So, now to the next topic in introductory algebraic number theory: ramification. This is a measure of how primes “split.”  (No, definitely wrong word there…)

e and f 

Fix a Dedekind domain {A} with quotient field {K}; let {L} be a finite separable extension of {K}, and {B} the integral closure of {A} in {L}. We know that {B} is a Dedekind domain.

(By the way, I’m now assuming that readers have been following the past few posts or so on these topics.)

Given a prime {\mathfrak{p} \subset A}, there is a prime {\mathfrak{P} \subset B} lying above {\mathfrak{p}}. I hinted at the proof in the previous post, but to save time and avoid too much redundancy I’ll refer interested readers to this post.

Now, we can do a prime factorization of {\mathfrak{p}B \subset B,} say {\mathfrak{p}B = \mathfrak{P}_1^{e_1} \dots \mathfrak{P}_g^{e_g}}. The primes {\mathfrak{P}_i} contain {\mathfrak{p}B} and consequently lie above {\mathfrak{p}}. Conversely, any prime of {B} containing {\mathfrak{p}B} must lie above {\mathfrak{p}}, since if {I} is an ideal in a Dedekind domain contained in a prime ideal {P}, then {P} occurs in the prime factorization of {I} (to see this, localize and work in a DVR).

In particular, only finitely many primes of {B} lie above a given prime of {A}

Definition 1 If {\mathfrak{P} \subset B} lies above {\mathfrak{p} \subset A}, we write {e_{\mathfrak{P}/\mathfrak{p}}} for the number of times {\mathfrak{P}} occurs in the prime factorization of {\mathfrak{p}B}. We call this the ramification index.We let {f_{\mathfrak{P}/\mathfrak{p}}} be the degree of the field extension {A/\mathfrak{p}\rightarrow B/\mathfrak{P}}. This is called the residue class degree. 


The ramification index {e} has an interpretation in terms of discrete valuations. Let {\left \lvert \cdot \right \rvert} be the absolute value on {K} corresponding to the prime {\mathfrak{p}} and, by abuse of notation, {\left \lvert \cdot \right \rvert} its extension to {L} corresponding to {\mathfrak{P}}. Then

\displaystyle e_{\mathfrak{P}/\mathfrak{p}} = ( \left \lvert L^* \right \rvert: \left \lvert K^* \right \rvert ).\ \ \ \ \ (1)

 This is because if {\pi_L} is a uniformizer of {\mathfrak{P}} (so that {\left \lvert \pi_L \right \rvert} generates the cyclic group {\left \lvert L^* \right \rvert}), and {\pi_K} at {\mathfrak{p}}, then {\pi_L^e/\pi_K \in L^*} is a unit with respect to the absolute value {\left \lvert \cdot \right \rvert}, or in the discrete valuation ring {B_{\mathfrak{P}}}.

A basic fact about {e} and {f} is that they are multiplicative in towers; that is, if {L \subset M} is a finite separable extension, {C} the integral closure in {L}, {\mathfrak{Q} \subset C} a prime lying over {\mathfrak{P} \subset B} which lies over {\mathfrak{p} \subset A}, we have:

\displaystyle e_{\mathfrak{Q}/\mathfrak{p}} = e_{\mathfrak{Q}/\mathfrak{P}} e_{\mathfrak{P}/\mathfrak{p}}, \quad f_{\mathfrak{Q}/\mathfrak{p}} = f_{\mathfrak{Q}/\mathfrak{P}} f_{\mathfrak{P}/\mathfrak{p}}.

The assertion about {e} follows from (1), and that about {f} by the multiplicativity of degrees of field extensions. The degree {n := [L:K]} is also multiplicative in towers for the same reason. There is a similarity.  

Proposition 2 For {\mathfrak{p} \subset A}, we have 

\displaystyle \sum_{\mathfrak{P} \mid \mathfrak{p}} e_{\mathfrak{P}/\mathfrak{p}} f_{\mathfrak{P}/\mathfrak{p}} = n.


Indeed, we may replace {A} with {S^{-1}A} and {B} with {S^{-1}B}, where {S := A - \mathfrak{p}}. Localization preserves integral closure, and the localization of a Dedekind domain is one too (unless it is a field). Finally, {e} and {f} are stable under localization, which follows from the definitions.

In this case, {A} is assumed to be a DVR, hence a PID. Thus {B} is a torsion-free, hence free, finitely generated {A}-module. Since {L = K \otimes_A B} is free over {K} of rank {n}, the rank of {B} over {A} is {n} too. Thus {B/\mathfrak{p} B} is a vector space over {A/\mathfrak{p}} of rank {n}. I claim that this rank is also {\sum ef}.

Indeed, let the factorization be {\mathfrak{p}B = \mathfrak{P}_1^{e_1} \dots \mathfrak{P}_g^{e_g}}. Now we have for {i \neq j}, {\mathfrak{P}_i + \mathfrak{P}_j = B}, so taking high powers yields {1 \in \mathfrak{P}_i^{e_i} + \mathfrak{P}_j^{e_j} }.

By the remainder theorem below, we have

\displaystyle B/\mathfrak{p} B = \prod_{i=1}^g B/\mathfrak{P}_i^{e_i}

as rings. We need to compute the dimension of each factor as an {A/\mathfrak{p}}-vector space. Now we have

\displaystyle \mathfrak{P}_i^{e_i} \subset \mathfrak{P}_i^{e_i-1} \subset \dots \subset \mathfrak{P}_i \subset B

and since {\mathfrak{P}_i} is principal (see below) all the successive quotients are isomorphic to {B/\mathfrak{P}_i}, which has dimension {f_{\mathfrak{P}_i/\mathfrak{p}}}. So counting dimensions gives the proof.  

The remainder theorem and a consequence  

The remainder theorem below was known for the integers for thousands of years, but its modern form is elegant. 

Theorem 3 (Chinese Remainder Theorem) Let {A} be a ring and {I_i, 1 \leq i \leq n} be ideals with {I_i + I_j = A} for {i \neq j}. Then the homomorphism

\displaystyle A \rightarrow \prod_i A/I_i

is surjective with kernel {I_1 \dots I_n}.


First we tackle surjectivity.

If {n = 1} the assertion is trivial. If {n=2}, then say {I_1 + I_2 = A}. We then have {1 = \xi_1 + \xi_2} where {\xi_1 \in I_2, \xi_2 \in I_2}. So, if we fix {a,b \in A} and want to choose {\xi\in A} with

\displaystyle \xi \equiv a \ \mod(I_1), \quad \xi \equiv b \ \mod(I_2),

the natural choice is {\xi := a \xi_1 + b \xi_2}.

For higher {n}, it will be sufficient to prove that each vector with all zeros except for one 1 in the product occurs in the image. By symmetry, we need only show that there is {a \in A} such that {a \equiv 1 \ \mod( I_1) } while {a \in I_j} for {j \geq 2}.

Since {1 \in I_1 + I_j } for {j \geq 2},

\displaystyle 1 \in \prod_{j \geq 2} (I_1 + I_j) \subset I_1 + I_2 \dots I_n,

i.e. {I_1 + I_2 \dots I_n = A}. Now apply the {n=2} assertion to {I_1, I_2 \dots I_n} to find {a} satisfying the conditions.

Now for the kernel assertion: we must show {\bigcap I_i = \prod I_i}. For two ideals {I_1, I_2} it follows because one chooses {\xi_1 \in I_1, \xi_2 \in I_2} as above, and notes that

\displaystyle (I_1 \cap I_2) = (I_1 \cap I_2)\xi_1 + (I_1 \cap I_2)\xi_2 \subset (I_1 I_2) + (I_2 I_2).

Then one uses induction and the fact that {I_1 + I_2 \dots I_n = A} as above, etc.

As a corollary, we get a criterion for when a Dedekind domain is principal:

Theorem 4 A Dedekind domain {A} with finitely many prime ideals is principal.


To do this, we need only show that each prime is principal. Let the (nonzero) primes be {\mathfrak{p}_1, \dots, \mathfrak{p}_k}; we show {\mathfrak{p}_1} is principal. Choose a uniformizer {\pi \in A} at {\mathfrak{p}_1}, i.e. {\mathfrak{p}_1 A_{\mathfrak{p}_1} = (\pi)}. Now choose {x \in A} such that

\displaystyle x \equiv \pi \ \mod(\mathfrak{p}_1)^2, \quad x \equiv 1 \ \mod(\mathfrak{p}_j) \ \mathrm{for} \ j \geq 2.

Then {(x)} and {\mathfrak{p}_1} have equal orders at all primes, hence are equal.

We tacitly used this theorem above: there {S^{-1}B} has only finitely many prime ideals, which are the localizations of {\mathfrak{P}_1, \dots, \mathfrak{P}_g}, so is principal.