i]$$. Now $$|S/2S| = 4$$ and $$(1-i)$$ properly contains $$2S$$ so $$|S/(1-i)|$$ 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 $$f(3S|3)=2$$.

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

$\array { e(U|P)&=&e(U|Q)e(Q|P) \\ f(U|P)&=&f(U|Q)f(Q|P) }$

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 $$f = f(Q|p)$$. 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

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

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

3. 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

$R / I J \rightarrow R/I \times R/J$

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

$\|p S\| = { \prod_{i=1}^r \|Q_i\|^{e_i} } = { \prod_{i=1}^r (p^{f_i})^{e_i} }$

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 $$e_i = e(P_i |p), f_i = f(P_i|p)$$. 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

$\|p S\| = \prod \|P_i S\|^{e_i} = \prod \|P_i\|^{n_i e_i} =\prod (p^{f_i})^{n_i e_i}$

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

$\|P S\| = \prod \|Q_i\|^{e_i} = \prod \|P\|^{f_i e_i}$

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

$\|N T\| = \prod_\sigma \|\sigma (\alpha)T\| = \|\alpha T\|^n$

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:

$\|\langle 1 - \omega \rangle \|^n = \|\langle 1-\omega\rangle^n\| = \| p\mathbb{Z}[\omega]\| = p^n$

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

$\array { x = 0 \pmod{Q'} \\ x = 1 \pmod{\sigma(Q)} \textrm{ for all } \sigma\in G }$

Let $$\alpha \in S$$ be such a solution. Then

$N^L_K(\alpha) \in R \cap Q' = P$

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 $$e(Q|P) = e(Q'|P)$$ and $$f(Q|P)=f(Q'|P)$$.

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

$P S = \sigma(P S) = Q_i^{e_1} \sigma(Q_2)^{e_2}... \sigma(Q_r)^{e_r}$

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 $$e(Q|P)\gt 1$$ 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 $$p|disc(R)$$.

Proof: Let $$P$$ be a prime of $$R$$ lying over $$p$$ with $$e(P|p) \gt 1$$. 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 $$\mathbb{C}$$.

Ben Lynn blynn@cs.stanford.edu 💡