Higher ramification groups

This post is based on Marcus’s Number Fields. More specifically, it is based on a series of exercises following chapter 4.

Recall the definition of the intertia group of a prime \frak P in \mathcal O_L lying over a prime \frak p in \mathcal O_K (L/K is a Galois extension of number fields) – it’s the set of all \sigma\in G=\mathrm{Gal}(L/K) such that, for all \alpha\in L, we have \sigma(\alpha)\equiv\alpha\pmod{\frak P}. We now generalize this group.

Definition: In setting as above, we define the nth ramification group E_n to be the set of all \sigma\in G such that \sigma(\alpha)\equiv\alpha\pmod{\frak P^{n+1}}. The groups E_n,n>1 are called the higher ramification groups.

It is straightforward to see that D\geq E=E_0\geq E_1\geq\dots, all the subgroups are normal in D and their intersection is trivial. The structure of groups E_n can be somewhat complicated, but the groups E_{n-1}/E_n are particularly simple:

Proposition 1: E/E_1 is isomorphic to a subgroup of (\mathcal O_L/\frak P)^\times.

Proof: Fix \pi\in\frak P\setminus\frak P^2. We can then factor (\pi) as \frak P I with \frak P,I relatively prime. Taking any \sigma\in E we can find, by Chinese remainder theorem, a solution to x\equiv\sigma(\pi)\pmod{\frak P^2},x\equiv 0\pmod I. Because \sigma\in E,\sigma(\pi)\in\frak P, so x\in\frak P I=\pi\mathcal O_L, so x=\alpha_\sigma\pi for some \alpha_\sigma\in\mathcal O_L. In particular, \sigma(\pi)\equiv\alpha_\sigma\pi\pmod{\frak P^2}. Also, \alpha_\sigma is well-defined modulo \frak P: If \alpha_\sigma\pi\equiv\sigma(\pi)\equiv \alpha'\pi\pmod{\frak P^2}, then \frak P^2\mid (\alpha_\sigma-\alpha')\pi, so \alpha_\sigma\equiv\alpha'\pmod{\frak P}.

Thus we have defined a mapping \sigma\mapsto\alpha_\sigma, and clearly \alpha_{\sigma\tau}\equiv\alpha_\sigma\alpha_\tau,\alpha_{\mathrm{id}}\equiv 1\pmod{\frak P}, in particular – this map is a homomorphism into \alpha_\sigma\in(\mathcal O_L/\frak P)^\times. To show that this it induces the desired isomorphism we need to show that its kernel is E_1, which will easily follow if we show that if \sigma(\pi)\equiv \pi\pmod{\frak P^2}, then \sigma(\alpha)\equiv\alpha\pmod{\frak P^2}, i.e. \sigma\in E_1. We will prove something more general:

Lemma 1: For \sigma\in E and \pi\in\frak P\setminus\frak P^2, if \sigma(\pi)\equiv\pi\pmod{\frak P^{n+1}}, then \sigma\in E_n.

Proof of the lemma: We will proceed by induction on n. This is immediate for n=0. Suppose now \sigma(\pi)\equiv\pi\pmod{\frak P^{n+1}},n>0. In particular, \sigma(\pi)\equiv\pi\pmod{\frak P^n}, so \sigma\in E_{n-1}. Therefore \sigma(\alpha)\equiv\alpha\pmod{\frak P^n} for all \alpha\in\mathcal O_L, so \sigma(\pi\alpha)\equiv\sigma(\pi)\sigma(\alpha)\equiv\pi\sigma(\alpha)\equiv\pi\alpha\pmod{\frak P^{n+1}} (for last congruence, recall \pi\in\frak P). So \sigma(\alpha)\equiv\alpha\pmod{\frak P^{n+1}} for all \alpha\in(\pi).

Now we show the congruence for \alpha\in\frak P. Let (\pi)=\frak P I (as in the proof of the proposition). Choose \beta\equiv 1\pmod P,\beta\equiv 0\pmod I. Then \alpha\beta\in(\pi), so \alpha\beta\equiv\sigma(\alpha\beta)\equiv\sigma(\alpha)\sigma(\beta)\equiv\sigma(\alpha)\beta\pmod{\frak P^{n+1}} by above and since \sigma(\beta)\equiv\beta\pmod{\frak P^n}. But \beta is a unit modulo \frak P, hence modulo \frak P^{n+1}, so \sigma(\alpha)\equiv\alpha\pmod{\frak P^{n+1}}.

At the same time, every conguence class modulo \frak P has an element which is fixed by \sigma, and indeed, by every element of E. By result from my previous post, \mathcal O_L/\frak P is a trivial extension of \mathcal O_{L_E}/\frak P_E, so every congruence class modulo \frak P has a representative in \mathcal O_{L_E}, and by definition these are fixed by elements of E. So every \alpha\in\mathcal O_L can be written as \beta+\gamma,\beta\in L_E,\gamma\in\frak P, so that \sigma(\alpha)=\sigma(\beta)+\sigma(\gamma)=\beta+\sigma(\gamma)\equiv\beta+\gamma\equiv\alpha\pmod{\frak P^{n+1}}. Therefore \sigma\in E_n. \square

Hence, as we said, E_1 is the kernel of constructed homomorphism, which therefore is an isomorphism of E/E_1 onto its image, which is a subgroup of (\mathcal O_L/\frak P)^\times. \square

In a quite similar way we can prove the following result:

Proposition 2: E_{n-1}/E_n,n>1 is isomorphic to a subgroup of the additive group \mathcal O_L/\frak P.

Proof: Let, as before, \pi\in\frak P\setminus\frak P^2. Take any \sigma\in E_{n-1}. Writing (again) (\pi)=\frak P I, choose x\equiv\pi\equiv\sigma(\pi)\pmod{\frak P^n}, x\equiv\pi\pmod{I^n}. Then x-\pi\in \frak P^n I^n=(\pi^n), so \sigma(\pi)\equiv\pi+x\equiv\pi+\alpha_\sigma\pi^n\pmod{\frak P^{n+1}} for some \alpha_\sigma\in\mathcal O_L. Like in the previous proposition, we easily see that \alpha_\sigma is uniquely defined modulo \frak P and \alpha_{\sigma\tau}\equiv\alpha_\sigma+\alpha_\tau\pmod{\frak P}. This gives us a homomorphism, and from the lemma we easily find that its homomorphism is E_n, so that we get the desired isomorphism from E_{n-1}/E_n to a subgroup of \mathcal O_L/\frak P. \square

A quite immediate corollary is the following.

Theorem 1: Groups D,E,E_n are solvable.

Proof: We consider the chain of normal subgroups D\trianglerighteq E\trianglerighteq E_1\trianglerighteq\dots. D/E is isomorphic to the Galois group of the finite field (\mathcal O_L/\frak P)^\times, E/E_1 is isomorphic to a subgroup of the multiplicative group of this field and E_{n-1}/E_n is isomorphic to a subgroup of its additive group. All of these are abelian, and the chain eventually terminates (eventually E_n are trivial), so all the groups in the chain are solvable. \square

Definition: Suppose a prime \frak p in \mathcal O_K ramifies in \mathcal O_L and let e=e(\frak P/\frak p) and p be a prime in \mathbb Z lying under \frak p. We say that \frak p wildly ramifies if p\mid e, and we say that it tamely rafimites otherwise.

The terminology above might seem unmotivated, but hopefully it is at least in part clarified by the following theorem.

Theorem 2: If a prime is ramified, then it’s tamely ramified iff all the higher ramification groups are trivial. Moreover, E_1 is a Sylow p-subgroup of E.

Proof: Since E_1/E_2,E_2/E_3,\dots are isomorphic to subgroups of \mathcal O_L/\frak P, which is a p-group, their sizes are powers of p. Hence |E_1|=|E_1/E_2|\cdot|E_2/E_3|\cdot\dots is a power of p, i.e. E_1 is a p-group. On the other hand, |E/E_1|\mid|(\mathcal O_L/\frak P)^\times| is indivisible by p, so E_1 must be the Sylow p-subgroup of E. In particular, it’s nontrivial iff p\mid |E|=e. \square

The next result will turn out to be rather useful later.

Proposition 3: Suppose D/E_1 is abelian. The embedding from the proof of proposition 1 actually sends E/E_1 into \mathcal O_K/\frak p.

Proof: Suppose \sigma\in E and \sigma(\pi)=\alpha_\sigma\pi\pmod{\frak P^2}. First we note that this implies, in a way similar to the first two paragraphs of the proof of lemma 1, that \sigma(\beta)\equiv\alpha_\sigma\beta\pmod{\frak P^2} for all \beta\in\frak P.

Abelianness of D/E_1 implies that, for any other \tau\in D, \sigma^{-1}\tau\sigma\tau^{-1}\in E_1, so \tau\sigma\tau^{-1}(\alpha)\equiv\sigma(\alpha)\pmod{\frak P^2} for all \alpha\in\mathcal O_L. Taking \alpha=\pi and noting \tau^{-1}(\pi)\in\frak P this gives \alpha_\sigma\pi\equiv\sigma(\pi)\equiv\tau\sigma(\tau^{-1}(\pi))\equiv\tau(\alpha_\sigma\tau^{-1}(\pi))\equiv\tau(\alpha_\sigma)\pi\pmod{\frak P^2}, therefore \alpha_\sigma\equiv\tau(\alpha_\sigma)\pmod{\frak P}. Since D maps surjectively onto Galois group of (\mathcal O_L/\frak P)/(\mathcal O_K/\frak p), this group acts trivially on \alpha\pmod{\frak P}, so \alpha\in\mathcal O_K \frak p. \square.

Higher ramification groups, especially the last proposition, will turn out to be very useful in a proof of Kronecker-Weber theorem, which will be the subject of an upcoming blog post.


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