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.

## HL

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