I am a beginner student of Algebraic Number Theory and I am starting to learn ramification theory (of global fields). My question asks for motivation for a definition I was given.

Let $K$ be an algebraic number field, $\mathcal{O}_{K}$ its ring of integers, $L/K$ a Galois extension and $\mathcal{O}_{L}$ the integral closure of $\mathcal{O}_{K}$ in $L$.

I know that the group $G=Gal(L/K)$ acts transitively on the set of prime ideals $\mathfrak{P}_{i}$ of $\mathcal{O}_{L}$ above a prime $\mathfrak{p}$ of $\mathcal{O}_{K}$ and it's just a natural thing to consider the decomposition group (of one of these ideals) $G^{Z}(\mathfrak{P})=\{\sigma\in G\:|\:\sigma(\mathfrak{P})=\mathfrak{P}\}$, which is the stabilizer of $\mathfrak{P}$ under this action.

Now, in the paper I am following, together with the decomposition group, it was defined the group \begin{equation} G^{T}(\mathfrak{P})=\{\sigma\in G\:|\:\sigma(\alpha)\equiv\alpha\mod \mathfrak{P}\:\:\forall\alpha\in\mathcal{O}_{L}\}, \end{equation}

and this one I want to understand better.

I was told that each element of $G^{Z}(\mathfrak{P})$ induces an automorphism in the quotient $\mathcal{O}_{L}/\mathfrak{P}$, which is pretty reasonable. This $G^{T}(\mathfrak{P})$ looks like the subgroup of elements of $G^{Z}(\mathfrak{P})$ that induce the identity in the quotient $\mathcal{O}_{L}/\mathfrak{P}$. From my spying on other books and papers, i recognize this group as the so called $\textbf{Inertia group}$.

My question is basically:

What does the Inertia group tells us? When we look at the index $(G:G^{Z}(\mathfrak{P}))$, it gives us a notion of "how many primes did $\mathfrak{p}$ split into in $\mathcal{O}_{L}$". What about the inertia group? What is its meaning? And it is something as natural as considering the stabilizer of a group action?

  • $\begingroup$ Are you familiar with the inertial degree $f$? If $\mathfrak{q}$ in $\mathcal{O}_L$ lies over $\mathfrak{p}$ in $\mathcal{O}_K$, then we define $f(\mathfrak{q}| \mathfrak{p}) = (\mathcal{O}_L/\mathfrak{q} : \mathcal{O}_K/ \mathfrak{p})$. $\endgroup$
    – Marcus M
    Sep 10, 2015 at 16:52
  • $\begingroup$ Yes. And i know about the fundamental identity. But I don't know exactly how is the inertial degree related to the inertia group, even though I can imagine it being the galois group of the extension of residue fields or something like this. $\endgroup$
    – Shoutre
    Sep 10, 2015 at 17:08
  • $\begingroup$ Okay, I'll write up something quick that'll explain it a bit. $\endgroup$
    – Marcus M
    Sep 10, 2015 at 17:31

1 Answer 1


Let $\mathfrak{q}$ be a prime of $\mathcal{O}_L$ lying over $\mathfrak{p}$, a prime in $\mathcal{O}_K$.

Let $E$ be the inertial group, i.e. $$ E = \{\sigma \in G : \sigma (\alpha) \equiv \alpha \mod \mathfrak{q} \}.$$

Let $D$ be the decomposition group, i.e. $$ D = \{\sigma \in G : \sigma(\mathfrak{q}) = \mathfrak{q} \}. $$

Moreover, let $L_E$ and $L_D$ denote the fixed fields of $E$ and $D$ respectively, and let $\mathfrak{q}_E = \mathfrak{q} \cap L_E$ and define $\mathfrak{q}_D$ in a similar manner. It'll turn out that you have the following: \begin{align*} (L:L_E) &= e\\ (L_E:L_D) &= f\\ (L_D:K) &= r \end{align*} where $r$ is the number of primes that lie over $\mathfrak{p}$.

Moreover, you'll have that $\mathfrak{q}$ is totally ramified over $\mathfrak{q}_E$, i.e. $e( \mathfrak{q} | \mathfrak{q}_E ) = e$ and $f(\mathfrak{q} | \mathfrak{q}_E ) = 1$.

We also have $e( \mathfrak{q}_E | \mathfrak{q}_F ) = 1$ and $f(\mathfrak{q}_E | \mathfrak{q}_F ) = f$.

Finally, we have $e( \mathfrak{q}_F | \mathfrak{p} ) = 1$ and $f(\mathfrak{q}_F | \mathfrak{p} ) = 1$.

Now, let $K'$ be an arbitrary intermediate field, i.e. $K \leq K' \leq L$ and let $\mathfrak{p}' = \mathfrak{q} \cap K'$. The following is taken from Daniel Marcus' Number Fields: we then have

  • $L_D$ is the largest intermediate field $K'$ s.t. $e(\mathfrak{p}' | \mathfrak{p}) = f(\mathfrak{p}' | \mathfrak{p}) = 1$.

  • $L_D$ is the smallest $K'$ s.t. $\mathfrak{q}$ is the only prime lying over $\mathfrak{p}'$.

  • $L_E$ is the largest $K'$ s.t. $e(\mathfrak{p}' | \mathfrak{p}) = 1$.

  • $L_E$ is the smallest $K'$ s.t. $\mathfrak{q}$ is totally ramified over $\mathfrak{p}'$.

In summary, $E$ and $D$ allow us to decompose the splitting into subfields where each step ($r,e,$ and $f$) occurs in a separate place


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .