We shall establish relations between degrees, inertia degrees and ramification indices involved in decomposition and inertia field of a given prime in Galois extension. This is based on Marcus’s Number Fields and online notes by R. Ash. It follows a very similar appoach to, but is not based on, the one which can be found in this blog post by Sander Mack-Crane. For this blog post, understanding of Galois theory and basic facts about number fields is necessary.

First we fix some notation. Let $$L/K$$ be a Galois extension of number fields with Galois group $$G$$ and degree $$n$$. For a subgroup $$H\leq G$$ we denote by $$L_H$$ its fixed field For any intermediate field $$F$$ we let $$\mathcal O_F$$ be its ring of integers. If $$\frak P$$ is a prime lying over $$\frak p$$ in some field extension $$F_1/F_2$$, we denote by $$e(\frak P/\frak p)$$ the corresponding ramification index (i.e. the exponent of $$\frak P$$ in factorization of $$\frak p\mathcal O_{F_2}$$) and by $$f(\frak P/\frak p)$$ the inertia degree (i.e. the degree of finite field extension $$[(\mathcal O_{F_2}/\frak P):(\mathcal O_{F_1}/\frak p)]$$). We have the following facts, which we don’t prove here:

Theorem 1: If $$\frak p$$ is a prime in $$\mathcal O_K$$ and $$\frak P_1,\frak P_2$$ are two primes in $$\mathcal O_L$$ lying over $$\frak p$$, then for some $$\sigma\in G$$ we have $$\frak P_2=\sigma(\frak P_1)$$, hence $$e(\frak P_1/\frak p)=e(\frak P_2/\frak p),f(\frak P_1/\frak p)=f(\frak P_2/\frak p)$$. Denoting these common values by $$e,f$$ we moreover have $$efg=n$$, where $$g$$ is the number of distinct primes in $$\mathcal O_L$$ lying over $$\frak p$$.

Theorem 2: $$e$$ and $$f$$ are multiplicative in towers, i.e. for a (not necessarily Galois) extensions tower $$F_1/F_2/F_3$$ and primes $$\frak p_1\supseteq\frak p_2\supseteq p_3$$ in respective integer rings, then $$e(\frak p_3/\frak p_1)=e(\frak p_3/\frak p_2)e(\frak p_2/\frak p_1),f(\frak p_3/\frak p_1)=f(\frak p_3/\frak p_2)f(\frak p_2/\frak p_1)$$.

From now on, fix a prime $$\frak p$$ in $$\mathcal O_K$$ lying under a prime $$\frak P$$ in $$\mathcal O_L$$. For a subgroup $$H\leq G$$ we denote by $$\frak P_H$$ the prime of $$O_{L_H}$$ lying between $$\frak p$$ and $$\frak P$$. We focus our attention on two subgroups of $$G$$:

Definition: We define the decomposition group $$D$$ of $$\frak P$$ to be the set of $$\sigma\in G$$ preserving $$\frak P$$ (i.e. $$\sigma(\frak P)=\frak P$$). We also define the inertia group $$E$$ of $$\frak P$$ to be the set of $$\sigma\in G$$ preserving each congruence class modulo $$\frak P$$ (i.e. $$\sigma(\alpha)\equiv\alpha\pmod{\frak P}$$).

It’s clear that $$D$$ is a subgroup of $$G$$ and $$E$$ is a subgroup of $$D$$. Our goal is to determine, for any two of the fields $$K,L_D,L_E,L$$ the degree of the extension, ramification index and inertia degree

It follows from theorem 1 that $$G$$ acts transitively on the set of all primes lying over $$\frak p$$, and $$D$$ can be seen as the stabilizer of $$\frak P$$ for this action. By orbit-stabilizer theorem, $$|G|=|D|g$$ so, again by theorem 1, $$|D|g=efg,|D|=ef$$.

Since $$L/L_D$$ is clearly Galois, we can apply theorem 1 to it as well to get $$ef=|D|=e’f’$$ for $$e’=e(\frak P/\frak P_D),f’=f(\frak P/\frak P_D)$$, where there is no $$g$$ term, since $$D$$ fixes $$\frak P$$, so there can be no other prime lying over $$\frak P_D$$. But clearly $$e’\leq e,f’\leq f$$ (e.g. because of multiplicativity in towers), so $$e’=e,f’=f$$. Thus $$e(\frak P_D/\frak p)=f(\frak P_D/\frak p)=1$$, so, in particular, $$\mathcal O_{L_D}/\frak P_D=\mathcal O_K/\frak p$$.

Since every element of $$D$$ preserves $$\frak P$$, we can view $$\sigma\in D$$ as permuting the congruence classes modulo $$\frak P$$, i.e. as a permutation $$\overline{\sigma}$$ of $$\mathcal O_L/\frak P$$. Indeed, it’s clearly seen to be an automorphism of $$\mathcal O_L/\frak P$$. Moreover, since $$\sigma$$ fixes $$K$$ pointwise, $$\overline{\sigma}$$ fixes $$\mathcal O_K/\frak p$$, so we have a natural homomorphism from $$D$$ to $$\overline{G}=\mathrm{Gal}((\mathcal O_L/\frak P)/(\mathcal O_K/\frak p))$$. From definition we see that $$E$$ is precisely the kernel of this homomorphism. So we have an injective homomorphism from $$D/E$$ into $$\overline{G}$$.

Proposition: The above homomorphism is an isomorphism.

Proof: We only need to show that it’s surjective. We will show that for any $$\tau\in\overline G$$ there is a $$\sigma\in D$$ such that $$\overline\sigma=\tau$$. For that, let $$\theta\in \mathcal O_L$$ be such that $$\overline\theta$$ is a generator of multiplicative group of $$\mathcal O_L/\frak P$$. Then any automorphism in $$\overline G$$ is determined by its value at $$\overline\theta$$. So we only need to find $$\sigma\in D$$ such that $$\overline{\sigma(\theta)}=\tau(\overline\theta)$$.
Let $$h(x)$$ be the minimal polynomial of $$\theta$$ over $$\mathcal O_{L_D}$$ and let $$k(x)$$ be the minimal polynomial of $$\overline\theta$$ over $$\mathcal O_{L_D}/\frak P_D=\mathcal O_K/\frak P$$. Then $$\overline{h(\theta)}=\overline 0$$, so $$k(x)\mid\overline{h(x)}$$. Since $$k(x)=\prod_{\tau\in\overline G}(x-\tau(\overline\theta))$$, $$x-\tau(\overline\theta)\mid\overline{h(x)}$$ for all $$\tau\in\overline G$$. But $$h(x)=\prod(x-\sigma(\theta))$$, product running over (not necessarily all) $$\sigma\in D$$, so $$(x-\tau(\overline\theta))\mid\overline{h(x)}=\prod(x-\overline{\sigma(\theta)})$$. In particular, $$x-\tau(\overline{\theta})=x-\overline{\sigma(\theta)}$$ for some $$\sigma\in D$$, so $$\tau(\overline{\theta})=\overline{\sigma(\theta)}$$. $$\square$$

Hence $$D/E\cong\overline G$$. Therefore, $$\frac{|D|}{|E|}=|D/E|=|\overline G|=f$$, so $$|E|=\frac{|D|}{f}=\frac{ef}{f}=e$$.

As $$E\subseteq D$$, every element of $$E$$ fixes $$\frak P$$, so there can’t be any more primes lying over $$\frak P_E$$, since $$L/L_E$$ is Galois. By theorem 1 we have $$e=|E|=e(\frak P/\frak P_E)f(\frak P/\frak P_E)$$. But every automorphism in $$E$$ acts trivially on $$\mathcal O_L/\frak P$$. But from the proposition $$E$$ maps onto the automorphism group of $$(\mathcal O_L/\frak P)/(\mathcal O_{L_E}/\frak P_E)$$. Therefore the automorphism group, and hence this extension, must be trivial, i.e. $$f(\frak P/\frak P_E)=1,e(\frak P/\frak P_E)=e$$.

At this point, by repeatedly using multiplicativity in towers, we can easily get the following result.

Theorem: Let $$e=e(\frak P/\frak p),f=(\frak P/\frak p)$$ and $$g$$ as in theorem 1.

• $$[L:L_E]=e$$, $$e(\frak P/\frak P_E)=e$$, $$f(\frak P/\frak P_E)=1$$,
• $$[L_E:L_D]=f$$, $$e(\frak P_E/\frak P_D)=1$$, $$f(\frak P_E/\frak P_D)=f$$,
• $$[L_D:K]=g$$, $$e(\frak P_D/\frak p)=1$$, $$f(\frak P_D/\frak p)=1$$.

Remark: Note that although primes lying over $$\frak p$$ behave in many ways the same thanks to theorem 1, the definition of $$D,E$$ does depend on which $$\frak P$$ we choose. Hence, for example, it needn’t be true that for some other prime $$\frak P’$$ in $$\mathcal O_L$$ lying over $$\frak p$$ we have $$e(\frak P’_D/\frak p)=1$$.