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 💡