## e, f, and the remainder theorem September 12, 2009

Posted by Akhil Mathew in algebra, algebraic number theory, commutative algebra, number theory.
Tags: , ,

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.