## Abstraction

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 $$n$$th 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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1. #### HL

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